No Free Lunch: Non-Asymptotic Analysis of Prediction-Powered Inference


Abstract

Prediction-Powered Inference (PPI) is a popular strategy for combining gold-standard and possibly noisy pseudo-labels to perform statistical estimation. Prior work has shown an asymptotic “free lunch” for PPI++, an adaptive form of PPI, showing that the asymptotic variance of PPI++ is always less than or equal to the variance obtained from using gold-standard labels alone. Notably, this result holds regardless of the quality of the pseudo-labels. In this work, we demystify this result by conducting an exact finite-sample analysis of the estimation error of PPI++ on the mean estimation problem. We give a “no free lunch” result, characterizing the settings (and sample sizes) where PPI++ has provably worse estimation error than using gold-standard labels alone. Specifically, PPI++ will outperform if and only if the correlation between pseudo- and gold-standard is above a certain level that depends on the number of labeled samples (\(n\)). In some cases our results simplify considerably: For Gaussian data, for instance, the correlation must be at least \(1/\sqrt{n - 2}\) in order to see improvement. More broadly, by providing exact non-asymptotic expressions for the variance of PPI++ under sample splitting, we aim to empower practitioners to transparently reason about the benefits of PPI++ in specific applications. In experiments, we illustrate that our theoretical findings hold on real-world datasets.

1 Introduction↩︎

Mean estimation is an ever-present problem: Medical researchers wish to understand the prevalence of disease, machine learning engineers want to understand the average performance of their models, and so on. Often, practitioners have access to a large set of unlabeled examples, e.g., clinical notes that may or may not record the presence of a disease, images with unknown labels, etc. A typical approach is to gather (expensive) gold-standard labels for a small number of randomly selected examples, and take a simple average of the relevant metric across them.

In many scenarios, it is possible to construct cheap pseudo-labels using machine learning models or other heuristics. For instance, large language models (LLMs) and vision-language models (VLMs) can achieve reasonable “zero shot” performance on a variety of tasks [1]. These pseudo-labels have the potential to improve estimation: For instance, if they were perfect substitutes for gold-standard labels, they would dramatically expand our sample size without introducing bias. In fact, it is already becoming a routine practice to use pseudo-labels from LLMs in “LLM-as-Judge” evaluations [2].

However, the quality of these pseudo-labels is often unknown a priori, and the naive approach of treating them as true standard labels can lead to erroneous conclusions. Prediction-Powered Inference (PPI) [3] proposes an estimator that is unbiased for any set of pseudo-labels, by using the small labeled dataset (on which pseudo-labels can also be obtained) to estimate and correct for any pseudo-label bias. While PPI is unbiased, it is not guaranteed to improve variance / statistical efficiency, particularly when the pseudo-labels have weak correlation with the true labels. To this end, [4] proposed Power-Tuned PPI (PPI++), which uses the labeled dataset to infer the pseudo/true label correlation and discount the pseudo-labels if this correlation is small. Informed by analysis showing that the (asymptotic) variance of PPI++ is never greater than the variance of the classical estimator that only uses labeled examples, they note (emphasis added)

Figure 1: An illustrative example of how our analytical results can guide practical decision-making. For any user-specified belief regarding the quality of pseudo-labels, we can determine analytically whether or not PPI++ (with cross-fitting, see 5) will improve over classical estimation or PPI. Specifying the quality of pseudo-labels F can be done in the binary setting using three parameters: The prevalence P(F = 1) and the positive / negative predictive value P(Y = 1 \mid F = 1), P(Y = 0 \mid F = 0). For any value of these three parameters, and the sample size of labels n, we can compute the variance of PPI & PPI++ relative to the variance of the classical approach. For modest sample sizes, there are distinct regions (color coded) where each method performs best. Given a priori practitioner beliefs about plausible accuracy of the pseudo-label (the “belief region”), this visualization allows for an informed choice as to which method to use: At n=20 (left), there is no dominant method, while at n=50 (right), PPI++ is superior across the belief region.

[our] methods automatically adapt to the quality of available predictions, yielding easy-to-compute confidence sets…that always improve on classical intervals using only the labeled data. [4]

However, this guarantee is asymptotic in nature, applying in the regime where the number of labeled examples becomes large, while the highest-value use of PPI++ comes precisely when labels are scarce. This leaves an open question: If the labeled sample is already too small to give a reliable estimate of the mean, when can we trust its assessment of the pseudo-labels?

We resolve this question by providing an exact non-asymptotic expression for the finite-sample variance of PPI++ under a sample-splitting implementation, and in the regime where the unlabeled dataset is taken to be arbitrarily large. Our expression holds for any joint distribution of pseudo- and true labels that has bounded fourth moments.

This analysis yields practical insights. First, our analysis produces a “no free lunch” result: PPI++ pays a price attributable to errors in estimating the correlation between pseudo- and true labels. If this correlation is small, then the costs of PPI++ outweigh the benefits. From our general results, we derive expressions for the minimum number of labeled samples that are required for PPI++ to improve upon the classical estimator. While our results apply to any distribution, they simplify considerably for Gaussian and binary labels. For Gaussian labels, the correlation must be larger than \((n-2)^{-1/2}\) for PPI++ to improve over the classical estimator, where \(n\) is the labeled sample size.1

Second, our results allow for direct calculations of the benefit / downside of using PPI vs PPI++ vs.classical estimators, depending on the assumed quality of pseudo labels and available size of labeled samples. Thus, practitioners can operationalize our framework into a recommended course of action by applying broad assumptions of the region in which the quality of their pseudo-labels lie. This process is described in 1. However, absent any assumption of the quality of the pseudo-labels, our method does not purport to identify the best performing method: that would fall into the same free-lunch trap that we caution against. Our conclusion is that PPI++ is a useful tool, but requires practitioners to implicitly assume that their pseudo-labels are “good enough”, a notion we define precisely in this work. Our overall contributions can be summarized as follows:

  • In 4.1, we analyze PPI++ in the Gaussian setting, and provide interpretable bounds on the correlation/sample size required for PPI++ to outperform the classical estimator that only uses the labeled sample.

  • In 4.2, we analyze PPI++ with sample splitting and cross-fitting, for general joint pseudo/true label distributions. We provide an exact expression for the variance of this unbiased estimator with finite labeled samples, and showing that similar intuitions hold as in the Gaussian case.

  • In 4.3, we return to PPI++ without sample splitting, showing that it has two undesirable properties: It is biased, and standard variance-estimation approaches can produce overly optimistic estimates of its variance.

  • In 5, we illustrate our findings on real data, highlighting settings where the use of PPI++ hurts variance, and show that PPI++ without sample splitting tends to draw overly narrow confidence intervals.

2 Related Work↩︎

The term “Prediction-Powered Inference” was coined by [3], who introduced the method in the context of both mean estimation and more general inference tasks. Recent work has proposed various extensions to PPI and PPI++, including their use with multiple estimands using additional shrinkage [5], their use with cross-fitting of predictors [6], and their use in the context of active statistical inference [7]. Others have proposed the use of PPI for application in clinical trials [8], causal inference [9], social science research [10], and the efficient evaluation of AI systems [11], [12].

PPI shares similarities with doubly-robust estimators in causal inference, as noted by [7], and can be seen as a special case of the Augmented Inverse Propensity Weighted (AIPW) estimator [13]. Through this lens, the unbiased nature of PPI (even when pseudo-labels are uninformative) is mirrored by the consistency of AIPW estimators under incorrect outcome-regression models. The core idea of PPI++ (taking linear combinations of unbiased estimators to minimize variance) dates back at least as far as [14]. However, the theoretical analysis of these estimators is often limited to potentially loose non-asymptotic error bounds [15] or study of their asymptotic efficiency. A distinct but related problem arises in the study of combining biased and unbiased estimators, where the bias of the former can be estimated from data. In that setting, a similar “no free lunch” phenomenon has been observed, where these methods improve asymptotic variance for any fixed bias, but perform poorly when the bias is comparable in magnitude to \(1/\sqrt{n}\) [16][18]. While our analysis strategy and problem setting differs substantially, our core result shows a conceptually similar point, that PPI++ tends to under-perform the classical estimator when the correlation between pseudo-labels and true labels is smaller than \(\approx 1 / \sqrt{n}\).

PPI++ can also be seen as a control variate method, where an unbiased estimator using \(n\) samples is additively combined with a mean-zero, correlated control variable to reduce variance. In the control variate literature, it is well known that estimating the optimal coefficient for the control variate from the same \(n\) samples can lead to variance inflation. For instance, [19] give a variance-inflation factor for Gaussian data which corresponds precisely to the threshold for improvement that we recover in our “warm up” on the Gaussian setting in 4.1 for the single-sample PPI++ estimator with infinite \(N\). We go further, however, in providing thresholds in the Gaussian setting for both a cross-fit variant of PPI++ and the finite \(N\) case. Sample-splitting is also proposed as a “remedy” in that literature for bias and over-optimistic variance estimates that can arise from re-using the same sample [20], analogous to the cross-fitting variant of PPI++ we analyze. However, to our knowledge, prior work characterizes variance only under the simplified setting of Gaussian data or asymptotically, while our main contribution (4.2) is an exact, distribution-free finite-sample expression.

Close connections exist between PPI, PPI++, and estimators from the survey sampling literature. In particular, as noted by [21], the PPI estimator for the mean is equivalent to the difference estimator, and the PPI++ estimator is equivalent to the generalized regression estimator (GREG) [22], [23]. Both estimators arise from a similar motivation (in survey sampling) to the motivation of PPI: The availability of auxiliary information (i.e., covariates) on many unsurveyed (i.e., unlabeled) units, and the use of predictive models to incorporate that information has a long history [24].

However, despite the depth of literature on the topic of similar estimators across causal inference, control variate methods, and survey sampling, analysis of performance is often asymptotic in nature, with the characterization of the exact finite-sample variance (or mean squared estimation error) perceived to be intractable. For instance, [25], a textbook on survey sampling estimators, states (see Chapter 9.4., page 203) that “It is not possible to accurately calculate the expectation and variance of the regression estimator” before introducing standard approximations. We perform that seemingly intractable exact calculation in our work, using the insight that this analysis becomes tractable precisely in the setting that motivates PPI++, where the model generating pseudo-labels is fixed, and an extremely large unlabeled dataset is available.

The theoretical characterization of PPI++ in other related work is largely limited to characterizing upper bounds on the benefits. [26] studies the limits of the benefits of PPI++, but focus on establishing upper bounds on the theoretical benefit, in the (optimistic) scenario where the optimal weight is chosen exactly. Likewise, [27], whose proposed method is largely identical to PPI++ in the mean estimation setting, do not consider finite-sample behavior of their estimator beyond establishing that the bias is \(O(1/n)\).

Also of interest is Cross-PPI [6] (distinct from the cross-fitting variant of PPI++ discussed in this work) which uses a portion of the data to train a predictor. We find there is no need to “demystify” the performance claim of Cross-PPI, since [6] already gives some sense of the trade-offs that come with poor pseudo-labels. In contrast, PPI++ stands out for its claim to “always” do no worse than classical, even if the predictions are poor. That said, if there existed a version of Cross-PPI with a similar “power-tuning” component, we believe our core insight (no free lunch) would carry over.

3 Setup and Preliminaries↩︎

Notation: We use upper-case letters to denote a random variable (e.g., \(Y\)) and lower-case letters to denote a particular value (e.g., \(y\)). We use \(P_Y\) to denote the distribution of a variable \(Y\). We denote the mean of a random variable \(Y\) by \(\mu_y\), the variance by \(\sigma_{y}^2\), and the covariance between two random variables \(Y, X\) by \(\sigma_{yx}\) or \(\sigma_{xy}\). Given \(n\) observations \(\{x_i,y_i\}_{i=1}^n\) of two random variables \(X, Y\), the empirical covariance is given by \(\hat{\sigma}_{xy} = \ch(X,Y) = \frac{1}{n-1}\sum_{i=1}^n(x_i - \bar{x}_n)(y_i - \bar{y}_n)\) where \(\bar{x}_n\) and \(\bar{y}_n\) are the sample means of \(x\) and \(y\) respectively over \(n\) samples.

Problem Setting: We focus on the mean estimation problem: Our goal is to estimate the expectation of a variable (or “label”) \(Y\), denoted \(\theta \mathrel{\vcenter{:}}= \mathbb{E}[Y]\).2 We have access to a dataset of size \(n\) (typically small), containing both covariates \(X\) and true labels \(Y\). Meanwhile, we have access to a dataset of size \(N\) (typically much larger), which only contains unlabeled examples \(X\). In this setting, we make the standard assumption that both samples are drawn from a common distribution. This assumption holds in the common scenario where a random subset of the entire dataset is chosen for labeling. We assume the ability to obtain cheap pseudo-labels \(F\) on both the labeled and unlabeled datasets. One way to obtain such labels is via a function \(f(X)\) (e.g., querying an LLM) that can be queried to get pseudo-labels, which motivates our use of upper-case \(F\) to denote these pseudo-labels. The classical estimator simply takes the average of the labeled examples to estimate the mean.

Definition 1 (Classical Estimator). Let \(\{y_i\}_{i=1}^n\) be \(n\) samples drawn independently from \(P_Y\), where the mean of \(Y\) is given by \(\theta \mathrel{\vcenter{:}}= \mathbb{E}[Y]\) and the variance by \(\vary \mathrel{\vcenter{:}}= \mathbb{E}[(Y - \theta)^2]\). The classical estimator of \(\theta\) is the sample average \(\thetahc = \frac{1}{n}\sum_{i=1}^{n} y_i\) and is unbiased with variance \(\sigma_y^2 / n\).

Prediction Powered Inference: Prediction Powered Inference is a family of estimators that combine the labeled dataset of \(n\) samples \(\cD_n = \{y_i, f_i\}_{i=1}^n\) with the larger set of \(N\) pseudo-label samples \(\{f_j\}_{j=1}^N\). For mean estimation, the original  estimator takes the following form.

Definition 2 (Prediction Powered Inference (PPI) [3]). Let \(\cD_n\) denote \(n\) samples of \(F, Y\) drawn from \(P_{FY}\), and let \(\cD'_N\) denote \(N\) samples of \(F\) drawn from \(P_{F}\). The PPI estimator is then given by \(\thetah_{\text{PPI}} = \frac{1}{n}\sum_{i \in \cD_n} y_i + \frac{1}{N}\sum_{j \in \cD'_N}f_j - \frac{1}{n}\sum_{i \in \cD_n}f_i\)

Notice that  is unbiased for any \(f\). That is, \(\mathbb{E}[\thetahp] = \mathbb{E}[Y] + \mathbb{E}[F] - \mathbb{E}[F] = \mathbb{E}[Y]\). However, the variance of  depends on the variance of \(Y - F\), which can be worse than the variance of \(Y\) alone if \(F\) is not sufficiently correlated with \(Y\). Motivated by this concern, the  estimator [4] uses a data-driven parameter \(\lambda\) to weight the contribution of the pseudo-labels. For mean estimation, the  estimator takes the following form.

Definition 3 (Power Tuned PPI (PPI++) [4]). Let \(\cD_n, \cD'_N\) be defined as in  2. The PPI++ estimator for mean estimation is then given by the following \(\thetahp = \frac{1}{n}\sum_{i \in \cD_n} y_i + \hat{\lambda} \left(\frac{1}{N}\sum_{j \in \cD'_N}f_j - \frac{1}{n}\sum_{i \in \cD_n}f_i\right)\) with \(\lh = \ch(Y,F) / ((1+\frac{n}{N})\widehat{\text{Var}}(F))\)

Note that if \(\lambda = 0\), then  reduces to , and if \(\lambda = 1\), then  reduces to . [4] note that the variance-optimal choice of \(\lambda^*\) is given by \(\lambda^* = (\sigma_{fy} / ((1 + \frac{n}{N})\sigma_f^2)\), which motivates the “plug-in” estimator of \(\hat{\lambda}\) above. We make two observations: First, the optimal value \(\lambda^*\) increases with higher covariance between the pseudo-labels and the true labels (\(\sigma_{fy}\)), and with lower variance in the pseudo-labels (\(\sigma_{f}^2)\). Both quantities need to be estimated from data, but the covariance in particular needs to be estimated using labeled data: Notably, the true labels \(Y\) are used for both (a) estimating the mean, and (b) estimating the quality of the pseudo-labels.

In many practical applications, the unlabeled sample size \(N\) is extremely large, and in our work, we consider it to be effectively infinite, such that \(\sigma_{f}^2\) and \(\mu_f\) can be estimated with arbitrary precision, and the optimal tuning parameter \(\lambda^*\) is given by \(\sigma_{yf} / \sigma_{f}^2\). Notably, even when \(N\) is infinite,  still relies heavily on the small labeled sample of size \(n\), which in practice may be very small. We now describe two variants of , which vary in how they use the labeled data to obtain their estimate of the optimal \(\lambda\). The first approach is to use the entire labeled sample for both (a) estimating \(\hat{\lambda}\), and (b) estimating \(\hat{\theta}\), which we refer to as .

Definition 4 (Single-Sample PPI++, Infinite \(N\)). Let \(\cD_n\) denote \(n\) samples of \(F, Y\) drawn from \(P_{FY}\). The Single-Sample PPI++ estimate is given by \[\begin{align} \thetah_{\text{Single-PPI++}} & = \frac{1}{n}\sum_{i \in \cD_n} y_i + \lh\bigg(\mu_f - \frac{1}{n}\sum_{i \in \cD_n}f_i\bigg) \\ \lh & = \ch(Y,F; \cD_n) / \sigma_f^2 \end{align}\]

This estimator is simple, but the lack of sample splitting can complicate variance estimation and the construction of confidence intervals. For example, as we show in [prop:optimistic95variance95single95sample95ppi4343], when the same data are used to both (a) construct \(\hat{\lambda}\), which minimizes the empirical variance, and (b) estimate the empirical variance while treating \(\hat{\lambda}\) as fixed, the resulting variance estimates can be overly optimistic. Cross-fitting is a standard solution to this type of problem: splitting the labeled data, using one half to estimate \(\lambda\), and the other to construct \(\hat{\theta}\) using the chosen \(\hat{\lambda}\). This procedure is repeated by swapping the roles of each half and the two estimates are averaged to produce a final estimate.

Definition 5 (Cross-fit PPI++ Estimator, Infinite \(N\)). Let \(\cD^{(1)}_{n/2}, \cD^{(2)}_{n/2}\) denote two independent \(n/2\) samples of \(F, Y\) drawn from \(P_{FY}\), and \(\hat{\theta}^{(k)}\) the PPI++ Estimator which uses \(\cD^{(k)}_{n/2}\) to estimate \(\hat{\lambda}\) and \(\cD^{(k')}_{n/2}\) (\(k \neq k'\)) to estimate \(\thetah\), s.t., \[\begin{align} \thetah^{(k)} & \mathrel{\vcenter{:}}= \frac{1}{n/2} \sum_{i \in \cD^{(k')}_{n/2}} y_i + \lh^{(k)} \bigg(\muf -\frac{1}{n/2} \sum_{i \in \cD^{(k')}_{n/2}} f_i \bigg) \\ \lh^{(k)} & \mathrel{\vcenter{:}}= \ch(F, Y; \cD^{(k)}_{n/2}) / \varf, \end{align}\] and the Cross-fit PPI++ estimator is \[\thetah_{\text{CF-PPI++}} = (\hat{\theta}^{(1)} + \hat{\theta}^{(2)}) / 2\]

We note that the cross-fit estimator uses only \(n/2\) samples per-fold, and thus the estimate of the covariance \(\ch(f, Y; \cD^{(k)}_{n/2})\) is computed using \(n/2\) labels.

Remark 1 (Mean Estimation vs M-Estimation). We note that our results focus primarily on mean estimation, whereas [4] also consider more general convex M-estimation. Our core insights still apply in that setting, since the asymptotic claims of PPI / PPI++ for M-estimation still proceed by a “mean estimation variance reduction” argument. For instance, the main result (Theorem 1) of [4] proceeds in two steps: First, it is argued that their approach leads to reduced variance in estimation of the expected gradient (instead of considering \(Y, F\), they use the per-sample gradients \(\nabla \ell(X, Y)\) and \(\nabla \ell(X, F)\) calculated using \(Y\) and \(F\)). Second, they link reduced variance in the estimate of the expected gradient to variance in the estimated parameter. The steps involved in the second step are similar to those conventionally used in standard proofs of asymptotic normality for M-estimators. As a result, our results would apply directly to a “no-free lunch” in that first stage of the argument, with the caveat that the linkage between “variance in mean estimation of the gradient” and “variance in parameter estimation” is harder to characterize exactly in a non-asymptotic framework.

4 Theoretical Analysis↩︎

To understand the relative performance of the classical and PPI++ estimators, we focus on an exact characterization of \(\mathbb{E}[(\hat{\theta} - \theta)^2]\), the finite-\(n\) estimation error. Since both the  and  estimators are unbiased, analyzing the finite-\(n\) estimation error is equivalent to the finite-\(n\) variance \(\mathrm{Var}(\thetah) \mathrel{\vcenter{:}}= \mathbb{E}[(\hat{\theta} - \mathbb{E}[\hat{\theta}])^2]\).

Our main result ([theorem:cross-ppi-general-mse]) makes no substantive assumptions regarding the distribution of \(Y\) and \(F\). That said, we warm-up in 4.1 by analyzing the special case where labels and pseudo-labels are Gaussian, and the analysis / results are far simpler. In this setting, both the  and  estimators are both unbiased, and their variances are comparable. The behavior in this setting is qualitatively similar to the more general results we derive later on.

In 4.2, we move on to analyze the variance of the  estimator for general distributions, where it remains unbiased. We decompose the variance into four parts: The variance of the classical estimator, a term that represents the gain in efficiency (i.e., a decrease in variance) attributable to pseudo-label quality, a term that represents the loss in efficiency due to errors in estimating the quality of the pseudo-labels and an additional term that captures the dependence between the two folds of the cross-fit estimator. The gain term depends directly on the covariance between the labels and the pseudo-labels while the loss term depends on the error in estimating this covariance from the labeled samples. The core of our no-free lunch result is that if this efficiency loss in estimating the covariance from labeled data exceeds the gain offered by the covariance itself, then will perform worse than .

A complete specification of the finite-sample variance of the   estimator requires an analysis of the efficiency loss term. We motivate this analysis with the illustrative case when the pseudo-labels are completely random. Here, the variance of   is always greater than the variance of the classical estimator, by a multiplicative factor of \(O(1/n)\). Then, in 4.2.2, we characterize the variance of the  estimator for any joint distribution of gold-standard and pseudo-labels. Finally, in 4.3, we return to the discussion of the  estimator, which is more difficult to analyze theoretically but has some unappealing properties from the perspective of confidence interval construction: It is biased in finite samples, and estimates of the variance that treat \(\lambda\) as fixed can lead to misleading optimism. We provide complete proofs for all theoretical results in 7.

4.1 Warm Up: Gaussian Labels↩︎

We begin with the special case where \(Y\) and \(F\) are normally distributed, for two reasons. First, this special case allows for a unified analysis of both the  and the   estimators, and establishes the existence of settings where both estimators under-perform . Second, this special case provides simple expressions for the variance of each estimator that capture the fundamental “no free lunch” nature of the problem, yielding simple conditions where PPI++ outperforms the classical estimate. In the Gaussian setting, both estimators are unbiased, and so we consider their variance as a measure of performance.3

propositionMSEGaussian Let \(Y,F\) be jointly Gaussian random variables, and consider the ,  and   estimators which all make use of \(n\) labeled samples. Then, \[\begin{align} & \mathrm{Var}(\thetah_{\text{PPI++}}) \label{eq:gaussian-mse} \\ & = \mathrm{Var}(\thetah_{\text{Classical}})\bigg(1 + \frac{1}{c \cdot n -1} \bigg) - \frac{c \cdot n - 2}{n(c \cdot n - 1)}\cdot \frac{\sigma_{fy}^2}{\sigma_{f}^2} \nonumber \end{align}\tag{1}\] where \(c = 1\) for  and \(c = 1/2\) for . Note that \(\mathrm{Var}(\thetah_{\text{Classical}}) = \sigma_y^2 / n\)

[theorem:GaussianMSE] shows the exact variance of the PPI++ variants in the Gaussian setting. We see that the variance is reduced by a term that depends on the correlation between \(F\) and \(Y\). However, the variance is also inflated by a multiplicative factor in the first term, and hence we require the correlation to be greater than a certain level for the overall variance to be lower than that of the classical estimator. 1 makes it easy to arrive at this required level, which we present in the below corollary:

corollaryMSESingleSampleNormal Let \(Y,F\) be jointly Gaussian random variables, and consider the and  estimators which both make use of \(n\) labeled samples. Then, \[\begin{align} \mathrm{Var}(\thetah_{\text{PPI++}}) & < \mathrm{Var}(\thetah_{\text{Classical}}) \iff \frac{1}{\sqrt{c \cdot n-2}} < \abs{\frac{\sigma_{fy}}{\sigma_{y}\sigma_{f}}} \end{align}\] where \(c = 1\) for  and \(c = 1/2\) for .

[theorem:jointly-normal-mse-condition-all] provides a unifying perspective on the “no free lunch” phenomenon: If the correlation \(\rho_{yf} \mathrel{\vcenter{:}}= \sigma_{fy} / (\sigma_y \sigma_f)\) is non-zero, then pseudo-labels can improve our variance, but there is a minimum sample size at which this improvement kicks in.4 However, if the correlation is zero, then both PPI++ estimators always perform worse than the classical estimator. This conclusion follows from a direct application of [theorem:GaussianMSE], where plugging a covariance of zero \((\sigma_{fy} = 0)\) into 1 yields a variance of PPI++ that is \((1 + 1 / (c \cdot n - 1))\) times the variance of the classical estimator. Intuitively, this additional variance arises from estimation of \(\lambda\). If the covariance is zero, then \(\lambda^* = 0\), and this choice of \(\lambda\) would yield the same variance as the classical estimator. In practice, however, \(\hat{\lambda}\) is not exactly zero, leading to higher variance. As we will see in [theorem:cross-ppi-indpt-mse], the same gap arises for  under general distributions, and we discuss the intuition further in 4.2.1.

For the remainder of the paper, we consider the setting where \(N\) is taken to be arbitrarily large. However, we note that in the Gaussian setting, it is possible to give exact expressions even when \(N\) is finite.

propositionMSESingleSampleNormalFiniteN Under the same conditions as [theorem:jointly-normal-mse-condition-all], consider the  where \(N\) is finite. Then the condition for improvement becomes \[\begin{align} \mathrm{Var}(\thetah_{\text{Single-PPI++}}) & < \mathrm{Var}(\thetah_{\text{Classical}}) \iff \\ & \frac{1}{\sqrt{n - 2 - 8 \frac{n-1}{N-1}}} < \abs{\frac{\sigma_{fy}}{\sigma_{y}\sigma_{f}}} \end{align}\]

4.2 Finite-Sample Variance of Cross-fit PPI++↩︎

Before we present our main results, we briefly note the unbiasedness of .

propositionCrossfitBias The   estimator is unbiased, i.e., \(\mathbb{E}[\thetah_{\text{CF-PPI++}}] = \theta\).

This is a trivial result included for completeness, and for contrast with [prop:bias95of95single95sample], where we will characterize the non-zero bias of . The unbiasedness of  follows by noting that the estimate on each fold after sample-splitting is unbiased since \(\lh\) is fixed within a fold, and so the simple average is unbiased by linearity of expectations. We now proceed to focus on the finite sample variance of  for general distributions. We first present our main result.

theoremCrossfitGeneralMSE Given a sample size of \(n\) labels, the variance of the  estimator (5) is given by \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) = & \underbrace{\frac{\sigma_{y}^2}{n}}_{\mathrm{Var}{\thetah_c}} - \underbrace{\frac{\sigma_{fy}^2}{n\sigma_f^2}}_{\text{(Eff. Gain)}} + \underbrace{\frac{1}{n\sigma_f^2}\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}_{\text{(Eff. Loss)}} \nonumber \\&+ \underbrace{\frac{2}{n^2\sigma_f^4}(\sigma_{yf^2} - 2\sigma_{fy}\mu_f)^2}_{\text{Cross-term b/w folds}}\label{eq:xfit-ppi-mse} \end{align}\tag{2}\]

The first term is the variance of the classical estimator. The second term (Eff. Gain) captures the gain in efficiency from using a predictor \(f\) that is correlated with the true label \(Y\), and the third term (Eff. Loss) captures the loss of efficiency that arises from the expected error in estimating the covariance from finite samples. In [theorem:gen-covariance95estimation] we will characterize the loss term in 2 , which can be combined with [theorem:cross-ppi-general-mse] to give the exact variance of the  estimator. For now, 2 yields intuition for when  fails to provide benefit.

Corollary 1. \[\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2] > \sigma_{fy}^2 \implies \mathrm{Var}(\thetah_{\text{CF-PPI++}}) > \sigma_y^2/n\]

In other words, there is no free lunch, but rather an intuitive trade-off: If the error in estimating the covariance \(\sigma_{fy}\) is higher than the magnitude of \(\sigma_{fy}\), then  performs worse than using the \(n\) labeled examples alone.

In the following, we characterize the covariance estimation error \(\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]\). To build intuition, we first revisit the setting where \(Y, F\) are independent, and prove that (in general), PPI++ performs worse than classical in this setting. In 4.2.2 we then give an explicit characterization (in [theorem:gen-covariance95estimation]) of the covariance estimation error term.

4.2.1 Warm-up: Performance Gap with Random Pseudo-Labels↩︎

We revisit the case where \(Y\) and \(F\) are independent, but now allow for non-Gaussian distributions.

propositionIndependentMSE Given \(n\) samples \(\{y_i, f_i\}_{i=1}^n\) drawn from \(P_{FY}\), where \(Y\) and \(F\) are independent with bounded second moments, the variance of  is given by \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) & = \frac{\vary}{n}\left(1 + \frac{2}{n-2}\right) \end{align}\]

Note that this non-Gaussian result is identical to the analogous result in the Gaussian case (obtained by plugging in \(\sigma_{yf} = 0\) to 1 , where \(c = 1/2\) for ). [theorem:cross-ppi-indpt-mse] can help build intuition that bridges the non-asymptotic and asymptotic regimes: when pseudo-labels are independent of the true labels, the finite-sample variance of the  estimator is always greater than the variance of the classical estimator, even if the asymptotic variance is identical.

4.2.2 Finite-Sample Covariance Estimation Error with General Pseudo-Labels↩︎

Before introducing our main result, we introduce additional notation. For two variables (e.g., \(F\) and \(Y\)) and two integers \(a, b\), we use \(\sigma_{f^a y^b}\) to refer to the covariance between the random variable \(F^a\) and the random variable \(Y^b\). For a single variable \(Y\), we use \(\sigma^2_{Y^a}\) to denote the variance of \(Y^a\).

First, we characterize the error in covariance estimation in [theorem:gen-covariance95estimation], which, when plugged into [theorem:cross-ppi-general-mse], yields an analytical expression for the variance of in terms of terms involving the mean and covariance of \(Y, F\) and their higher-order moments.

lemmaGeneralCovarianceEstimation Let \(Y\) and \(F\) have bounded fourth-moments. Given \(n\) IID samples \(\{y_i, f_i\}_{i=1}^n\) drawn from \(P_{YF}\), we have \[\begin{align} \mathbb{E}[(\hat{\sigma}_{yf} - \sigma_{yf})^2] & = \frac{1}{n}\sigma_{f^2y^2} + \frac{1}{n-1}\sigma_f^2\sigma_y^2 - \frac{(n-2)}{n(n-1)}\sigma_{fy}^2 \\&- \frac{2}{n}[\sigma_{y^2f}\mu_f + \sigma_{f^2y}\mu_y - 2\sigma_{fy}\mu_f\mu_y] \end{align}\]

Note that while here the covariance estimate is formed from \(n\) IID samples, forms this estimate from only \(n/2\) samples assuming \(n\) total labels are available.

4.2.3 Analytical Expression for the Finite-Sample Variance↩︎

Combining [theorem:gen-covariance95estimation] with [theorem:cross-ppi-general-mse] provides an analytical expression for the exact variance of . While we do not write the entire expression here due to space constraints, we make the following observation: Of the four components of the variance in 2 , the first two are \(O(1/n)\), and the last two are \(O(1 / n^2)\), since the covariance estimation error in [theorem:gen-covariance95estimation] is \(O(1/n)\) \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) =& \frac{1}{n} \left(\sigma_{y}^2 - \frac{\sigma_{fy}^2}{\sigma_f^2}\right) \\ & + \frac{1}{n} \left(\frac{1}{\sigma_f^2} \overbrace{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}^{O(1/n) \text{ by \ref{theorem:gen-covariance95estimation}}}\right) \\ & + \frac{1}{n^2}\left(\frac{2}{\sigma_f^4}(\sigma_{yf^2} - 2\sigma_{fy}\mu_f)^2\right) \\ =& \frac{1}{n} \left(\sigma_{y}^2 - \frac{\sigma_{fy}^2}{\sigma_f^2}\right) + O\left(\frac{1}{n^2}\right) \end{align}\] The full expression for the variance, expanding the covariance estimation error term, is a simple corollary of [theorem:cross-ppi-general-mse] and [theorem:gen-covariance95estimation], and is provided in 2 in the Appendix. The decomposition above provides further intuition regarding the mismatch between asymptotic results, as established in e.g., [4], which only consider the \(O(1/n)\) term since the \(O(1/n^2)\) terms vanishes asymptotically, and our finite-sample results, which take the \(O(1/n^2)\) terms into account.

4.3 Bias and Over-Optimistic Variance Estimation of Single-Sample PPI++↩︎

We now turn to some less desirable properties of the  estimator. Practitioners conducting statistical inference are often concerned with the construction of confidence intervals. Typically, we construct asymptotically valid confidence intervals using the variance of a consistent estimator, and hence lower variance yields tighter intervals. While such intervals are generally not guaranteed to maintain nominal coverage in finite samples, introduces two additional challenges. First, while asymptotically consistent, it is biased in finite sample, and second, it can produce overly optimistic variance estimates. We first characterize the exact bias5 in [prop:bias95of95single95sample], and in [prop:optimistic95variance95single95sample95ppi4343] we illustrate that under mild conditions, standard methods for estimating the variance of  will necessarily yield estimates that are lower than the classical variance, even in scenarios (as illustrated in 4.1) where we know the MSE of  is higher than that of the classical estimator.

propositionBiasSingleSample The bias of the  defined in 4 is given by \(\mathbb{E}[\thetah_{\text{Single-PPI++}} - \theta] = \left(2 \sigma_{yf} \muf - \sigma_{yf^2}\right) / (n \sigma_f^2)\)

While in 4.2, we exactly characterized the variance of  , we omit a similar analysis of , since the latter case is far less tractable to analyze, owing to complex dependencies that arise from estimating \(\lambda\) on the same data used to form the final estimate. We focus instead on the typical approach to estimating the variance that practitioners use to derive confidence intervals, and give the following result on its overly optimistic nature.

In [prop:optimistic95variance95single95sample95ppi4343] we consider a variance estimation strategy that treats \(\lambda\) as fixed, and uses plug-in estimators to estimate each component of the variance (namely, \(\hat{\sigma}_y, \hat{\sigma}_f,\) and \(\hat{\sigma}_{yf}\)). In the appendix, we present a similar result for another common method of variance estimation, that similarly treats the estimated \(\lambda\) as fixed, and uses the empirical variance of \(y - \lambda f\) in \(\cD_n\) and \(\lambda f\) in \(\cD'_N\).

propositionOptimisticVarianceSSPPI The plug-in variance estimator for  given by \(\hat{\sigma}^2_{\text{Single-PPI++}} = (1/n) [\hat{\sigma}^2_y + \lambda^2\sigma^2_f - 2\lambda \hat{\sigma}_{fy}]\) reduces to \((1/n)[\hat{\sigma}^2_y - \hat{\sigma}^2_{fy}/\sigma^2_f]\), which is always less than the plug-in estimate of the classical variance \(\hat{\sigma}^2_y / n\).

As a result of [prop:optimistic95variance95single95sample95ppi4343], the asymptotic confidence intervals constructed for  will always be narrower than those of , even if the true variance is larger. In 5, we show empirically that these always tighter intervals can result in significant drops in coverage.

5 Experiments↩︎

a
b

Figure 2: \(\MSE\) diff: \(\MSE(\hat{\theta}_{\text{PPI++}}) - \MSE(\hat{\theta}_{\text{Classical}})\) vs. Sample Size on the Alphafold Dataset for PPI++ estimators with pseudo-labels of varying quality. Each line represents PPI++ with pseudo-labels of different correlations. The blue line uses the original set, which has strong predictive performance, but only improves estimation error at \(n \geq 20\) (a) and \(n \geq 10\) (b).. a — Cross-fit PPI++ Estimator, b — Single-Sample PPI++ Estimator

Figure 3: AlphaFold. Coverage (left) and interval width (right) for  vs.  using pseudo-labels from Alphafold (\mathrm{corr}=0.81). under-covers relative to and .

We now illustrate our theory with experiments on the Alphafold [29] dataset used in [4], using a similar experimental setup where we use bootstrapping to simulate multiple draws from a common distribution, and use the empirical mean of the (larger) original dataset set for establishing ground truth. Note that we focus on the low sample size regime where n \(<\) 50. See 8 for more details. See 9 for additional results on the galaxies [4] dataset.

In 2, we show the difference between MSE of PPI++ and classical as a function of sample size \(n\). We use \(\MSE\) to reflect the fact that while is unbiased,   is not. For \(\MSE\) difference, the region below \(0\) marks improvement. We consider pseudo-labels with various correlation levels (0.81 is the original set in Alphafold). The results are in line with our theoretical developments for , and illustrate that similar trends hold for . Using pseudo-labels with varying quality, the MSE either (a) starts off worse than classical, but starts to improve over classical past a certain ‘tipping point’ sample size, (b) starts off worse than classical, but closes the gap as dataset size increases, or (c) starts off better than classical and stays there. For all non-zero correlation sets, there exists some \(n\) after which PPI++ has improved \(\MSE\) over classical. We note that for the pseudo-label models shown here,  tends to achieve lower \(\MSE\) than .

In 3, we show the coverage (left) and average widths of intervals (right) constructed via and over \(n\). Here we present the results with the original pseudo-label set in Alphafold with correlation \(0.81\), while a further-noised set is presented in 10. We note that the method produces intervals that are narrower than those from the . However, these substantively tighter intervals come at the cost of inadequate coverage. For instance, for the original pseudo-labels (top), at \(n = 20\), the realized coverage with is 87%. In 10,  6, we see that both and show wider intervals as the correlation drops, and while widens enough to match the coverage of , still falls short in this regard.

6 Discussion↩︎

Having presented our main theoretical results, we now step back to summarize what these results mean for applied use of prediction-powered methods.

A Cautionary Note (No Free Lunch): The practical appeal of is clear: when pseudo-labels are informative, they can increase the effective sample size. Our work seeks to establish the same spirit for , guiding practitioners away from the pitfall of believing that pseudo-labels always help, even when they are close to uninformative.

An Optimistic Note: Our results show that the required dependence between pseudo-labels and true labels is not fixed, but depends on the number of labeled samples, and is not necessarily an onerous requirement: Taking the Gaussian case as a general rule-of-thumb, the correlation lower-bound of \((n/2 - 2)^{-1/2}\) for Cross-fit PPI++ may be reasonable to achieve for moderate \(n\). The key is that some assumption on pseudo-label performance is required for improvement, but such assumptions may be quite reasonable.

A Practical Note: Our work provides more than just a “no free lunch” result, or a “rule-of-thumb” for gauging the benefit of pseudo-labels. [theorem:cross-ppi-general-mse], combined with [theorem:gen-covariance95estimation], can be used to derive the exact relative variance of . While the full form depends on some higher-order covariances (e.g., between \(Y\) and \(F^2\)), the full expression can be computed analytically in many common settings, and used to construct visualizations like the one shown in 1 for the binary setting. Thus, if the practitioner is willing to posit a reasonable assumption on the quality of the pseudo-labels, and the number of labels they have, our framework can determine if there is an upside to use or and the magnitude of improvement.

Overall, our hope is that this work helps practitioners use techniques like PPI++ with confidence, making transparent the implicit assumptions required to see practical benefit.

Impact Statement↩︎

Given the availability and ease of use of LLMs and VLMs, it has never been easier to construct some predictor that appears to plausibly have "good enough" performance, even without using labeled data. Thus, pseudo-labels are already a routine and central part of the decision-making processes within many high-impact applications. For instance, LLM-as-judge evaluations in industry production pipelines [2], [30][32], LLMs as proxies for human responses in social science experimentation [33][35], the use of foundation models in randomized controlled trials [36], [37], clinical trials [8], PPI-based content moderation [38], [39], LLMs-as-peer-reviewers [40], [41], etc. While procedures like PPI++ are important tools, the promise of these methods to always turn these synthetic signals, however noisy they may be, into statistically valid inferences might tempt practitioners to throw in “the kitchen sink” of black-box pseudo labels, confident that it cannot hurt. Given the already widespread use of LLM-labels, such misplaced optimism could prove very costly. Our theoretical findings narrate a “no-free lunch” style cautionary tale: proof that there are practical settings where PPI++ can hurt and PPI++ confidence intervals fail to meet advertised rates of coverage. On the other hand, our theory provides a quantitative framework that converts the assumptions that a practitioner is willing to make about the quality of their pseudo-labels into a quantification of the magnitude of the upside, if any.

7 Proofs↩︎

In this appendix section, we include formal proofs for all the mathematical statements in the main paper. We structure the section as follows. After giving the full expression for the variance of Cross-Fit PPI++, we begin with an introduction of some useful lemmas on properties of moments, variances and covariances. Some lemmas are interesting in their own right while other lemmas are introduced solely to establish some results that recur in the workings of other proofs. We then consider the bias of and . We show that is unbiased while is biased and we derive this bias that we presented in [prop:bias95of95single95sample]. We then present a full proof of the finite sample variance of the and all the statements in 4.2 including conditions where has higher variance than 1. Following this, we construct the proof for the covariance estimation error presented in [theorem:gen-covariance95estimation]. With these proofs in place, we are able to apply them to the Gaussian setting to complete the proofs in 4.1. Finally, we provide proofs for the optimistic variance estimate of that we presented in [prop:optimistic95variance95single95sample95ppi4343].

7.1 Full Expression for the Variance of Cross-Fit PPI++↩︎

Corollary 2. Under the conditions of [theorem:cross-ppi-general-mse] and [theorem:gen-covariance95estimation], we have \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) =& \frac{1}{n} \left(\sigma_{y}^2 - \frac{\sigma_{fy}^2}{\sigma_f^2}\right) \\ & + \frac{1}{n \sigma_f^2} \left( \frac{1}{n/2}\sigma_{f^2y^2} + \frac{1}{n/2-1}\sigma_f^2\sigma_y^2 - \frac{(n/2-2)}{n/2(n/2-1)}\sigma_{fy}^2 - \frac{2}{n/2}[\sigma_{y^2f}\mu_f + \sigma_{f^2y}\mu_y - 2\sigma_{fy}\mu_f\mu_y] \right) \\ & + \frac{1}{n^2}\left(\frac{2}{\sigma_f^4}(\sigma_{yf^2} - 2\sigma_{fy}\mu_f)^2\right) \\ =& \frac{1}{n} \left(\sigma_{y}^2 - \frac{\sigma_{fy}^2}{\sigma_f^2}\right) + O\left(\frac{1}{n^2}\right) \end{align}\]

Proof. The statement follows by plugging in [theorem:gen-covariance95estimation] (using \(n/2\) samples) into [theorem:cross-ppi-general-mse]. ◻

7.2 Foreword and Useful Lemmas↩︎

Remark 2 (Use of subscripts). We often make use of subscripts \(y_i, f_j\), etc in two contexts: First, we use these subscripts in the standard context as part of a sum. For instance, we may write \(\mathbb{E}[\bar{y} \bar{f}] = \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[y_i \bar{f}]\). Second, we use these subscripts as short-hand to distinguish between variables corresponding to different samples. Continuing the previous example, we may write \(\mathbb{E}[\bar{y} \bar{f}] = \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[y_i \bar{f}] = \frac{1}{n} n \cdot \mathbb{E}[y_i \bar{f}]\). Here, we make implicit use of the fact that for any \(i \in \{1, \ldots, n\}\), the expectation \(\mathbb{E}[y_i \bar{f}]\) is identical by exchangeability. This notation is particularly useful when indicies differ. For instance \(\mathbb{E}[y_i \bar{f}] = \frac{1}{n} \mathbb{E}[y_i (f_i + \sum_{j \neq i} f_j)] = \frac{1}{n} \mathbb{E}[y_i f_i] + \frac{(n-1)}{n} \mathbb{E}[y_i f_j]\), where \(\mathbb{E}[y_i f_j]\) denotes the expectation of \(y_i\) (from some sample \(i\)) and \(f_j\) (drawn from a distinct sample \(j \neq i\)), such that \(\mathbb{E}[y_i f_j] = \mathbb{E}[y]\mathbb{E}[f]\) by independence, while \(\mathbb{E}[y_i f_i] = \mathbb{E}[yf]\).

Lemma 1 (Covariance Representation). The empirical covariance estimated from \(n\)-samples can be equivalently written \[\frac{1}{n-1} \sum_{i=1}^{n} (y_i - \bar{y}) (f_i - \bar{f}) = \frac{1}{n-1} \sum_{i=1}^{n} (y_i f_i - \bar{y} \bar{f})\]

Proof. \[\begin{align} \frac{1}{n-1} \sum_{i=1}^{n} (y_i - \bar{y}) (f_i - \bar{f}) & = \frac{1}{n-1} \sum_{i=1}^{n} (y_i f_i - y_i \bar{f} - f_i \bar{y} + \bar{y} \bar{f}) \end{align}\] and \(\sum_{i} y_i \bar{f} = n \bar{y} \bar{f} = \sum_{i} \bar{y} \bar{f}\), and similarly with \(f_i \bar{y}\). ◻

lemmaCrossTermExpectationGeneral Let \(Y\) and \(F\) be random variables with bounded second-moments. Suppose that we observe \(n\) IID samples \(\{y_i, f_i\}_{i=1}^n\) from the joint distribution of \(Y\) and \(F\). Let \(\bar{y}_n, \bar{f}_n\) denote the sample means of \(Y, F\) on the observed n-sample. Furthermore, let \(\lh\) be independent of \(\{y_i, f_i\}_{i=1}^n\) and computed on an independent \(n\)-sample. Then we have, \[\begin{align} \mathbb{E} \left[ \lh(\bar{y}_n - \theta)(\mu_f - \bar{f}_n)\right] & = -\frac{\sigma_{yf}^2}{n\sigma_f^2} \end{align}\]

Proof. \[\begin{align} & \mathbb{E} \left[ \lh(\bar{y}_n - \theta)(\mu_f - \bar{f}_n)\right] \\ & = \mathbb{E}[\lh]\mathbb{E}[(\bar{y}_n - \theta)(\mu_f - \bar{f}_n)] & \text{Independence of \lh} \\ & = \mathbb{E}[\lh]\mathbb{E}\left[\bar{y}_n (\mu_f - \bar{f}_n) - \theta (\mu_f - \bar{f}_n)\right] \\ & = \mathbb{E}[\lh]\mathbb{E}\left[\bar{y}_n (\mu_f - \bar{f}_n)\right] - \mathbb{E}[\lh]\mathbb{E}\left[\theta (\mu_f - \bar{f}_n)\right] & \text{Linearity of Expectation} \\ & = \mathbb{E}[\lh]\mathbb{E}\left[\bar{y}_n (\mu_f - \bar{f}_n)\right] - \mathbb{E}[\lh]\theta\mathbb{E}\left[ (\mu_f - \bar{f}_n)\right] & \text{\theta is constant} \\ & = \mathbb{E}[\lh]\mathbb{E}\left[\bar{y}_n (\mu_f - \bar{f}_n)\right] - \mathbb{E}[\lh]\theta(\mathbb{E}[\mu_f] - \mathbb{E}[\bar{f}_n]) & \text{Linearity of Expectation} \end{align}\] Since \(\mu_f = \mathbb{E}[\bar{f}_n]\), the second term vanishes. Further \(\lh = \ch(Y,F) / \varf\). where \(\ch(Y,F)\) is an unbiased estimate of the covariance between \(Y,F\). Therefore, \(\mathbb{E}[\lh] = \frac{\sigma_{fy}}{\varf}\). Plugging these in, \[\begin{align} \mathbb{E} \left[ \lh(\bar{y}_n - \theta)(\mu_f - \bar{f}_n)\right] & = \frac{\sigma_{fy}}{\varf}\mathbb{E}\left[\bar{y}_n (\mu_f - \bar{f}_n)\right] \\ & = \frac{\sigma_{fy}}{\varf}\left[\mathbb{E}[\bar{y}_n\mu_f] - \mathbb{E}[\bar{y}_n\bar{f}_n] \right] & \text{Linearity of Expectation} \end{align}\] Note that \(\mathbb{E}[\bar{y}_n \mu_f] = \mu_y \mu_f\), and \(\mathbb{E}[\bar{y}_n \bar{f}_n]\) can be written as \[\begin{align} \mathbb{E}\left[\bar{y}_n\bar{f}_n\right] & = \frac{1}{n^2} \mathbb{E}\left[\sum_{i} \sum_{j} y_i f_j\right] \\ & = \frac{1}{n^2} \mathbb{E}\left[\sum_{i = j} y_i f_j + \sum_{i \neq j} y_i f_j\right] \\ & = \frac{1}{n^2} (n \mathbb{E}[y_i f_j] + n(n-1) \mathbb{E}[y_i f_j]) \\ & = \frac{\sigma_{yf}}{n} + \frac{n}{n^2} \mu_f \mu_y + \frac{n(n-1)}{n^2}\mu_f \mu_y \\ & = \frac{\sigma_{fy}}{n} + \mu_f\mu_y \end{align}\] Where we have used the fact that in the expansion of \(\bar{y} \bar{f}\), there are \(n\) terms where \(i=j\), such that \(E[y_if_i] = \sigma_{fy} + \mu_f\mu_y\) by definition of covariance, and there are \(n(n-1)\) terms where \(i \neq j\), such that \(E[y_if_j] = \mu_f\mu_y\), since \(y_i \perp f_j\). Putting these together, we have \[\mathbb{E}[\bar{y}_n\mu_f] - \mathbb{E}[\bar{y}_n\bar{f}_n] = \mu_y\mu_f - \frac{\sigma_{fy}}{n} - \mu_f\mu_y = - \frac{\sigma_{fy}}{n}\] Therefore, \[\mathbb{E} \left[ \lh(\bar{y}_n - \theta)(\mu_f - \bar{f}_n)\right] = \frac{\sigma_{fy}}{\varf}\left[\mathbb{E}[\bar{y}_n\mu_f] - \mathbb{E}[\bar{y}_n\bar{f}_n] \right] = \frac{\sigma_{fy}}{\varf}[-\frac{\sigma_{fy}}{n}] = -\frac{\sigma_{fy}^2}{n\sigma_f^2}\] which completes the proof.

Note: if the sample means where computed using \(n' = c \cdot n\) samples, then we would simply get: \[\begin{align} \mathbb{E} \left[ \lh(\bar{y}_{n'} - \theta)(\mu_f - \bar{f}_{n'})\right] = -\frac{\sigma_{fy}^2}{n'\sigma_f^2} = -\frac{\sigma_{fy}^2}{cn\sigma_f^2} \end{align}\] ◻

Lemma 2.

Let \(\bar{f}, \bar{y}\) denote sample means computed on \(n\) IID samples.

We have it that \[\begin{align} \mathbb{E}[\hat{\sigma}_{yf} \bar{f}] & = \frac{1}{n} \left(\sigma_{yf^2} + (n-2) \sigma_{yf} \mu_f \right) \\ \mathbb{E}[\hat{\sigma}_{yf} \bar{y}] & = \frac{1}{n} \left(\sigma_{y^2 f} + (n-2) \sigma_{yf} \mu_y \right) \end{align}\]

Proof. We demonstrate for \(\mathbb{E}[\hat{\sigma}_{yf} \bar{f}]\), and the case with \(\bar{y}\) follows by symmetry. \[\mathbb{E}[\ch(y, f)\bar{f}] = \mathbb{E}\left[\frac{1}{n-1} \sum_{i=1}^{n} (y_i - \bar{y})(f_i - \bar{f}) \bar{f}\right] = \frac{1}{n-1} \sum_{i=1}^{n} \mathbb{E}[(y_i f_i - \bar{f} \bar{y}))\bar{f}]\] where have we used 1 and linearity of expectation. We can then use the fact that \[\bar{f} \bar{y} = \frac{1}{n^2} \sum_{j} \sum_{k} f_j y_k = \frac{1}{n^2} \left(\sum_{j = k} f_j y_k + \sum_{j \neq k} f_j y_k\right),\] to obtain \[\begin{align} \mathbb{E}[\ch(y, f)\bar{f}] & = \frac{1}{n-1} \sum_{i=1}^{n} \mathbb{E}[(y_i f_i - \bar{f} \bar{y}))\bar{f}] \nonumber \\ & = \frac{1}{n-1} \sum_{i=1}^{n} \mathbb{E}\left[\frac{1}{n^2} \left(n^2 (y_i f_i) - \sum_{j = k} f_j y_k - \sum_{j \neq k} f_j y_k\right)\bar{f}\right] \nonumber \\ & = \frac{1}{n-1} \sum_{i=1}^{n} \left(\frac{n(n-1)}{n^2} \mathbb{E}[y_i f_i \bar{f}] - \frac{n(n-1)}{n^2}\mathbb{E}[f_j y_k \bar{f}]\right) \nonumber \\ & = \frac{n}{n-1} \left(\frac{n(n-1)}{n^2} \mathbb{E}[y_i f_i \bar{f}] - \frac{n(n-1)}{n^2}\mathbb{E}[f_j y_k \bar{f}]\right) \nonumber \\ & = \left(\mathbb{E}[y_i f_i \bar{f}] - \mathbb{E}[f_j y_k \bar{f}]\right) \label{eq:lem95cov95prod95eq1} \end{align}\tag{3}\] We then consider each term in 3 separately \[\begin{align} \mathbb{E}[y_i f_i \bar{f}] & = \frac{1}{n} \left(\mathbb{E}[y_i f_i f_i] + \sum_{j \neq i} \mathbb{E}[y_i f_i f_j]\right) & & = \frac{1}{n} \mathbb{E}[y_i f_i f_i] + \frac{n-1}{n} \mathbb{E}[y_i f_i f_j] \\ \mathbb{E}[f_j y_k \bar{f}] & = \frac{1}{n} \left(\mathbb{E}[f_j y_k f_j] + \mathbb{E}[f_j y_k f_k] + \sum_{i \neq j \neq k} \mathbb{E}[f_j y_k f_i]\right) & & = \frac{1}{n} \left(\mathbb{E}[f_j y_k f_j] + \mathbb{E}[f_j y_k f_k] + (n-2) \mathbb{E}[f_j y_k f_i]\right) \end{align}\] where we use the subscripts \(i, j, k\) as described in 2. We then rewrite 3 as \[\begin{align} \mathbb{E}[y_i f_i \bar{f}] - \mathbb{E}[f_j y_k \bar{f}] & = \frac{1}{n} \left(\mathbb{E}[y f^2] + (n-1) \mathbb{E}[yf]\mathbb{E}[f] - \mathbb{E}[y]\mathbb{E}[f^2] - \mathbb{E}[f]\mathbb{E}[yf] - (n-2) \mathbb{E}[f]^2 \mathbb{E}[y]\right) \\ & = \frac{1}{n} \left(\sigma_{yf^2} + (n-1) \mathbb{E}[yf]\mathbb{E}[f] - \mathbb{E}[f]\mathbb{E}[yf] - (n-2) \mathbb{E}[f]^2 \mathbb{E}[y]\right) \\ & = \frac{1}{n} \left(\sigma_{yf^2} + (n-2) \mathbb{E}[yf]\mathbb{E}[f] - (n-2) \mathbb{E}[f]^2 \mathbb{E}[y]\right) \\ & = \frac{1}{n} \left(\sigma_{yf^2} + (n-2) \sigma_{yf} \mu_f \right) \end{align}\] which gives the desired result. ◻

While the following result is of interest in it’s own right (and makes up substantially the entire proof of [prop:bias95of95single95sample]), we give it here as a Lemma, as we also make use of it in the proof of [theorem:cross-ppi-general-mse].

lemmaBiasPPISingle Given \(\hat{\lambda}\) and \(\bar{f}\) estimated on the same sample, we have it that \[\mathbb{E}[(\hat{\lambda} (\bar{f} - \muf))] = \frac{1}{n \varf} \left(\sigma_{yf^2} - 2 \sigma_{yf} \mu_f \right) \nonumber\]

Proof. \[\begin{align} \mathbb{E}[(\hat{\lambda} (\bar{f} - \muf))] & = \frac{1}{\varf} \mathbb{E}[\ch(y, f)(\bar{f} - \muf)]\nonumber \\ & = \frac{1}{\varf} \mathbb{E}[\ch(y, f)\bar{f}] - \frac{\sigma_{yf} \muf}{\varf} \nonumber \\ & = \frac{1}{n \varf} \left(\sigma_{yf^2} + (n-2) \sigma_{yf} \mu_f \right) - \frac{\sigma_{yf} \muf}{\varf} \label{eq:use95cov95prod} \\ & = \frac{1}{n \varf} \left(\sigma_{yf^2} - 2 \sigma_{yf} \mu_f \right) \nonumber \end{align}\tag{4}\] Where 4 follows from 2. ◻

A well-known result in statistics is that for Gaussian data, the sample mean and the sample covariance matrix are independent. Here we give the formal statement and include the proof.

lemmaGaussianIndependence Given I.I.D. random sample \(X_1,...,X_n\) from a multivariate Gaussian distribution \(\mathcal{N}(\mu, \Sigma)\), then the sample mean \(\bar{X} = \frac{\sum_{i=1}^n X_i}{n}\) and the sample covariance matrix \(S = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar{X})(X_i - \bar{X})^T\) are independent.

Proof. Let \(D_i:= X_i - \bar X\). Since \(D_i\) is a linear combination of independent random normal vectors, \(D_i\) also follows a multivariate normal distribution.

Now let’s consider the covariance between \(D_i\) and \(\bar X\). \[\begin{align} \text{Cov}(D_i, \bar X) & = \text{Cov}(X_i - \bar X , \bar X) \\ &= \text{Cov}(X_i, \bar X) - \text{Cov}(\bar X, \bar X) \\ &= \text{Cov}(X_i, \frac{1}{n}\sum_{j=1}^n X_j) - \text{Cov}(\frac{1}{n}\sum_{j=1}^n X_j, \frac{1}{n}\sum_{j=1}^n X_j) \\& = \text{Cov}(X_i, \frac{1}{n}X_i) - \frac{1}{n^2}\sum_{j=1}^n\text{Cov}(X_j, X_j) \quad\quad X_j \perp X_i, \forall i\ne j \\& = \frac{1}{n}\Sigma - \frac{1}{n}\Sigma \\& = 0 \end{align}\] Since \(D_i, \bar X\) are Gaussian, \(D_i, \bar X\) are independent, \(\forall i\). Therefore \(S=\sum_{i=1}^n D_iD_i^T\) and \(\bar X\) are independent. ◻

7.3 is unbiased↩︎

While fairly simple to observe, we provide a proof for the unbiasedness of the estimator. This proof is intended to provide an intuition for the next section where we consider the bias of . We include this unbiasedness as a proposition

Proposition 1. Consider the as defined in 5. The expected value of this estimator is equal to the mean \(E[Y]\). That is,

\[\begin{align} \mathbb{E}[\thetah_{\text{CF-PPI++}}] = \theta \end{align}\] where \(\theta = \mathbb{E}[Y]\)

Proof. \[\begin{align} \thetah_{\text{CF-PPI++}} = \frac{\hat{\theta}_1 + \hat{\theta}_2}{2} \end{align}\] where \(\hat{\theta}_k\), as defined in 5, are the estimates formed on each fold of the data using the other fold for computing \(\lh\).

Therefore, \[\begin{align} \mathbb{E}[\thetah_{\text{CF-PPI++}}] = \mathbb{E}\left[\frac{\hat{\theta}_1 + \hat{\theta}_2}{2} \right] = \frac{\mathbb{E}\left[\hat{\theta}_1 + \hat{\theta}_2\right]}{2} = \mathbb{E}[\hat{\theta}_{k}] \end{align}\] since \(\mathbb{E}[\hat{\theta}_1] = \mathbb{E}[\hat{\theta}_2]\) which we denote \(\mathbb{E}[\hat{\theta}_{k}]\).

Now, \[\begin{align} \hat{\theta}_k = \bar{y}_n + \lh(\muf - \bar{f}_n) \end{align}\]

Then, \[\begin{align} \mathbb{E}[\hat{\theta}_{k}] = \mathbb{E}[\bar{y}_n] + \mathbb{E}[\lh] \times (\muf - \mathbb{E}[\bar{f}_n]) \end{align}\]

where we have used that \(\lh\) is independent of \(\bar{f}_n\), combined with the linearity of expectations.

Now, since \(\bar{y}_n = \mathbb{E}[Y] = \theta\) and \(\bar{f}_n = \mathbb{E}[F] = \muf\), we have

\[\begin{align} \mathbb{E}[\hat{\theta}_{k}] = \theta + \mathbb{E}[\lh] \times (\muf - \muf) = \theta \end{align}\] ◻

The key to the unbiasedness of is the independence of \(\lh\) from the data used to compute sample means \(\bar{f}_n\). Since this independence is not applicable to , we cannot factor \(\lh\) into its own expectation leading to a finite-sample bias as we shall prove next.

7.4 Proof of [prop:bias95of95single95sample]: Bias of↩︎

Proof. The as defined in 4 is \[\begin{align} \thetah_{\text{Single-PPI++}} & = \frac{1}{n}\sum_{i \in \cD_n} y_i + \lh\left(\mu_f - \frac{1}{n}\sum_{i \in \cD_n}f_i\right) \\ \lh & = \frac{\ch(Y,F; \cD_n)}{\sigma_f^2} \end{align}\]

We have, \[\begin{align} \mathbb{E}[\thetah_{\text{Single-PPI++}}] = \frac{1}{n}\mathbb{E}\left[\sum_{i \in \cD_n} y_i\right] + \mathbb{E}\left[\lh\left(\mu_f - \frac{1}{n}\sum_{i \in \cD_n}f_i\right)\right] \end{align}\] where we have used the linearity of expectations. Applying this linearity into the sum over \(y_i\), we obtain

\[\begin{align} \mathbb{E}[\thetah_{\text{Single-PPI++}}] = \frac{1}{n}\sum_{i \in \cD_n} \mathbb{E}[y_i] + \mathbb{E}\left[\lh\left(\mu_f - \frac{1}{n}\sum_{i \in \cD_n}f_i\right)\right] \end{align}\]

Since, \(E[Y_i] = E[Y] = \theta\), we get \[\begin{align} \mathbb{E}[\thetah_{\text{Single-PPI++}}] = \theta + \mathbb{E}\left[\lh\left(\mu_f - \frac{1}{n}\sum_{i \in \cD_n}f_i\right)\right] \end{align}\]

Thus the bias, becomes, \[\begin{align} \mathbb{E}[\thetah_{\text{Single-PPI++}} - \theta] = \mathbb{E}\left[\lh\left(\mu_f - \bar{f}_n\right)\right] \end{align}\]

In the case of , at this stage, we were able to separate \(\lh\) from the difference in expectation of mean and sample mean, reducing the bias to zero. Here, \(\lh\) and \(\bar{f}_n\) are estimated on the sample making them dependent. However, from [lemma:bias95ppi95single] we have,

\[\begin{align} \mathbb{E}\left[\lh\left(\mu_f - \frac{1}{n}\sum_{i \in \cD_n}f_i\right)\right] = \frac{1}{n \varf} \left(2 \sigma_{yf} \mu_f - \sigma_{yf^2}\right) \end{align}\]

which completes the proof. ◻

7.5 [theorem:cross-ppi-general-mse]: Finite-Sample Variance of Estimator↩︎

Proof. The structure of the proof is as follows. We first derive the variance in terms of the variance of the PPI++ estimate formed from a single-fold. Then, we derive the variance of a single-fold. Putting the two together completes the proof of [theorem:cross-ppi-general-mse].

7.5.1 Variance in terms of variance of single-fold estimate \(\hat{\theta}_k\)↩︎

Let \(\cD_1, \cD_2\) denote our two dataset splits. Then we have that \(\hat{\theta}_1\) is derived using \(\cD_1\) for \(\hat{\lambda}\) and \(\cD_2\) for the point estimate, and vice versa for \(\hat{\theta}_2\). The variance of the estimate is given by, \[\begin{align} \frac{\hat{\theta}_1 + \hat{\theta}_2}{2} - \theta & = \frac{\hat{\theta}_1 - \theta + \hat{\theta}_2 - \theta}{2}\nonumber \\ \mathbb{E}\left[\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2} - \theta\right)^2\right] & = \frac{1}{4} \mathbb{E}[(\hat{\theta}_1 - \theta + \hat{\theta}_2 - \theta)^2]\nonumber \\ & = \frac{1}{4} \mathbb{E}\left[(\hat{\theta}_1 - \theta)^2 + 2 (\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta) + (\hat{\theta}_2 - \theta)^2\right] \nonumber \\ & = \frac{1}{4} \left(\mathrm{Var}(\hat{\theta}_1) + \mathrm{Var}(\hat{\theta}_2) + 2 E\left[(\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta)\right]\right)\nonumber \\ & = \frac{1}{2} \mathrm{Var}(\hat{\theta}_1) + \frac{1}{2} \mathbb{E}\left[(\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta)\right] \label{eq:mse95cross95fit95main} \end{align}\tag{5}\]

where we have used \(\mathbb{E}[\hat{\theta}_\text{CF-PPI++}] = \theta\) in writing the variance definition on the left hand side.

The first term is intuitive: The cross-fit estimator uses twice the samples of the estimate on just each fold, and so it has half the variance. The second term arises due to correlations in the errors of each cross-fit term. To analyze this term, we use subscripts to denote different datasets that serve as the source of different terms, e.g., \(\bar{y}_1\) is the average of \(y\) in the first split, and \(\bar{y}_2\) is the average in the second split. Then the cross term is given by \[\begin{align} \mathbb{E}[(\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta)] & = \mathbb{E}[((\bar{y}_2 - \theta) + \hat{\lambda}_1 (\bar{f}_2 - \muf))((\bar{y}_1 - \theta) + \hat{\lambda}_2 (\bar{f}_1 - \muf))]\nonumber \\ & = \mathbb{E}[(\bar{y}_2 - \theta)(\bar{y}_1 - \theta)]\nonumber \\ & \quad + \mathbb{E}[(\bar{y}_2 - \theta)(\hat{\lambda}_2(\bar{f}_1 - \muf))]\nonumber \\ & \quad + \mathbb{E}[(\bar{y}_1 - \theta)(\hat{\lambda}_1(\bar{f}_2 - \muf))]\nonumber \\ & \quad + \mathbb{E}[(\hat{\lambda}_1 (\bar{f}_2 - \muf))(\hat{\lambda}_2 (\bar{f}_1 - \muf))]\nonumber \\ & = \mathbb{E}[(\bar{y}_2 - \theta)]\mathbb{E}[(\bar{y}_1 - \theta)] & \bar{y}_1 \indep \bar{y}_2\nonumber \\ & \quad + \mathbb{E}[\hat{\lambda}_2 (\bar{y}_2 - \theta)]\mathbb{E}[\bar{f}_1 - \muf] & \bar{y}_2, \hat{\lambda}_2 \indep \bar{f}_1 \nonumber \\ & \quad + \mathbb{E}[\hat{\lambda}_1 (\bar{y}_1 - \theta)]\mathbb{E}[\bar{f}_2 - \muf] & \bar{y}_1, \hat{\lambda}_1 \indep \bar{f}_2 \nonumber \\ & \quad + \mathbb{E}[(\hat{\lambda}_1 (\bar{f}_1 - \muf))]\mathbb{E}[(\hat{\lambda}_2 (\bar{f}_2 - \muf))] \label{eq:cross95term95last95term} \end{align}\tag{6}\] Note that the first three terms are zero, since they include mean-zero terms (e.g., \(\mathbb{E}[\bar{y} - \theta] = 0, \mathbb{E}[\bar{f} - \muf] = 0\), etc). The fourth term is given by [lemma:bias95ppi95single], which we recall below:

However, the above lemma assumes the sample means are computed on an \(n\)-sample. In the case of , the sample means are computed on \(n' = n/2\) samples. Therefore,

\[\begin{align} \mathbb{E}[(\hat{\lambda} (\bar{f} - \muf))] = \frac{1}{(n/2) \varf} \left(\sigma_{yf^2} - 2 \sigma_{yf} \mu_f \right) = \frac{2}{n \varf} \left(\sigma_{yf^2} - 2 \sigma_{yf} \mu_f \right) \end{align}\]

Plugging this expression back into 6 , we get the value of the cross term \(\mathbb{E}[(\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta)]\). \[\begin{align} \mathbb{E}[(\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta)] & = \mathbb{E}[(\hat{\lambda}_1 (\bar{f}_1 - \muf))]\mathbb{E}[(\hat{\lambda}_2 (\bar{f}_2 - \muf))] & \text{From~\ref{eq:cross95term95last95term}} \\ & = \frac{4}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2 \end{align}\] Plugging this expression back into 5 gives us the overall MSE of the cross-fit estimator. \[\begin{align} \mathbb{E}\left[\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2} - \theta\right)^2\right] & = \frac{1}{2} \mathrm{Var}(\hat{\theta}_1) + \frac{1}{2} \mathbb{E}\left[(\hat{\theta}_1 - \theta)(\hat{\theta}_2 - \theta)\right] & \text{From~\ref{eq:mse95cross95fit95main}} \\ & = \frac{1}{2} \left(\mathrm{Var}(\hat{\theta}_1) + \frac{4}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2\right) \end{align}\] ◻

7.5.2 Variance of a single-fold estimate \(\hat{\theta}_1\)↩︎

Having expressed the variance in terms of the variance of a single-fold estimate, we now characterize the variance of this single-fold estimate.

Consider the estimate on a single-fold from 5, from which we have it that the variance can be decomposed as follows, (using \(E[\thetah_k] = \theta\))

\[\label{eq:ppi-split-mse} (\thetah_{k} - \theta) = \underbrace{(\bar{y}_{n/2} - \theta)}_{A} + \lh \underbrace{\left(\muf - \frac{1}{n/2} \sum_{i \in \cD'_{n/2}} f_i \right)}_{D} = A + \lh D\tag{7}\] Such that the variance is given by \[\begin{align} \mathrm{Var}(\thetah_{k}) & = \mathbb{E}[(\thetah_k - \theta)^2] \\ & = \mathbb{E}[(A + \lh D)^2] & \text{by \ref{eq:ppi-split-mse}} \\ & = \mathbb{E}[A^2 + 2A \lh D + {\lh}^{2} D^2] \\ & = \mathbb{E}[A^2] + 2\mathbb{E}[A \lh D] + \mathbb{E}[{\lh}^{2} D^2] & \text{by Linearity of Expectation} \end{align}\] We can observe that \[\mathbb{E}[A^2] = \mathbb{E}[(\bar{y}_{n/2} - \theta)^2] = \frac{\sigma_y^2}{n/2} = \frac{2\sigma_y^2}{n}\]

Consider the second term: \[\begin{align} 2\mathbb{E}[A \lh D] = 2\mathbb{E}[\lh (\bar{y}_{n/2} - \theta)(\mu_f - \bar{f}_{n/2})] \end{align}\] Recall that in each split-estimate, \(\lh\) is estimated on an independent split and is therefore independent of the other terms in the expectation. This allows us to apply [lemma:gen-cross-expect] to obtain \[\begin{align} 2\mathbb{E}[A \lh D] = 2\mathbb{E}[\lh (\bar{y}_{n/2} - \theta)(\mu_f - \bar{f}_{n/2})] = -2\frac{\sigma_{fy}^2}{(n/2)\sigma_f^2} = -4\frac{\sigma_{fy}^2}{n\sigma_f^2} \end{align}\] where we have accordingly applied \(n/2\) where the lemma uses \(n\). It is also useful to note that while \(\lh\) depends on the covariance estimate which also depends on the number of samples it is computed on, the expectation of the covariance estimate is independent of \(n\) and is just \(\sigma_{fy}\). Since only the expectation of \(\lh\) appears in the lemma, there is no need to adjust additionally for this.

Now consider the third term: Since \(\lh\) is independent of \(D\), we have it that \[\begin{align} \mathbb{E}[{\lh}^{2} D^2] & = \mathbb{E}[\lh^2]\mathbb{E}[(\mu_f - \bar{f}_{n/2})^2] \\ & = \frac{\mathbb{E}[\hat{\sigma}_{fy}^2]}{\sigma_f^4} \cdot \frac{\sigma_{f}^2}{n/2} & \lh & \mathrel{\vcenter{:}}= \frac{\hat{\sigma}_{fy}}{\varf} \\ & = \frac{2\mathbb{E}[\hat{\sigma}_{fy}^2]}{n \sigma_f^2} \end{align}\] where we use the definition of \(\hat{\lambda}\) and the fact that \(\mathbb{E}[(\bar{f} - \mu_f)^2] = \sigma_f^2 / (n/2)\) by definition of \(\bar{f}\). Putting these together, we get \[\begin{align} \mathrm{Var}(\thetah_k) & = \frac{2\sigma_y^2}{n} -4\frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{2\mathbb{E}[\hat{\sigma}_{fy}^2]}{n \sigma_f^2} \\ & = \frac{2\sigma_y^2}{n} -\frac{2\sigma_{fy}^2}{n\sigma_f^2} + \frac{2\mathbb{E}[\hat{\sigma}_{fy}^2]}{n\sigma_f^2} - \frac{2\sigma_{fy}^2}{n\sigma_f^2} \\ & = \frac{2\sigma_y^2}{n} -\frac{2\sigma_{fy}^2}{n\sigma_f^2} + \frac{2\mathbb{E}[\hat{\sigma}_{fy}^2 - \sigma_{fy}^2]}{n\sigma_f^2} \\ & = \frac{2\sigma_y^2}{n} -\frac{2\sigma_{fy}^2}{n\sigma_f^2} + \frac{2\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} \end{align}\] where the last line follows from the fact that \(\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2] = \mathbb{E}[\hat{\sigma}_{fy}^2 - 2\hat{\sigma}_{fy} \sigma_{fy} + \sigma_{fy}^2] = \mathbb{E}[\hat{\sigma}_{fy}^2 - \sigma_{fy}^2]\), from the fact that \(\mathbb{E}[\hat{\sigma}_{fy}] = \sigma_{fy}\).

7.5.3 Finite-Sample Variance of↩︎

Having characterized the variance of a single-fold estimate, we can plug this into the overall variance of the to complete the proof as follows:

\[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) = \mathbb{E}\left[\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2} - \theta\right)^2\right] & = \frac{1}{2} \left(\mathrm{Var}(\hat{\theta}_1) + \frac{4}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2\right) \end{align}\]

\[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) & = \frac{1}{2} \left( \frac{2\sigma_y^2}{n} -\frac{2\sigma_{fy}^2}{n\sigma_f^2} + \frac{2\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} + \frac{4}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2\right) \\ & = \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} + \frac{2}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2 \end{align}\]

7.6 Proof of : Condition for Worse Performance of↩︎

With the above theorem, it is easy to prove 1 which we restate here.

Corollary 3. \(\mathrm{Var}(\thetah_{\text{CF-PPI++}}) > \sigma_y^2/n\) if \(\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2] > \sigma_{fy}^2\) i.e., if the mean-squared error in estimating the covariance \(\sigma_{fy}\) is higher than the squared covariance \(\sigma_{yf}^2\).

From [theorem:cross-ppi-general-mse], we have \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) = \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} + \frac{2}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2 \end{align}\]

Further, since \(\frac{2}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2 \geq 0\), we have

\[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) - \frac{\sigma_y^2}{n} \geq \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} - \frac{\sigma_{fy}^2}{n\sigma_f^2} \end{align}\]

Therefore if \(\frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} > \frac{\sigma_{fy}^2}{n\sigma_f^2}\), we get \(\mathrm{Var}(\thetah_{\text{CF-PPI++}}) > \frac{\sigma_y^2}{n}\)

7.7 [theorem:cross-ppi-indpt-mse]: Variance of with Independent Pseudo-Labels↩︎

Motivated by the desire to build intuition, we first consider the case when \(Y,F\) are independent, and present a proof in that scenario. Using [lemma:indp-covariance95estimation] in combination with [theorem:cross-ppi-general-mse] allows us to prove [theorem:cross-ppi-indpt-mse].

lemmaCovarianceEstimation Given \(n\) samples \(\{y_i, f_i\}_{i=1}^n\) drawn iid from \(P_{FY}\), where \(Y\) and \(F\) are independent with bounded second moments, then \[\label{eq:cov95mse95independent} \mathbb{E}\left[ (\hat{\sigma}_{yf} - \sigma_{yf})^2 \right] = \frac{\vary \varf}{n - 1},\tag{8}\]

Proof. By definition, we can write that \[\hat{\sigma}_{fy}(Y, F) = \frac{1}{n-1} \sum_{i=1}^n (y_i - \bar{y})(f_i - \bar{f}).\] where \(\bar{y} = {n}^{-1} \sum_{i=1}^{n} y_i\) and similar for \(\bar{f}\). Taking the square gives us \[\begin{align} \hat{\sigma}_{fy}^2 & = \left( \frac{1}{n-1} \sum_{i=1}^n (y_i - \bar{y})(f_i - \bar{f}) \right)^2 \\ & = \frac{1}{(n-1)^2} \sum_{i=1}^n \sum_{j=1}^n (y_i - \bar{y})(f_i - \bar{f})(y_j - \bar{y}) (f_j - \bar{f}) \end{align}\] and by linearity of expectation, we therefore have it that \[\mathbb{E}[\hat{\sigma}_{fy}^2] = \frac{1}{(n-1)^2} \sum_{i=1}^n \sum_{j=1}^n \mathbb{E}\left[ (y_i - \bar{y})(f_i - \bar{f})(y_j - \bar{y})(f_j - \bar{f}) \right] \\None\]

Since we have independence of \(F, Y\) (and hence, joint independence \((F_1, \ldots, F_n) \perp (Y_1, \ldots, Y_n)\)), we can treat the expectations over terms involving \(F,Y\) separately, giving us \[\label{eq:cov95independent95minor950} \mathbb{E}[\hat{\sigma}_{fy}^2] = \frac{1}{(n-1)^2} \sum_{i=1}^n \sum_{j=1}^n \mathbb{E}\left[ (y_i - \bar{y})(y_j - \bar{y})\right] \mathbb{E}\left[(f_i - \bar{f})(f_j - \bar{f}) \right]\tag{9}\] where \[\label{eq:cov95independent95minor} \mathbb{E}[(y_i - \bar{y})(y_j - \bar{y})] = \begin{cases} \sigma_y^2 \left(1 - \frac{1}{n}\right) & \text{if } i = j, \\ - \frac{\sigma_y^2}{n} & \text{if } i \ne j. \end{cases}\tag{10}\] While 10 is a standard result, it can be proven as follows: First, we distribute terms and use linearity of expectation to write \[\begin{align} \mathbb{E}[(y_i - \bar{y})(y_j - \bar{y})] & = \mathbb{E}[y_i y_j] - \mathbb{E}[y_i \bar{y}] - \mathbb{E}[y_j \bar{y}] + \mathbb{E}[\bar{y}^2] \end{align}\] and we can then observe that each term can be written as \[\begin{align} \mathbb{E}[y_i y_j] & = \begin{cases} \sigma_y^2 + \mu_y^2 & \text{if } i = j \\ \mu_y^2 & \text{if } i \ne j \end{cases} \\ \mathbb{E}[y_i \bar{y}] & = \frac{1}{n} \left( \mathbb{E}[y_i^2] + (n-1)\mu_y^2 \right) = \frac{1}{n} (\sigma_y^2 + n\mu_y^2) \\ \mathbb{E}[\bar{y}^2] & = \frac{1}{n^2} \left( n(\sigma_y^2 + \mu_y^2) + n(n-1)\mu_y^2 \right) = \frac{\sigma_y^2}{n} + \mu_y^2 \end{align}\] We can then consider each case in 10 as follows:

Case 1: \(i = j\) \[\begin{align} \mathbb{E}[(y_i - \bar{y})^2] & = (\sigma_y^2 + \mu_y^2) - 2 \cdot \frac{1}{n}(\sigma_y^2 + n\mu_y^2) + \left( \frac{\sigma_y^2}{n} + \mu_y^2 \right) = \sigma_y^2 \left(1 - \frac{1}{n} \right) \end{align}\] Case 2: \(i \ne j\) \[\begin{align} \mathbb{E}[(y_i - \bar{y})(y_j - \bar{y})] & = \mu_y^2 - 2 \cdot \frac{1}{n} (\sigma_y^2 + n\mu_y^2) + \left( \frac{\sigma_y^2}{n} + \mu_y^2 \right) = -\frac{\sigma_y^2}{n} \end{align}\] which yields 10 , with a similar result holding for \(F\). Returning to 9 , we can observe that there are \(n\) terms where \(i = j\) and \(n(n-1)\) terms where \(i \neq j\). Therefore, we obtain \[\begin{align} \mathbb{E}[\hat{\sigma}_{fy}^2] & = \frac{1}{(n-1)^2} \left[ n \cdot \sigma_y^2 \left(1 - \frac{1}{n} \right) \cdot \sigma_f^2 \left(1 - \frac{1}{n} \right) + n(n-1) \cdot \left(-\frac{\sigma_y^2}{n}\right) \cdot \left(-\frac{\sigma_f^2}{n}\right) \right] \\ & = \frac{1}{(n-1)^2} \left[ n \cdot \vary\varf \frac{(n-1)^2}{n^2} + \vary\varf\frac{n(n-1)}{n^2}\right] \\ & = \frac{n(n-1)}{n^2(n-1)^2} \vary\varf \left[n - 1 + 1 \right] \\ & = \frac{\vary \varf}{n-1} \end{align}\] which gives the desired result.

Finally, it is worth noting that if we estimate the covariance from \(n' = c \cdot n\) samples, then the estimation error becomes, \[\begin{align} \mathbb{E}[\hat{\sigma}_{fy}^2] = \frac{\vary \varf}{cn-1} \end{align}\] ◻

With [lemma:indp-covariance95estimation] in hand, we now prove [theorem:cross-ppi-indpt-mse]. Note that we could have also written this proof using the more general result of [theorem:gen-covariance95estimation] instead of [lemma:indp-covariance95estimation].

Proof. From [theorem:cross-ppi-general-mse], we have after applying zero covariance between \(Y, F\), \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) = \frac{\sigma_y^2}{n} + \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} \end{align}\] since any term involving a covariance between powers of \(F\) and powers of \(Y\) reduces to zero in the independent case.

Now, we have from [lemma:indp-covariance95estimation] that \[\begin{align} \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} = \frac{1}{n\sigma_f^2} \cdot \frac{\vary \varf}{(n/2)-1} \end{align}\]

where we have adjusted the covariance estimation error to use \(n/2\) samples as in the case of a single-fold of cross-fit. This gives us the desired result: \[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) & = \frac{\sigma_y^2}{n} + \frac{1}{n\sigma_f^2} \cdot \frac{\vary \varf}{(n/2)-1} \\ & = \frac{\sigma_y^2}{n} + \frac{2}{n(n-2)}\cdot\vary \\ & = \frac{\vary}{n} \left(1 + \frac{2}{n-2}\right) \end{align}\] ◻

7.8 [theorem:cross-ppi-general-mse]: Covariance Estimation Error General↩︎

Proof. Noting that \(\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2] = \mathbb{E}[\hat{\sigma}_{fy}^2] - \sigma_{fy}^2\), we focus on first term. \[\begin{align} \mathbb{E}[\hat{\sigma}_{fy}^2] & = \frac{1}{(n-1)^2} \sum_{i} \sum_{j} \mathbb{E}[(y_i f_i - \bar{y} \bar{f})(y_j f_j - \bar{y} \bar{f})] & \text{by~\ref{lem:rewrite95covariance}} \nonumber \\ & = \frac{1}{(n-1)^2} \sum_{i} \sum_{j} \mathbb{E}[(y_i f_i y_j f_j - y_i f_i \bar{y} \bar{f} - y_j f_j \bar{y} \bar{f} + (\bar{y} \bar{f})^2)] \nonumber \\ & = \frac{1}{(n-1)^2} \left(\sum_{i} \sum_{j} \mathbb{E}[y_i f_i y_j f_j] - 2n^2 \mathbb{E}[y_i f_i \bar{y} \bar{f}] + n^2 \mathbb{E}[\bar{y} \bar{f} \bar{y} \bar{f}]\right) \nonumber \\ & = \frac{1}{(n-1)^2} \left(n \mathbb{E}[y_i f_i y_i f_i] + n(n-1) \mathbb{E}[y_i f_i y_j f_j] - 2n^2 \mathbb{E}[y_i f_i \bar{y} \bar{f}] + n^2 \mathbb{E}[\bar{y} \bar{f} \bar{y} \bar{f}]\right) \label{eq:gen95cov95simple951} \end{align}\tag{11}\] where we use \(\{(y_i, f_i), (y_j, f_j)\}\) as shorthand for any pair of samples \((y, f)\) that are independently drawn, as discussed in 2. We can further note that \[\begin{align} \mathbb{E}[\bar{y} \bar{f} \bar{y} \bar{f}] & = \mathbb{E}[y_i \bar{f} \bar{y} \bar{f}] = \frac{1}{n} \mathbb{E}[y_i f_i \bar{y} \bar{f}] + \frac{n-1}{n} \mathbb{E}[y_i f_j \bar{y} \bar{f}] \end{align}\] which we can combine with 11 to obtain \[\begin{align} & \mathbb{E}[\hat{\sigma}_{fy}^2]\nonumber \\ & = \frac{1}{(n-1)^2} \left(n \mathbb{E}[y_i f_i y_i f_i] + n(n-1) \mathbb{E}[y_i f_i y_j f_j] - 2n^2 \mathbb{E}[y_i f_i \bar{y} \bar{f}] + n^2 \left(\frac{1}{n} \mathbb{E}[y_i f_i \bar{y} \bar{f}] + \frac{n-1}{n} \mathbb{E}[y_i f_j \bar{y} \bar{f}]\right)\right) \nonumber \\ & = \frac{1}{(n-1)^2} \left(n \mathbb{E}[y_i f_i y_i f_i] + n(n-1) \mathbb{E}[y_i f_i y_j f_j] - (2n^2 - n) \mathbb{E}[y_i f_i \bar{y} \bar{f}] + n(n-1) \mathbb{E}[y_i f_j \bar{y} \bar{f}]\right)\nonumber \\ & = \frac{1}{(n-1)^2} \left(n \mathbb{E}[y^2 f^2] + n(n-1) \mathbb{E}[yf]^2 - (2n^2 - n) \mathbb{E}[y_i f_i \bar{y} \bar{f}] + n(n-1) \mathbb{E}[y_i f_j \bar{y} \bar{f}]\right)\nonumber \\ & = \frac{1}{(n-1)^2} \left(n \mathbb{E}[y^2 f^2] + n(n-1) \mathbb{E}[yf]^2 - n^2 \mathbb{E}[y_i f_i \bar{y} \bar{f}] - n(n-1) (\mathbb{E}[y_i f_i \bar{y} \bar{f}] -\mathbb{E}[y_i f_j \bar{y} \bar{f}]\right)\label{eq:gen95cov95simple952} \end{align}\tag{12}\] We can further note that \[\begin{align} & \mathbb{E}[y_i f_i \bar{y} \bar{f}] = \frac{1}{n} \mathbb{E}[y_i f_i (y_i + \sum_{j\neq i} y_j) \bar{f}] \nonumber \\ & = \frac{1}{n} \left(\mathbb{E}[y_i^2 f_i \bar{f}] + (n-1) \mathbb{E}[y_i f_i y_j \bar{f}]\right) \nonumber \\ & = \frac{1}{n^2} \left(\mathbb{E}[y_i^2 f_i (f_i + \sum_{j \neq i} f_j)] + (n-1) \mathbb{E}[y_i f_i y_j (f_i + f_j + \sum_{k \neq (i, j)} f_k )]\right) \nonumber \\ & = \frac{1}{n^2} \left(\mathbb{E}[y^2 f^2] + (n-1) \mathbb{E}[y^2 f]\mathbb{E}[f] + (n-1) \mathbb{E}[yf^2]\mathbb{E}[y] + (n-1) \mathbb{E}[yf]^2 + (n-1)(n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f]\right) \label{eq:gen95cov95simple953} \end{align}\tag{13}\] and similarly, \[\begin{align} & \mathbb{E}[y_i f_j \bar{y} \bar{f}] = \frac{1}{n} \mathbb{E}[y_i f_j (y_i + y_j + \sum_{k\neq (i, j)} y_k) \bar{f}] \\ & = \frac{1}{n} \left(\mathbb{E}[y_i^2 f_j \bar{f}] + \mathbb{E}[y_i f_j y_j \bar{f}] + (n-2) \mathbb{E}[y_i f_j y_k \bar{f}]\right) \\ & = \frac{1}{n^2} \left(\mathbb{E}[y_i^2 f_j (f_i + f_j + \sum_{k \neq (i, j)} f_k) ] + \mathbb{E}[y_i f_j y_j (f_i + f_j + \sum_{k \neq (i, j)} f_k)] + (n-2) \mathbb{E}[y_i f_j y_k (f_i + f_j + f_k + \sum_{l \neq (i, j, k)} f_l)]\right) \\ & = \frac{1}{n^2} \bigg(\mathbb{E}[y^2 f] \mathbb{E}[f] + \mathbb{E}[y^2]\mathbb{E}[f^2] + (n-2) \mathbb{E}[y^2]\mathbb{E}[f]^2 \\ & \qquad \qquad + \mathbb{E}[y f]^2 + \mathbb{E}[yf^2]\mathbb{E}[y] + (n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] \\ & \qquad \qquad + (n-2) \bigg(\mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] + \mathbb{E}[f^2]\mathbb{E}[y]^2 + \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] + (n-3) \mathbb{E}[y]^2 \mathbb{E}[f]^2\bigg) \bigg) \end{align}\] which gives us that the last term of 12 is given by \[\begin{align} & n(n-1) (\mathbb{E}[y_i f_i \bar{y} \bar{f}] - \mathbb{E}[y_i f_j \bar{y} \bar{f}]) \\ & = \frac{n(n-1)}{n^2} \bigg(\mathbb{E}[y^2 f^2] + (n-1) \mathbb{E}[y^2 f]\mathbb{E}[f] + (n-1) \mathbb{E}[yf^2]\mathbb{E}[y] + (n-1) \mathbb{E}[yf]^2 + (n-1)(n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] \\ & \qquad - \bigg(\mathbb{E}[y^2 f] \mathbb{E}[f] + \mathbb{E}[y^2]\mathbb{E}[f^2] + (n-2) \mathbb{E}[y^2]\mathbb{E}[f]^2 \\ & \qquad \qquad + \mathbb{E}[y f]^2 + \mathbb{E}[yf^2]\mathbb{E}[y] + (n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] \\ & \qquad \qquad + 2 (n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] + (n-2) \mathbb{E}[f^2]\mathbb{E}[y]^2 + (n-2) (n-3) \mathbb{E}[y]^2 \mathbb{E}[f]^2\bigg)\bigg) \\ & = \frac{n(n-1)}{n^2} \bigg( [ \mathbb{E}[y^2 f^2] - \mathbb{E}[y^2]\mathbb{E}[f^2] ] + (n-2) [ \mathbb{E}[y^2 f]\mathbb{E}[f] - \mathbb{E}[y^2]\mathbb{E}[f]\mathbb{E}[f] ] + (n-2) [ \mathbb{E}[yf^2]\mathbb{E}[y] - \mathbb{E}[y]\mathbb{E}[f^2]\mathbb{E}[y] ] \\ & \qquad \qquad + (n-2) (\mathbb{E}[yf]^2 - \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f]) + (n-2)(n-3) [ \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f] - \mathbb{E}[y]\mathbb{E}[f]\mathbb{E}[y]\mathbb{E}[f] ]\bigg) \end{align}\] Noting that \(\sigma_{ab} = \mathbb{E}[ab] - \mathbb{E}[a]\mathbb{E}[b]\), we can then rewrite the above as \[\begin{align} & n(n-1) (\mathbb{E}[y_i f_i \bar{y} \bar{f}] - \mathbb{E}[y_i f_j \bar{y} \bar{f}]) \nonumber \\ & = \frac{n(n-1)}{n^2} \bigg( \sigma_{y^2 f^2} + (n-2) \sigma_{y^2 f} \mu_f + (n-2) \sigma_{yf^2} \mu_y + (n-2) \mathbb{E}[yf] \sigma_{yf} + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg) \label{eq:gen95cov95simple954} \end{align}\tag{14}\] and we can combine 13 14 with 12 to obtain \[\begin{align} & \mathbb{E}[\hat{\sigma}_{fy}^2] \\ & = \frac{1}{(n-1)^2} \bigg(n \mathbb{E}[y^2 f^2] + n(n-1) \mathbb{E}[yf]^2 & \text{(From Eq.~\ref{eq:gen95cov95simple952})} \\ & \qquad \qquad - n^2 \mathbb{E}[y_i f_i \bar{y} \bar{f}] - n(n-1) (\mathbb{E}[y_i f_i \bar{y} \bar{f}] -\mathbb{E}[y_i f_j \bar{y} \bar{f}]\bigg) \\ & = \frac{1}{(n-1)^2} \bigg(n \mathbb{E}[y^2 f^2] + n(n-1) \mathbb{E}[yf]^2 \\ & \qquad - \bigg(\mathbb{E}[y^2 f^2] + (n-1) \mathbb{E}[y^2 f]\mathbb{E}[f] + (n-1) \mathbb{E}[yf^2]\mathbb{E}[y] & \text{(From Eq.~\ref{eq:gen95cov95simple953})} \\ & \qquad \qquad + (n-1) \mathbb{E}[yf]^2 + (n-1)(n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f]\bigg) \\ & \qquad - \frac{n(n-1)}{n^2} \bigg( \sigma_{y^2 f^2} + (n-2) \sigma_{y^2 f} \mu_f & \text{(From Eq.~\ref{eq:gen95cov95simple954})} \\ & \qquad \qquad + (n-2) \sigma_{yf^2} \mu_y + (n-2) \mathbb{E}[yf] \sigma_{yf} + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{1}{(n-1)^2} \bigg( (n - 1) \mathbb{E}[y^2 f^2] + (n-1) (n-1) \mathbb{E}[yf]^2 & \text{(Collect terms)} \\ & \qquad - \bigg((n-1) \mathbb{E}[y^2 f]\mathbb{E}[f] + (n-1) \mathbb{E}[yf^2]\mathbb{E}[y] + (n-1)(n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f]\bigg) \\ & \qquad - \frac{n(n-1)}{n^2} \bigg( \sigma_{y^2 f^2} + (n-2) \sigma_{y^2 f} \mu_f \\ & \qquad \qquad + (n-2) \sigma_{yf^2} \mu_y + (n-2) \mathbb{E}[yf] \sigma_{yf} + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{1}{n(n-1)} \bigg(n \mathbb{E}[y^2 f^2] + n(n-1) \mathbb{E}[yf]^2 & \text{(Multiply by \frac{n}{n})}\\ & \qquad - \bigg(n\mathbb{E}[y^2 f]\mathbb{E}[f] + n\mathbb{E}[yf^2]\mathbb{E}[y] + n(n-2) \mathbb{E}[yf]\mathbb{E}[y]\mathbb{E}[f]\bigg) \\ & \qquad - \bigg(\sigma_{y^2 f^2} + (n-2) \sigma_{y^2 f} \mu_f \\ & \qquad \qquad + (n-2) \sigma_{yf^2} \mu_y + (n-2) \mathbb{E}[yf] \sigma_{yf} + (n-2)(n-3)\sigma_{yf} \mu_y \mu_f\bigg)\bigg) \end{align}\] To further simplify, we expand terms like \(\mathbb{E}[y^2 f] = \sigma_{y^2 f} + \mathbb{E}[y^2] \mathbb{E}[f] = \sigma_{y^2 f} + (\sigma_{y}^2 + \muy^2) \mu_f\) and so on. \[\begin{align} & = \frac{1}{n(n-1)} \bigg(n\sigma_{y^2 f^2} + n (\sigma_y^2 + \muy^2)(\sigma_f^2 + \muf^2) + n(n-1) (\sigma_{yf} + \muf \muy)(\sigma_{yf} + \muf \muy) \\ & \qquad - \bigg(n \sigma_{y^2 f} \muf + n (\sigma_{y}^2 + \muy^2)\muf^2 + n \sigma_{yf^2} \muy + n (\sigma_{f}^2 + \muf^2) \muy^2 + n(n-2) (\sigma_{yf} + \muf \muy) \muf \muy\bigg) \\ & \qquad - \bigg(\sigma_{y^2 f^2} + (n-2) \sigma_{y^2 f} \mu_f + (n-2) \sigma_{yf^2} \mu_y + (n-2) (\sigma_{yf} + \muf \muy) \sigma_{yf} + (n-2)(n-3)\sigma_{yf} \mu_y \mu_f\bigg)\bigg) \end{align}\] We then expand and collect similar terms \[\begin{align} & = \frac{1}{n(n-1)} \bigg({\color{blue} n\sigma_{y^2 f^2}} + n ({\color{blue} \sigma_y^2 \sigma_f^2} + \sigma_y^2 \muf^2 + \sigma_f^2 \muy^2 + \muy^2 \muf^2) + n(n-1) (\sigma_{yf}^2 + 2 \sigma_{yf} \muf \muy + \muf^2 \muy^2) \\ & \qquad - \bigg(n \sigma_{y^2 f} \muf + n (\sigma_{y}^2 \muf^2 + \muf^2 \muy^2) + n \sigma_{yf^2} \muy + n (\sigma_{f}^2 \muy^2 + \muf^2 \muy^2) + n(n-2) (\sigma_{yf} \muf \muy + \muf^2 \muy^2) \bigg) \\ & \qquad - \bigg({\color{blue} \sigma_{y^2 f^2}} + (n-2) \sigma_{y^2 f} \mu_f + (n-2) \sigma_{yf^2} \mu_y + (n-2) (\sigma_{yf}^2 + \sigma_{yf} \muf \muy) + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{\sigma_{y^2 f^2}}{n} + \frac{\sigma_y^2 \sigma_f^2}{n-1} + \frac{1}{n(n-1)} \bigg(n (\sigma_y^2 \muf^2 + \sigma_f^2 \muy^2 + \muy^2 \muf^2) + n(n-1) ({\color{blue} \sigma_{yf}^2 } + 2 \sigma_{yf} \muf \muy + \muf^2 \muy^2) \\ & \qquad - \bigg(n \sigma_{y^2 f} \muf + n (\sigma_{y}^2 \muf^2 + \muf^2 \muy^2) + n \sigma_{yf^2} \muy + n (\sigma_{f}^2 \muy^2 + \muf^2 \muy^2) + n(n-2) (\sigma_{yf} \muf \muy + \muf^2 \muy^2) \bigg) \\ & \qquad - \bigg((n-2) \sigma_{y^2 f} \mu_f + (n-2) \sigma_{yf^2} \mu_y + (n-2) ({\color{blue} \sigma_{yf}^2} + \sigma_{yf} \muf \muy) + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{\sigma_{y^2 f^2}}{n} + \frac{\sigma_y^2 \sigma_f^2}{n-1} + \left(1 - \frac{n-2}{n(n-1)}\right) \sigma_{yf}^2 + \frac{1}{n(n-1)} \bigg(n (\sigma_y^2 \muf^2 + \sigma_f^2 \muy^2 + {\color{red} \muy^2 \muf^2)} + n(n-1) (2 \sigma_{yf} \muf \muy + {\color{red} \muf^2 \muy^2})\\ & \qquad - \bigg(n \sigma_{y^2 f} \muf + n (\sigma_{y}^2 \muf^2 + {\color{red} \muf^2 \muy^2}) + n \sigma_{yf^2} \muy + n (\sigma_{f}^2 \muy^2 + {\color{red} \muf^2 \muy^2}) + n(n-2) (\sigma_{yf} \muf \muy + {\color{red} \muf^2 \muy^2}) \bigg) \\ & \qquad - \bigg((n-2) \sigma_{y^2 f} \mu_f + (n-2) \sigma_{yf^2} \mu_y + (n-2) (\sigma_{yf} \muf \muy) + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{\sigma_{y^2 f^2}}{n} + \frac{\sigma_y^2 \sigma_f^2}{n-1} + \left(1 - \frac{n-2}{n(n-1)}\right) \sigma_{yf}^2 + \frac{1}{n(n-1)} \bigg(n ({\color{red} \sigma_y^2 \muf^2 + \sigma_f^2 \muy^2}) + n(n-1) (2 \sigma_{yf} \muf \muy)\\ & \qquad - \bigg(n \sigma_{y^2 f} \muf + n {\color{red} \sigma_{y}^2 \muf^2} + n \sigma_{yf^2} \muy + n {\color{red} \sigma_{f}^2 \muy^2} + n(n-2) (\sigma_{yf} \muf \muy) \bigg) \\ & \qquad - \bigg((n-2) \sigma_{y^2 f} \mu_f + (n-2) \sigma_{yf^2} \mu_y + (n-2) (\sigma_{yf} \muf \muy) + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{\sigma_{y^2 f^2}}{n} + \frac{\sigma_y^2 \sigma_f^2}{n-1} + \left(1 - \frac{n-2}{n(n-1)}\right) \sigma_{yf}^2 + \frac{1}{n(n-1)} \bigg(n(n-1) (2 \sigma_{yf} \muf \muy)\\ & \qquad - \bigg(n {\color{blue} \sigma_{y^2 f} \muf} + n {\color{blue} \sigma_{yf^2} \muy } + n(n-2) (\sigma_{yf} \muf \muy) \bigg) \\ & \qquad - \bigg((n-2) {\color{blue} \sigma_{y^2 f} \mu_f} + (n-2) {\color{blue} \sigma_{yf^2} \mu_y } + (n-2) (\sigma_{yf} \muf \muy) + (n-2)(n-3) \sigma_{yf} \mu_y \mu_f\bigg)\bigg) \\ & = \frac{\sigma_{y^2 f^2}}{n} + \frac{\sigma_y^2 \sigma_f^2}{n-1} + \left(1 - \frac{n-2}{n(n-1)}\right) \sigma_{yf}^2 - \frac{1}{n} (2\sigma_{y^2 f} \muf + 2\sigma_{yf^2} \muy ) \\ & \qquad + \frac{1}{n(n-1)} \bigg( [ 2n(n-1) - n(n-2) - (n-2) - (n-2)(n-3) ] (\sigma_{yf} \muf \muy) \bigg)\\ & = \frac{\sigma_{y^2 f^2}}{n} + \frac{\sigma_y^2 \sigma_f^2}{n-1} + \left(1 - \frac{n-2}{n(n-1)}\right) \sigma_{yf}^2 - \frac{2}{n} (\sigma_{y^2 f} \muf + \sigma_{yf^2} \muy - 2 \sigma_{yf} \muf \muy ) \addtocounter{equation}{1}\label{eq:gen95cov95simple95final} \end{align}\tag{15}\] where in the last line we use the fact that \[\begin{align} &\frac{1}{n(n-1)} [ 2n(n-1) - n(n-2) - (n-2) - (n-2)(n-3) ] \\ &= \frac{1}{n(n-1)} [{\color{red}2n^2} - 2n - {\color{red}n^2} + 2n - n + 2 - ({\color{red}n^2} - 5n + 6)] \\ &= \frac{1}{n(n-1)} [- {\color{red}2n} + {\color{red}2n} - n + 2 + 5n - 6] \\ &= \frac{1}{n(n-1)} [- n + 2 + 5n - 6] \\ &= \frac{1}{n(n-1)} [4(n - 1)] \\ &= \frac{4}{n} \end{align}\] Subtracting \(\sigma_{yf}^2\) from the final expression for \(\mathbb{E}[\hat{\sigma}_{yf}^2]\) in 15 , to obtain \(\mathbb{E}[\hat{\sigma}^2_{yf}] - \sigma^2_{yf} = \mathbb{E}[(\hat{\sigma}_{yf} - \sigma_{yf})^2]\), gives the desired result. ◻

7.9 4.1 Proofs for the Gaussian Setting↩︎

We begin with some notes on special properties of Gaussians. We then prove the results in 4.1 for and then follow it up with the proofs for the portion of the statements.

For jointly normal random variables \(X\) and \(Y\) with means \(\mu_x\), \(\mu_y\), variances \(\sigma_x^2\), \(\sigma_y^2\), and covariance \(\sigma_{xy}\), we have the following identities: \[\begin{align} \mathbb{E}[X^2Y^2] & = \mu_x^2\mu_y^2 + \sigma_x^2\mu_y^2 + \sigma_y^2\mu_x^2 + \sigma_x^2\sigma_y^2 + 2\sigma_{xy}^2 + 4\sigma_{xy}\mu_x\mu_y \\ \mathbb{E}[X^2Y] & = \mu_x^2\mu_y + \sigma_x^2\mu_y + 2\sigma_{xy}\mu_x \\ \mathbb{E}[XY^2] & = \mu_x\mu_y^2 + \sigma_y^2\mu_x + 2\sigma_{xy}\mu_y \end{align}\] Now consider the following expression for the expected value of the squared sample covariance obtained from [theorem:gen-covariance95estimation]: \[\begin{align} \mathbb{E}\left[\hat{\sigma}_{fy}^2\right] & = \frac{1}{n}\sigma_{f^2y^2} + \frac{1}{n-1}\sigma_f^2\sigma_y^2 + \left[\frac{n-1}{n} + \frac{1}{n(n-1)}\right]\sigma_{fy}^2 \\ & + \frac{1}{n}[-2\sigma_{y^2f}\mu_f - 2\sigma_{f^2y}\mu_y + 4\sigma_{fy}\mu_f\mu_y] \end{align}\] Using these identities, we can simplify the covariance terms in our expression: \[\begin{align} \sigma_{f^2y^2} & = \mathbb{E}[F^2Y^2] - \mathbb{E}[F^2]\mathbb{E}[Y^2] \\ & = (\mu_f^2\mu_y^2 + \sigma_f^2\mu_y^2 + \sigma_y^2\mu_f^2 + \sigma_f^2\sigma_y^2 + 2\sigma_{fy}^2 + 4\sigma_{fy}\mu_f\mu_y) \\ & \quad - (\mu_f^2 + \sigma_f^2)(\mu_y^2 + \sigma_y^2) \\ & = \mu_f^2\mu_y^2 + \sigma_f^2\mu_y^2 + \sigma_y^2\mu_f^2 + \sigma_f^2\sigma_y^2 + 2\sigma_{fy}^2 + 4\sigma_{fy}\mu_f\mu_y \\ & \quad - \mu_f^2\mu_y^2 - \mu_f^2\sigma_y^2 - \sigma_f^2\mu_y^2 - \sigma_f^2\sigma_y^2 \\ & = 2\sigma_{fy}^2 + 4\sigma_{fy}\mu_f\mu_y \end{align}\] Similarly for the other covariance terms: \[\begin{align} \label{eq:special-gaussian} \sigma_{f^2y} & = \mathbb{E}[F^2Y] - \mathbb{E}[F^2]\mathbb{E}[Y] \\ & = (\mu_f^2\mu_y + \sigma_f^2\mu_y + 2\sigma_{fy}\mu_f) - (\mu_f^2 + \sigma_f^2)\mu_y \\ & = \mu_f^2\mu_y + \sigma_f^2\mu_y + 2\sigma_{fy}\mu_f - \mu_f^2\mu_y - \sigma_f^2\mu_y \\ & = 2\sigma_{fy}\mu_f \end{align}\tag{16}\] and \[\begin{align} \sigma_{y^2f} & = \mathbb{E}[Y^2F] - \mathbb{E}[Y^2]\mathbb{E}[F] \\ & = (\mu_y^2\mu_f + \sigma_y^2\mu_f + 2\sigma_{fy}\mu_y) - (\mu_y^2 + \sigma_y^2)\mu_f \\ & = \mu_y^2\mu_f + \sigma_y^2\mu_f + 2\sigma_{fy}\mu_y - \mu_y^2\mu_f - \sigma_y^2\mu_f \\ & = 2\sigma_{fy}\mu_y \end{align}\] Now we substitute these expressions into our original equation: \[\begin{align} \mathbb{E}[\hat{\sigma}_{fy}^2] & = \frac{1}{n}(2\sigma_{fy}^2 + 4\sigma_{fy}\mu_f\mu_y) + \frac{1}{n-1}\sigma_f^2\sigma_y^2 \\ & \quad + \left[\frac{n-1}{n} + \frac{1}{n(n-1)}\right]\sigma_{fy}^2 \\ & \quad + \frac{1}{n}[-2(2\sigma_{fy}\mu_y)\mu_f - 2(2\sigma_{fy}\mu_f)\mu_y + 4\sigma_{fy}\mu_f\mu_y] \end{align}\] Simplifying the terms with \(\mu_f\) and \(\mu_y\): \[\begin{align} \frac{1}{n}(4\sigma_{fy}\mu_f\mu_y - 4\sigma_{fy}\mu_y\mu_f - 4\sigma_{fy}\mu_f\mu_y + 4\sigma_{fy}\mu_f\mu_y) = 0 \end{align}\] So these terms cancel out completely. Now we simplify the terms with \(\sigma_{fy}^2\): \[\begin{align} \frac{2\sigma_{fy}^2}{n} + \left[\frac{n-1}{n} + \frac{1}{n(n-1)}\right]\sigma_{fy}^2 & = \sigma_{fy}^2\left[\frac{2}{n} + \frac{n-1}{n} + \frac{1}{n(n-1)}\right] \\ & = \sigma_{fy}^2\left[\frac{n+1}{n} + \frac{1}{n(n-1)}\right] \\ & = \sigma_{fy}^2\left[\frac{(n+1)(n-1) + 1}{n(n-1)}\right] \\ & = \sigma_{fy}^2\left[\frac{n^2-1+1}{n(n-1)}\right] \\ & = \sigma_{fy}^2\left[\frac{n^2}{n(n-1)}\right] \\ & = \sigma_{fy}^2\left[\frac{n}{n-1}\right] \end{align}\] Therefore, for jointly normal random variables \(F\) and \(Y\), the Covariance Estimation Error becomes: \[\begin{align} \mathbb{E}[\hat{\sigma}_{fy}^2] - \sigma_{fy}^2 = \frac{n}{n-1}\sigma_{fy}^2 + \frac{1}{n-1}\sigma_f^2\sigma_y^2 - \sigma_{fy}^2 = \frac{1}{n-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) \end{align}\] or \[\begin{align} \label{eq:gauss-cov-err} \mathbb{E}[(\hat{\sigma}_{fy}-\sigma_{fy})^2] = \frac{n}{n-1}\sigma_{fy}^2 + \frac{1}{n-1}\sigma_f^2\sigma_y^2 - \sigma_{fy}^2 = \frac{1}{n-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) \end{align}\tag{17}\]

We are now ready to prove the results in 4.1. We begin with a proof for the Cross-fit PPI++ portion of the results. Then we treat the Single-sample case.

7.9.1 Proofs for the estimator, Gaussian↩︎

We now include an additional lemma for use,

Lemma 3 (Variance between the Folds, Gaussian). In the case of jointly Gaussian variables, \(F,Y\) the variance between the folds for vanishes.

Proof. We have from 16 , that for Gaussians, \[\begin{align} \sigma_{f^2y} = 2\sigma_{fy}\mu_f \implies \sigma_{f^2y} - 2\sigma_{fy}\mu_f = 0 \end{align}\]

The variance between the folds, from [theorem:cross-ppi-general-mse], specifically, 2 shows this difference squared is in fact the cross term. This completes the proof of this lemma. ◻

That the variance between the folds is zero has the following consequences for

  1. 1 becomes a necessary condition and not just a sufficient condition. It was previously sufficient since the variance between the folds added an additional positive term to the variance of the estimator. This term is now zero making the loss \(>\) gain for worse performance from a necessary condition.

  2. This is now a reversible condition. That is, if \(\mathbb{E}[(\hat{\sigma}_{fy}-\sigma_{fy})^2] < \sigma_{fy}^2\) then performs better than 1.

We now prove the condition for improvement using in the Gaussian case.

Proof of [theorem:jointly-normal-mse-condition-all]: Condition for Performant Cross-fit PPI++, Gaussian

Proof. While 17 expresses the covariance estimation error when the covariance is estimated using \(n\) samples, uses only \(n/2\) samples to estimate the covariance. In this case, the covariance estimation error becomes,

\[\begin{align} \label{eq:gauss-cov-err-crossfit} \mathbb{E}[(\hat{\sigma}_{fy}-\sigma_{fy})^2] = \frac{1}{n/2-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) = \frac{2}{n-2}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) \end{align}\tag{18}\]

Now 1 gives us the condition for \(\mathrm{Var}\) of the estimator to be higher than \(\mathrm{Var}\) of the estimator as

\(\mathrm{Var}(\thetah_{\text{CF-PPI++}}) > \sigma_y^2 / n\) when \(\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2] > \sigma_{fy}^2\). For the Gaussian case, substituting 18 in this, we get the condition for higher \(\mathrm{Var}\) of as \[\begin{align} \frac{2}{n-2}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) & > \sigma_{fy}^2 \\ 2\sigma_{fy}^2 + 2\sigma_f^2\sigma_y^2 & > n\sigma_{fy}^2 - 2\sigma_{fy}^2 \\ 2\sigma_f^2\sigma_y^2 & > (n-4)\sigma_{fy}^2 \end{align}\] Or, the condition for lower variance is given by, \[\begin{align} (n-4)\sigma_{fy}^2 & > 2\sigma_f^2\sigma_y^2 \\ \frac{\sigma_{fy}^2}{\sigma_f^2\sigma_y^2} & > \frac{2}{(n-4)} \\ \lvert \frac{\sigma_{fy}}{\sigma_f\sigma_y} \rvert & > \frac{\sqrt{2}}{\sqrt{n-4}} \\ \lvert \frac{\sigma_{fy}}{\sigma_f\sigma_y}\rvert &> \frac{1}{\sqrt{n/2 -2}} \\ \lvert \frac{\sigma_{fy}}{\sigma_f\sigma_y}\rvert &> \frac{1}{\sqrt{cn -2}} \\ \end{align}\]

which completes the proof of [theorem:jointly-normal-mse-condition-all] ◻

Proof of [theorem:GaussianMSE]: Variance of Cross-fit PPI++, Gaussian

Our covariance estimation error can be directly used in [theorem:cross-ppi-general-mse] to obtain the variance in the Gaussian case.

Proof. Consider 2 . Now, from 3 we know that its cross-term is zero. Further, from 18 we obtain the covariance estimation error for the case. Substituting these and simplifying should land the desired result. Thus, from 2 , we have

\[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) = \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} + \frac{2}{n^2 \sigma_{f}^4} \left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2 \end{align}\],

and we know from 3, we have \(\left(\sigma_{yf^2} - 2 \sigma_{yf} \muf \right)^2 = 0\).

Further, from 18 , we get \(\mathbb{E}[(\hat{\sigma}_{fy}-\sigma_{fy})^2] = \frac{2}{n-2}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2)\)

Substituting, we obtain,

\[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) &= \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{1}{n\sigma_f^2}\frac{2}{n-2}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) \\ &= \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2}.\frac{n-2}{n-2} + \frac{1}{n\sigma_f^2}\frac{2}{n-2}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2) \\ &= \frac{\sigma_y^2}{n} - \frac{\left(n - 2 -2\right)\sigma_{fy}^2}{n(n-2)\sigma_f^2} + \frac{2\sigma_y^2}{n(n-2)} \\ &= \frac{\sigma_y^2}{n}\left(1 + \frac{2}{n-2}\right) - \frac{n-4}{n(n-2)}\sigma_{fy}^2 \\ \end{align}\]

To obtain the form that we want,

\[\begin{align} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) &= \frac{\sigma_y^2}{n}\left(1 + \frac{2}{n-2}\right) - \frac{n-4}{n(n-2)}\sigma_{fy}^2 \\ &= \frac{\sigma_y^2}{n}\left(1 + \frac{1}{n/2 - 1}\right) - \frac{2(n/2 - 2)\sigma_{fy}^2}{2n(n/2-1)} \\ &= \frac{\sigma_y^2}{n}\left(1 + \frac{1}{n/2 - 1}\right) - \frac{(n/2 - 2)\sigma_{fy}^2}{n(n/2-1)} \end{align}\]

Setting \(c = 1/2\), we obtain \[\begin{align} \label{eq:gauss-mse-crossfit-proof} \mathrm{Var}(\thetah_{\text{CF-PPI++}}) &= \frac{\sigma_y^2}{n}\left(1 + \frac{1}{n/2 - 1}\right) - \frac{(n/2 - 2)\sigma_{fy}^2}{n(n/2-1)} \\ &= \frac{\sigma_y^2}{n}\left(1 + \frac{1}{cn - 1}\right) - \frac{(cn - 2)\sigma_{fy}^2}{n(cn-1)} \end{align}\tag{19}\] which completes the proof. ◻

Finally, the variance when \(F, Y\) are independent Gaussians follows trivially from the above as follows.

Those complete the proofs for in the Gaussian case. We now turn to the proof of portion of the statements from 4.1.

7.9.2 Proofs for the estimator, Gaussian↩︎

Given that we don’t have a generalized expression for the variance of the , our proof involves deriving this variance from first principles. The reason we are able to do so in this Gaussian setting lies in [lemma:gaussian-indpt]. The independence of the sample covariance and the sample means results in \(\lh\) being independent of \((\mu_f - \bar{f}_n)\) which dramatically simplifies the analysis. The rest of the proof mechanics is quite similar to our proof of [theorem:cross-ppi-general-mse] in 7.5. Then we apply the conditions for improvement using this derived variance.

Consider the variance of the estimator: We have it that the errors in our estimator can be decomposed as follows \[(\thetahp - \theta) = \underbrace{(\bar{y}_{n} - \theta)}_{A} + \lh \underbrace{\left(\muf - \frac{1}{n} \sum_{i \in \cD_n} f_i \right)}_{D} = A + \lh D\] Such that the mean squared error is given by \[\begin{align} \MSE(\thetahp) & = \mathbb{E}[(\thetahp - \theta)^2] \\ & = \mathbb{E}[(A + \lh D)^2] \\ & = \mathbb{E}[A^2 + 2A \lh D + {\lh}^{2} D^2] \\ & = \mathbb{E}[A^2] + 2\mathbb{E}[A \lh D] + \mathbb{E}[{\lh}^{2} D^2] & \text{by Linearity of Expectation} \end{align}\] We can observe that \[\mathbb{E}[A^2] = \mathbb{E}[(\bar{y}_{n} - \theta)^2] = \frac{\sigma_y^2}{n}\] is \(\mathrm{Var}(\thetahc)\) the classical variance. Now we consider the second term: \[\begin{align} 2\mathbb{E}[A \lh D] = 2\mathbb{E}[\lh (\bar{y}_{n} - \theta)(\mu_f - \bar{f}_{n})] \end{align}\] From [lemma:gaussian-indpt] we have that since \(Y,F\) are gaussian, \(\lh\) is independent of \(\bar{y}_{n}, \bar{f}_{n}\). Further, since \(\theta, \mu_f\) are non-random, we have that \(\lh\) is also independent of them. Overall, this gets us that \(\lh\) is independent of \((\bar{y}_{n} - \theta)(\mu_f - \bar{f}_{n})\). This allows us to apply [lemma:gen-cross-expect] with \(n\) samples to obtain \[\begin{align} 2\mathbb{E}[A \lh D] = 2\mathbb{E}[\lh (\bar{y}_{n} - \theta)(\mu_f - \bar{f}_{n})] = -2\frac{\sigma_{fy}^2}{n\sigma_f^2} = -\frac{2\sigma_{fy}^2}{n\sigma_f^2} \end{align}\] Now consider the third term: \(\mathbb{E}[{\lh}^{2} D^2]\), once again using [lemma:gaussian-indpt], \(\lh\) is independent of \(D\), we get \(\mathbb{E}[{\lh}^{2} D^2] = \mathbb{E}[\lh^2]\mathbb{E}[(\mu_f - \bar{f}_{n})^2]\) \[\begin{align} \lh = \frac{\hat{\sigma}_{fy}}{\varf} \end{align}\] where \(\hat{\sigma}_{fy}\) is an unbiased estimate of the covariance \(\sigma_{fy}\). Therefore, \[\begin{align} \mathbb{E}[\lh^2] = \frac{\mathbb{E}[\hat{\sigma}_{fy}^2]}{\sigma_f^4} \end{align}\] Further, consider, \(\mathbb{E}[(\bar{f}_{n} - \mu_f)^2]\). Since \(\mu_f\) is the expected value of \(\bar{f}_{n}\), this term reduces to the variance of \(\bar{f}_{n}\) which is \(\frac{\sigma_f^2}{n}\). Putting these together, we get \[\begin{align} \MSE(\thetahp) & = \frac{\sigma_y^2}{n} - \frac{2\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[\hat{\sigma}_{fy}^2]}{\sigma_f^4}\cdot\frac{\sigma_f^2}{n} \nonumber \\ & = \frac{\sigma_y^2}{n} -\frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[\hat{\sigma}_{fy}^2]}{n\sigma_f^2} - \frac{\sigma_{fy}^2}{n\sigma_f^2} \nonumber \\ & = \MSE(\thetahc) -\frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} \label{eq:mse95gaussian95single95sample95interm} \end{align}\tag{20}\]

Therefore, the variance of in the Gaussian case takes exactly the same form as . However, one important difference is that uses all \(n\) available samples for estimating the covariance thus reducing \(\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]\).

With this variance derived, we can easily prove the required theorems by simply combining it with the covariance estimation error that we derived.

Proof of [theorem:GaussianMSE]: Variance of , Gaussian

Proof. The covariance estimation error from 17 is \[\mathbb{E}[(\hat{\sigma}_{fy}-\sigma_{fy})^2] = \frac{1}{n-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2)\] Combining with 20 , we obtain \[\begin{align} \MSE(\thetahp) &= \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{\mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]}{n\sigma_f^2} \\ &= \frac{\sigma_y^2}{n} - \frac{\sigma_{fy}^2}{n\sigma_f^2} + \frac{1}{n-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2)\frac{1}{n\sigma_f^2} \\ &= \frac{\sigma_y^2}{n}\left(1 + \frac{1}{n-1}\right) - \frac{\sigma_{fy}^2}{n\sigma_f^2}\left(1 - \frac{1}{n-1}\right) \\ &= \frac{\sigma_y^2}{n}\left(1 + \frac{1}{n-1}\right) - \frac{\sigma_{fy}^2}{n\sigma_f^2}\left(\frac{n - 2 }{n-1}\right) \end{align}\]

If we set \(c = 1\), we obtain, \[\begin{align} \MSE(\thetahp) = \frac{\sigma_y^2}{n}\left(1 + \frac{1}{cn-1}\right) - \frac{\sigma_{fy}^2}{n\sigma_f^2}\left(\frac{cn - 2 }{cn-1}\right) \end{align}\] which completes the proof. ◻

Further, setting \(\sigma_{fy}\) to \(0\), we obtain \[\begin{align} \MSE(\thetahp) = \frac{\sigma_y^2}{n}\left(1 + \frac{1}{cn-1}\right) \end{align}\] which completes the proof for the estimator.

Finally, we prove the condition for improvement,

Proof of [theorem:jointly-normal-mse-condition-all], Condition for variance improvement estimator

Proof. Now, \(\MSE(\thetahp)\) is lower when \[\sigma_{fy}^2 > \mathbb{E}[(\hat{\sigma}_{fy} - \sigma_{fy})^2]\] This comes directly from 20 The covariance estimation error from 17 is \[\mathbb{E}[(\hat{\sigma}_{fy}-\sigma_{fy})^2] = \frac{1}{n-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2)\] Therefore the required condition becomes, \[\sigma_{fy}^2 > \frac{1}{n-1}(\sigma_{fy}^2 + \sigma_f^2\sigma_y^2)\] giving us, \[\begin{align} (n-2) \sigma_{fy}^2 & > \sigma_f^2\sigma_y^2 \\ \lvert \frac{\sigma_{fy}}{\sigma_f\sigma_y} \rvert & > \frac{1}{\sqrt{n - 2}} \end{align}\] ◻

7.9.3 Proof of the Finite-N Case↩︎

We now consider the setting where \(N\) is finite.

Proof. Suppose we have \((y_1, f_1), \dots, (y_n, f_n) \sim P_{FY}\), where \(P_{FY}\) is a jointly Gaussian distribution with correlation \(\rho_{fy}\). Let \(\bar{y}_n\) denote the sample mean of \(Y\) on this n-sample and \(\bar{f}_n\) denote the sample mean of \(f\) on this n-sample. Suppose similarly that we also observe an independent N-samples of \(F\) from \(P\), \(f_1, \dots, f_N \sim P_{FY}\). Similarly define the sample mean as \(\bar{f}_N\).

The \(\MSE\) of Single-Sample PPI++ can be written as,

\[\begin{align} \label{eq:ppi-gaussian-mse} \mathbb{E}[(\thetahp - \theta)^2] = \mathbb{E}[(\thetahc - \theta)^2] + 2\underbrace{\mathbb{E}[\lh(\thetahc - \theta)(\bar{f}_N - \bar{f}_n)]}_{T1} + \underbrace{\mathbb{E}[\lh^2(\bar{f}_N - \bar{f}_n)^2]}_{T2} \end{align}\tag{21}\]

We use the following helper lemmas:

lemmagaussiansamplecov Let \(Y,F\) be jointly normal. Suppose we observe \(n\) samples \(\{y_i,f_i\}_{i=1}^{n}\). Then the empirical covariance \(\hat{\sigma}_{fy}\) is statistically independent of the sample mean \(\bar{y}_n\) and \(\bar{f}_n\).

A special case of [lemma:gaussian-sample-cov-indpt] occurs when \(Y = F\), we state this here in the context of the \(N\) sample.

lemmagaussiansamplevar Let \(F\) be a random variable following a Gaussian distribution. Suppose we observe \(N\) samples \(\{f_i\}_{i=1}^{N}\). Then the empirical variance \(\hat{\sigma}_f^2\) is statistically independent of the sample mean \(\bar{f}_N\).

Informally, these lemmas help us buy the statistical independence of \(\lh\) from the difference in sample means \(\bar{f}_N - \bar{f}_n\). Notice that the numerator in \(\lh\) is composed of \(\hat{\sigma}_{fy}\) which is trivially independent of \(\bar{f}_N\) but now also independent of \(\bar{f}_n\) using [lemma:gaussian-sample-cov-indpt]. Further, the denominator in \(\lh\) contains \(\hat{\sigma}_f^2\) which from  [lemma:gaussian-sample-var-indpt] is independent of the sample mean difference. Therefore, \(\lh\) is completely independent of the sample mean difference. We state this formally,

lemmagaussianindptlambda If \(F,Y\) are jointly gaussian, then the estimate \(\lh\) defined in 4 is independent of \((\bar{f}_N - \bar{f}_n)\)

Let us now consider the term \(T1\) in 21 \[\begin{align} T_1 & = \mathbb{E}[\lh] \mathbb{E}[(\bar{y}_n - \theta)(\bar{f}_N - \bar{f}_n)] \\ & = \mathbb{E}[\lh] \mathbb{E}[(\bar{y}_n - \theta)(\bar{f}_N - \mu_f + \mu_f - \bar{f}_n)] \\ & = \mathbb{E}[\lh] \left( \mathbb{E}[(\bar{y}_n - \theta)(\bar{f}_N - \mu_f)] - \mathbb{E}[(\bar{y}_n - \theta)(\bar{f}_n - \mu_f)] \right) \\ & = -\mathbb{E}[\lh] \mathbb{E}[(\bar{y}_n - \theta)(\bar{f}_n - \mu_f)] \\ & = -\mathbb{E}[\lh] \frac{1}{n} \sigma_{fy} \\ & = -\mathbb{E}\left[\frac{\hat{\sigma}_{fy}}{(1 + n/N)\hat{\sigma}_f^2}\right] \frac{1}{n} \sigma_{fy} \\ & = -\mathbb{E}[\hat{\sigma}_{fy}] \mathbb{E}\left[\frac{1}{(1 + n/N)\hat{\sigma}_f^2}\right] \frac{1}{n} \sigma_{fy} \\ & = -\frac{\sigma_{fy}^2}{n(1 + n/N)} \mathbb{E}\left[\frac{1}{\hat{\sigma}_f^2}\right] \\ & = -\frac{\sigma_{fy}^2}{n(1 + n/N)} \cdot \frac{N - 1}{(N - 3)\sigma_f^2} \\ & = -\frac{N - 1}{n(1 + n/N)(N - 3)} \rho_{fy}^2 \sigma_y^2 \end{align}\] where we use the fact that \(\frac{1}{\hat{\sigma}_f^2}\) follows an inverse chi-square distribution, which implies \(\mathbb{E}\left[\frac{1}{\hat{\sigma}_f^2}\right] = \frac{N-1}{(N-3)\sigma_f^2}\), and in the last line we use the fact that \(\sigma^2_{fy} / \sigma^2_f = \rho_{fy}^2 \sigma_{y}^2\)

Now we consider \(T_2\). \[\begin{align} T_2 & = \mathbb{E}[\lh^2]\mathbb{E}[(\bar{f}_N - \bar{f}_n)^2] \\ & = \mathbb{E}[\lh^2] \left( \mathbb{E}[(\bar{f}_N - \mu_f)^2] + \mathbb{E}[(\bar{f}_n - \mu_f)^2] \right) \\ & = \mathbb{E}[\lh^2]\left( \frac{1}{N} + \frac{1}{n} \right)\sigma_f^2 \\ & = \mathbb{E}[\hat{\sigma}_{fy}^2] \mathbb{E}\left[ \frac{1}{(1 + n/N)^2 \hat{\sigma}_f^4} \right] \left( \frac{1}{N} + \frac{1}{n} \right) \sigma_f^2 \\ & = \mathbb{E}[\hat{\sigma}_{fy}^2] \frac{1}{(1 + n/N)^2} \cdot \frac{(N - 1)^2}{(N - 3)(N - 5)\sigma_f^4} \left( \frac{1}{n} + \frac{1}{N} \right) \sigma_f^2 \\ & = \mathbb{E}[\hat{\sigma}_{fy}^2] \cdot \frac{(N - 1)^2}{n(1 + n/N)(N - 3)(N - 5)\sigma_f^2} \\ & = \left( \frac{n}{n - 1} \sigma_{fy}^2 + \frac{1}{n - 1} \sigma_f^2 \sigma_y^2 \right) \cdot \frac{(N - 1)^2}{n(1 + n/N)(N - 3)(N - 5)\sigma_f^2} \\ & = \sigma_y^2 \left( \frac{n}{n - 1} \rho_{fy}^2 + \frac{1}{n - 1} \right) \cdot \frac{(N - 1)^2}{n(1 + n/N)(N - 3)(N - 5)} \end{align}\]

here again we use the fact that \(\frac{1}{\hat{\sigma}_f^4}\) follows an inverse chi-square distribution, which implies \(\mathbb{E}\left[\frac{1}{\hat{\sigma}_f^4}\right] = \frac{(N - 1)^2}{(N - 3)(N - 5)\sigma_f^4}\).

Now we have the MSE \[\begin{align} \mathbb{E}[(\thetahp - \theta)^2] & = \MSE(\thetahc) - 2 \frac{N-1}{n(1+n/N)(N-3)}\rho_{fy}^2\sigma_y^2 + \sigma_y^2\left( \dfrac{n}{n-1} \rho_{fy}^2 + \dfrac{1}{n-1} \right)\frac{(N-1)^2}{n(1 + n/N)(N-3)(N-5)} \\ & = \MSE(\thetahc) + \sigma_y^2 \frac{N-1}{n(1+n/N)(N-3)}\left(-2\rho_{fy}^2 + \frac{n(N-1)}{(n-1)(N-5)}\rho_{fy}^2 + \frac{N-1}{(n-1)(N-5)}\right) \\ & = \MSE(\thetahc) + \sigma_y^2 \frac{(N-1)^2}{n(n-1)(1+n/N)(N-3)(N-5)}\left(\frac{-2(n-1)(N-5)}{N-1}\rho_{fy}^2 + n\rho_{fy}^2 + 1\right) \\ & = \MSE(\thetahc) + \sigma_y^2 \frac{(N-1)^2}{n(n-1)(1+n/N)(N-3)(N-5)}\left(\frac{-2(n-1)(N-5) + n(N-1)}{N-1}\rho_{fy}^2 + 1\right) \\ & = \MSE(\thetahc) + \sigma_y^2 \frac{(N-1)^2}{n(n-1)(1+n/N)(N-3)(N-5)}\left(\frac{-2(nN - N - 5n + 5) + nN - n}{N-1}\rho_{fy}^2 + 1\right) \\ & = \MSE(\thetahc) + \sigma_y^2 \frac{(N-1)^2}{n(n-1)(1+n/N)(N-3)(N-5)}\left(1 - \frac{nN - 2N - 9n + 10}{N-1}\rho_{fy}^2\right) \\ & = \MSE(\thetahc) + \sigma_y^2 \frac{(N-1)^2}{n(n-1)(1+n/N)(N-3)(N-5)}\left[1 - \left(n - 2 - 8\frac{n - 1}{N - 1}\right)\rho_{fy}^2\right] \end{align}\]

Looking at the last term, we have that the condition becomes \[\begin{align} \rho_{fy}^2 (n - 2 - 8 \frac{n-1}{N-1}) &\geq 1 \\ \lvert \rho_{fy} \rvert &\geq \frac{1}{\sqrt{n - 2 - 8 \frac{n-1}{N-1}}} \end{align}\] ◻

7.10 Variance Estimates of↩︎

We now consider the variance estimate of the 4 estimator in two methods. The first method is based on the idea of \(\lambda\) as a minimum variance linear combination, the spirit of PPI++. Here, we compute the theoretical variance of the PPI++ estimator assuming \(\lambda\) is independent of the data, and then use all the required plug-ins, including \(\lh\). In the second method, we naively compute the direct empirical plug-in in the estimator.

7.10.1 [prop:optimistic95variance95single95sample95ppi4343]: Data Independent Plug-in↩︎

Proof. We first calculate the variance of the estimator assuming \(\lambda\) is fixed. Then we evaluate the result upon substituting the data-dependent plug-in \(\lh\) and plug-ins for other quantities involved.

lemmaVariancePPI The variance of the PPI++ estimator defined in 3 is given by \[\label{eq:variance-ppi4343} \sigma^2_{\thetahp} = \frac{\sigma_y^2}{n} + \lambda^2 \left( \frac{\sigma^2_{f}}{N} + \frac{\sigma^2_f}{n} \right) - 2\lambda \frac{\sigma_{fy}}{n}.\tag{22}\]

\[\thetahp = \frac{1}{n} \sum_{i=1}^{n} y_i + \lambda \left( \frac{1}{N} \sum_{i=1}^{N} f_i - \frac{1}{n} \sum_{i=1}^{n} f_i \right).\]

To compute its variance: \[\begin{align} \text{Var}(\thetahp) & = \text{Var} \left( \frac{1}{n} \sum_{i=1}^{n} y_i \right) + \lambda^2 \text{Var} \left( \frac{1}{N} \sum_{i=1}^{N} f_i \right) + \lambda^2 \text{Var} \left( \frac{1}{n} \sum_{i=1}^{n} f_i \right) \\ & \quad - 2\lambda \text{Cov} \left( \frac{1}{n} \sum_{i=1}^{n} y_i, \frac{1}{n} \sum_{i=1}^{n} f_i \right). \end{align}\]

Using the standard variance formulas for sample means: \[\begin{align} \text{Var} \left( \frac{1}{n} \sum_{i=1}^{n} y_i \right) & = \frac{\sigma_y^2}{n}, \\ \text{Var} \left( \frac{1}{N} \sum_{i=1}^{N} f_i \right) & = \frac{\sigma_f^2}{N}, \\ \text{Var} \left( \frac{1}{n} \sum_{i=1}^{n} f_i \right) & = \frac{\sigma_f^2}{n}, \\ \text{Cov} \left( \frac{1}{n} \sum_{i=1}^{n} Y_i, \frac{1}{n} \sum_{i=1}^{n} f(X_i) \right) & = \frac{\sigma_{fy}}{n}. \end{align}\]

Therefore, putting it all together, we get the variance when \(\lambda\) is fixed,

\[\sigma^2_{\thetahp} = \frac{\sigma_y^2}{n} + \lambda^2 \left( \frac{\sigma_f^2}{N} + \frac{\sigma_f^2}{n} \right) - 2\lambda \frac{\sigma_{fy}}{n}.\]

The plug-in estimate of the variance of the Single-Sample estimator considers the variance of the PPI++ estimator as given in 22 with plug-ins for all quantities (including \(\lambda\) computed from data).

Therefore, substitute plug-ins and use \(\lh = \frac{\hat{\sigma}_{fy}}{(1 + \frac{n}{N})\hat{\sigma_f}^2}\), gives us,

\[\hat{\sigma^2}_{\thetahp} = \frac{\hat{\sigma_y^2}}{n} - \frac{\left( \frac{\hat{\sigma}_{fy}}{n} \right)^2}{\frac{\hat{\sigma^2_f}}{N} + \frac{\hat{\sigma^2_f}}{n}}.\]

\[\hat{\sigma^2}_{\thetahp} = \frac{\hat{\sigma^2_y}}{n} - \frac{\left( \frac{\hat{\sigma}_{fy}}{n} \right)^2}{(\frac{1}{N} + \frac{1}{n}) \hat{\sigma^2_f}}.\]

\[\hat{\sigma^2}_{\thetahp} = \frac{\hat{\sigma_y^2}}{n} - \frac{\left( \frac{\hat{\sigma}_{fy}}{n} \right)^2}{(\frac{1}{N} + \frac{1}{n}) \hat{\sigma^2_f}}.\]

Further, if we let \(N \to \infty\), \(\hat{\sigma^2_f} \to \sigma_f^2\) and \(1/N \to 0\), giving us,

\[\hat{\sigma^2}_{\thetahp} = \frac{\hat{\sigma_y^2}}{n} - \frac{\hat{\sigma}_{fy}^2}{n\sigma_f^2}\]

Since, \(\frac{\hat{\sigma^2_y}}{n}\) is the estimate of the variance of the classical estimator, this shows that the estimate of the variance of the is always lower than its classical counterpart. ◻

7.10.2 Naive Plug-in↩︎

We now consider an alternate method for computing the empirical variance. Starting with the definition, assuming again that \(\lambda\) is fixed, we rewrite it as a sum of two independent terms.

\[\thetahp = (\bar{y} - \lambda\bar{f}_n) + \lambda\bar{f}_N\]

Now, consider the following standard method for estimating variance, by taking the empirical variance of \(\bar{y} - \lambda \bar{f}_n\) and the empirical variance of \(\lambda \bar{f}_N\), and adding up the empirical variance estimates, adjusted for sample size. The estimate of the variance of the former is given by \[\begin{align} \hat{\sigma}_{y - \lambda f}^2 & = \frac{1}{n-1} \sum_{i} ((y_i - \lambda f_i) - (\bar{y} - \lambda \bar{f}))^2 \\ & = \frac{1}{n-1} \sum_{i} ((y_i - \bar{y}) - (\lambda f_i - \lambda \bar{f}))^2 \\ & = \frac{1}{n-1} \sum_{i} (y_i - \bar{y})^2 - 2 \lambda (y_i - \bar{y})(f_i - \bar{f}) + \lambda^2 (f_i - \bar{f})^2 \\ & = \hat{\sigma}_y^2 + \lambda^2 \hat{\sigma}_f^2[n] - 2 \lambda \hat{\sigma}_{yf} \end{align}\] where \(\hat{\sigma}_y^2, \hat{\sigma}_{f}^2[n], \hat{\sigma}_{yf}\) are the empirical variances of \(Y\) and \(F\) and the empirical covariance respectively, where we note that \(\hat{\sigma}_f^2[n]\) is the empirical variance on \(n\) samples. Using the fact that \(\lambda \mathrel{\vcenter{:}}= \hat{\sigma}_{yf} / ((1 + n/N) \hat{\sigma}_f^2[N])\), we have it that \[\begin{align} \hat{\sigma}_{y - \lambda f}^2 & = \hat{\sigma}_y^2 + \lambda^2 \hat{\sigma}_f^2[n] - 2 \lambda \hat{\sigma}_{yf} \\ & = \hat{\sigma}_y^2 + \left(\frac{\hat{\sigma}_{yf}^2}{(1 + \frac{n}{N})^2 \hat{\sigma}_f^4[N]}\right) \hat{\sigma}_f^2[n] - 2 \left(\frac{\hat{\sigma}_{yf}}{(1 + \frac{n}{N}) \hat{\sigma}_f^2[N]}\right) \hat{\sigma}_{yf} \\ & = \hat{\sigma}_y^2 + \left(\frac{\hat{\sigma}_{yf}^2}{(1 + \frac{n}{N})^2 \hat{\sigma}_f^4[N]}\right)\hat{\sigma}_f^2[n] - 2 \left(\frac{\hat{\sigma}^2_{yf}}{(1 + \frac{n}{N}) \hat{\sigma}_f^2[N]}\right) \\ & = \hat{\sigma}_y^2 + \left(\frac{\hat{\sigma}_{yf}^2}{(1 + \frac{n}{N})^2 \hat{\sigma}_f^4[N]}\right)\hat{\sigma}_f^2[n] - 2 \hat{\sigma}_f^2[N] \left(1 - \frac{n}{N}\right) \left(\frac{\hat{\sigma}^2_{yf}}{(1 + \frac{n}{N})^2 \hat{\sigma}_f^4[N]}\right) \\ & = \hat{\sigma}_y^2 - \left(2 \hat{\sigma}_f^2[N] \left(1 - \frac{n}{N}\right) - \hat{\sigma}_f^2[n]\right)\left(\frac{\hat{\sigma}_{yf}^2}{(1 + \frac{n}{N})^2 \hat{\sigma}_f^4[N]}\right) \\ & = \hat{\sigma}_y^2 - \left(2 \hat{\sigma}_f^2[N] \left(1 - \frac{n}{N}\right) - \hat{\sigma}_f^2[n]\right) \hat{\lambda}^2 \end{align}\] Then the empirical variance of \(\hat{\sigma}_{\lambda f}^2\), computed on \(N\) samples, is given by \(\lambda^2 \hat{\sigma}_{f}^2[N]\), so that the total variance (dividing by \(n\) and \(N\) respectively, and then taking the sum), is given by \[\begin{align} \frac{\hat{\sigma}_{y - \lambda f}^2}{n} + \frac{\hat{\sigma}_{\lambda f}^2}{N} & = \frac{\hat{\sigma}_y^2}{n} - \frac{\left(2 \hat{\sigma}_f^2[N] \left(1 - \frac{n}{N}\right)\right) \hat{\lambda}^2}{n} + \frac{\hat{\lambda}^2\hat{\sigma}_f^2[n]}{n} + \frac{\hat{\lambda}^2 \hat{\sigma}_f^2[N]}{N} \end{align}\]

If \(N \to \infty\), then \(\frac{\hat{\sigma}_{\lambda f}^2}{N} \to 0\), \(1 - \frac{n}{N} \to 1\), \(\frac{\hat{\lambda}^2 \hat{\sigma}_f^2[N]}{N} \to 0\), and \(\sigma_f^2[N] \to \sigma_f^2\), and hence the empirical variance estimate becomes, \[\frac{\hat{\sigma}_y^2}{n} - \frac{2\hat{\lambda}^2\sigma_f^2}{n} + \frac{\hat{\lambda}^2\sigma_f^2[n]}{n}\] This is always less than the empirical estimate of the classical variance as long as the empirical variance of \(F\) on \(n\) samples is no greater than double the true variance.

8 Experiment Setup Details↩︎

In this section, we provide details of the experiment setup on Alphafold [29]. For both the \(\MSE\) as well as the coverage experiments, our results are bootstrapped over 50,000 draws. To closely match the theoretical intuition, we aim to study the setting where \(n\) is small and \(N\) is large. Therefore, after randomly sampling \(n\) (or \(2n\), for sample-splitting), we treat all the remaining data as unlabeled. Since the true labels are binary, we also use a binary noise model for the configurations that do not use the original predictor. This noise model is implemented by specifying \(\mathbb{P}(F = 1| Y = 1)\) and \(\mathbb{P}(F = 1 | Y = 0)\). Alternatively, this can be viewed as specifying the True Positive Rate and the False Positive Rate of the pseudo-labels. From a noisy label lens, this can be viewed as specifying the complete noise transition matrix. For each configuration, for each \(n\), we reset random states, ensuring full reproducibility. For all plug-in statistical quantities computed, we use ddof = 1 for unbiasedness.

9 Additional Experiments: Galaxies Dataset↩︎

In this section, we extend our experiments to the galaxies dataset used in [4]. Similar insights hold as in 5. Our experiment apparatus is the same as described in 8. We run \(50,000\) draws using small-regime of \(n\) or \(2n\) and all the remaining data as unlabeled. We use a similar binary noise model. We show our results in 4 5.

a
b

Figure 4: Relative \(\MSE\) vs. Sample Size on the Galaxies Dataset for PPI++ estimators with black-box models \(f\) of varying quality. Y-axis gives the difference \(\MSE(\hat{\theta}_{\text{PPI++}}) - \MSE(\hat{\theta}_{\text{Classical}})\), such that lower (negative) values imply improvement over the classical estimator. Each line represents PPI++ with a different black-box model \(f\). The blue line uses the original model, which has strong predictive performance, but only improves estimation error at \(n \geq 12\) (left) and \(n \geq 6\) (right).. a — Cross-fit PPI++ Estimator, b — Single-Sample PPI++ Estimator

a

b

Figure 5: Comparison of Coverage and Interval Width for  and , using different models \(f\) on the Galaxies dataset. () Original pseudo-label model \(f\), where Single-Sample PPI++ has lower coverage than either Cross-fit PPI++ or the classical estimator. () A modified model with lower correlation, where similar trends hold..

10 Additional Experiments: Coverage and Interval Width, Alphafold Noised↩︎

In this section, we present the coverage and interval width results on Alphafold, similar to 3. We note that in this lowered correlation setting, while and widen their interval width to improve coverage, matches the coverage of but still falls short. The experiment details are as presented in 5 and 8.

Figure 6: AlphaFold. Coverage (left) and interval width (right) for  vs.  using noised pseudo-labels from Alphafold (\mathrm{corr}=0.48). under-covers relative to and .

References↩︎

[1]
A. Radford et al., “Learning transferable visual models from natural language supervision,” arXiv preprint (2103.00020v1), Feb. 2021, [Online]. Available: http://arxiv.org/abs/2103.00020v1.
[2]
H. A. Rahmani et al., “Judging the judges: A collection of llm-generated relevance judgements,” arXiv preprint arXiv:2502.13908, 2025.
[3]
A. N. Angelopoulos, S. Bates, C. Fannjiang, M. I. Jordan, and T. Zrnic, “Prediction-powered inference,” Science, vol. 382, no. 6671, pp. 669–674, 2023.
[4]
A. N. Angelopoulos, J. C. Duchi, and T. Zrnic, “Ppi++: Efficient prediction-powered inference,” arXiv preprint arXiv:2311.01453, 2023.
[5]
S. Li and N. Ignatiadis, “Prediction-powered adaptive shrinkage estimation,” arXiv preprint arXiv:2502.14166, 2025.
[6]
T. Zrnic and E. J. Candès, “Cross-prediction-powered inference,” Proceedings of the National Academy of Sciences, vol. 121, no. 15, p. e2322083121, 2024.
[7]
T. Zrnic and E. J. Candès, “Active statistical inference,” arXiv preprint (2403.03208v2), Mar. 2024, [Online]. Available: http://arxiv.org/abs/2403.03208v2.
[8]
P.-E. Poulet, M. Tran, S. T. du Montcel, B. Dubois, S. Durrleman, and B. Jedynak, “Prediction-powered inference for clinical trials,” medRxiv, 2025.
[9]
I. Demirel, A. Alaa, A. Philippakis, and D. Sontag, “Prediction-powered generalization of causal inferences,” arXiv preprint arXiv:2406.02873, 2024.
[10]
D. Broska, M. Howes, and A. van Loon, “The mixed subjects design: Treating large language models as potentially informative observations,” Sociological Methods & Research, p. 00491241251326865, 2024.
[11]
P. Boyeau, A. N. Angelopoulos, N. Yosef, J. Malik, and M. I. Jordan, “Autoeval done right: Using synthetic data for model evaluation,” arXiv preprint arXiv:2403.07008, 2024.
[12]
K. Gligoric, T. Zrnic, C. Lee, E. Candes, and D. Jurafsky, “Can unconfident LLM annotations be used for confident conclusions?” in Proceedings of the 2025 conference of the nations of the americas chapter of the association for computational linguistics: Human language technologies (volume 1: Long papers), 2025, pp. 3514–3533, [Online]. Available: https://aclanthology.org/2025.naacl-long.179/.
[13]
H. Bang and J. M. Robins, “Doubly robust estimation in missing data and causal inference models,” Biometrics, vol. 61, pp. 962–973, Dec. 2005, doi: 10.1111/j.1541-0420.2005.00377.x.
[14]
J. M. Bates and C. W. J. Granger, “The combination of forecasts,” Operations Research, vol. 20, p. 451, Dec. 1969, doi: 10.2307/3008764.
[15]
F. Lavancier and P. Rochet, “A general procedure to combine estimators,” Computational statistics \& data analysis, vol. 94, pp. 175–192, 2016.
[16]
S. Yang, C. Gao, D. Zeng, and X. Wang, “Elastic integrative analysis of randomised trial and real-world data for treatment heterogeneity estimation,” Journal of the Royal Statistical Society Series B: Statistical Methodology, vol. 85, pp. 575–596, Jul. 2023, doi: 10.1093/jrsssb/qkad017.
[17]
S. Chen, B. Zhang, and T. Ye, “Minimax rates and adaptivity in combining experimental and observational data,” arXiv preprint (2109.10522), 2021.
[18]
M. Oberst, A. D'Amour, M. Chen, Y. Wang, D. Sontag, and S. Yadlowsky, “Understanding the risks and rewards of combining unbiased and possibly biased estimators, with applications to causal inference,” arXiv preprint (2205.10467v2), May 2022, [Online]. Available: http://arxiv.org/abs/2205.10467v2.
[19]
S. S. Lavenberg and P. D. Welch, “A perspective on the use of control variables to increase the efficiency of monte carlo simulations,” Management Science, vol. 27, pp. 322–335, Mar. 1981, doi: 10.1287/mnsc.27.3.322.
[20]
B. L. Nelson, “Control variate remedies,” Operations Research, vol. 38, pp. 974–992, Dec. 1990, doi: 10.1287/opre.38.6.974.
[21]
R. Mozer, “PPI is the difference estimator: Recognizing the survey sampling roots of prediction-powered inference,” arXiv preprint (2603.19160v1), Mar. 2026, [Online]. Available: http://arxiv.org/abs/2603.19160v1.
[22]
C. M. Cassel, C. E. Särndal, and J. H. Wretman, “Some results on generalized difference estimation and generalized regression estimation for finite populations,” Biometrika, vol. 63, pp. 615–620, 1976, doi: 10.1093/biomet/63.3.615.
[23]
C.-E. Särndal, B. Swensson, and J. H. Wretman, “The weighted residual technique for estimating the variance of the general regression estimator of the finite population total,” Biometrika, vol. 76, pp. 527–537, 1989, doi: 10.1093/biomet/76.3.527.
[24]
F. J. Breidt and J. D. Opsomer, “Model-assisted survey estimation with modern prediction techniques,” Statistical Science, vol. 32, May 2017, doi: 10.1214/16-sts589.
[25]
Y. Tille, Sampling and estimation from finite populations. Wiley & Sons, Incorporated, John, 2020.
[26]
F. E. Dorner, V. Y. Nastl, and M. Hardt, “Limits to scalable evaluation at the frontier: LLM as judge won’t beat twice the data,” arXiv preprint (2410.13341v2), Oct. 2024, [Online]. Available: http://arxiv.org/abs/2410.13341v2.
[27]
A. T. Chaganty, S. Mussman, and P. Liang, “The price of debiasing automatic metrics in natural language evaluation,” arXiv preprint (1807.02202v1), Jul. 2018, [Online]. Available: http://arxiv.org/abs/1807.02202v1.
[28]
B. Eyre and D. Madras, “Regression for the mean: Auto-evaluation and inference with few labels through post-hoc regression,” in Proceedings of the 42nd international conference on machine learning, 2025, vol. 267, pp. 15596–15612, [Online]. Available: https://proceedings.mlr.press/v267/eyre25a.html.
[29]
J. Jumper et al., “Highly accurate protein structure prediction with AlphaFold,” nature, vol. 596, no. 7873, pp. 583–589, 2021.
[30]
K. Zhou, J. M. Giorgi, P. Mani, P. Xu, D. Liang, and C. Tan, “From feedback to checklists: Grounded evaluation of AI-generated clinical notes,” in Proceedings of the 2025 conference on empirical methods in natural language processing: Industry track, 2025, pp. 1485–1499.
[31]
D. Li et al., “Preference leakage: A contamination problem in llm-as-a-judge, 2025,” URL https://arxiv. org/abs/2502.01534.
[32]
L. Zheng et al., “Judging llm-as-a-judge with mt-bench and chatbot arena,” Advances in neural information processing systems, vol. 36, pp. 46595–46623, 2023.
[33]
L. Hewitt, A. Ashokkumar, I. Ghezae, and R. Willer, “Predicting results of social science experiments using large language models,” Preprint, 2024.
[34]
B. S. Manning, K. Zhu, and J. J. Horton, “Automated social science: Language models as scientist and subjects,” National Bureau of Economic Research, 2024.
[35]
J. R. Anthis et al., “Llm social simulations are a promising research method,” arXiv preprint arXiv:2504.02234, 2025.
[36]
D. M. Goldenholz, S. R. Goldenholz, S. Habib, and M. B. Westover, “Inductive reasoning with large language models: A simulated randomized controlled trial for epilepsy,” Epilepsy Research, vol. 211, p. 107532, 2025.
[37]
H. Lai et al., “Assessing the risk of bias in randomized clinical trials with large language models,” JAMA Network Open, vol. 7, no. 5, pp. e2412687–e2412687, 2024.
[38]
H. Waldetoft, J. Torgander, and M. Magnusson, “Prediction-powered estimators for finite population statistics in highly imbalanced textual data: Public hate crime estimation,” arXiv preprint arXiv:2505.04643, 2025.
[39]
F. Ludwig, K. Dolos, A. Alves-Pinto, and T. Zesch, “Unraveling the dynamics of semi-supervised hate speech detection: The impact of unlabeled data characteristics and pseudo-labeling strategies,” in Findings of the association for computational linguistics: EACL 2024, 2024, pp. 1974–1986.
[40]
K. Tyser et al., “Ai-driven review systems: Evaluating llms in scalable and bias-aware academic reviews,” arXiv preprint arXiv:2408.10365, 2024.
[41]
A. Goldberg et al., “Usefulness of LLMs as an author checklist assistant for scientific papers: NeurIPS’24 experiment,” arXiv preprint arXiv:2411.03417, 2024.

  1. As discussed in [theorem:jointly-normal-mse-condition-all], this threshold is for an implementation without sample splitting. In the Gaussian setting with sample-splitting and cross-fitting, the threshold is slightly higher, at \((n/2 - 2)^{-1/2}\).↩︎

  2. We use \(\theta\) here to align with notation in the PPI literature, but note that \(\theta\) and \(\mu_y\) are the same quantity.↩︎

  3. The setting of Gaussian labels is unique: The empirical covariance of normally distributed random variables is independent of their sample means. This property reduces the dependence between the estimate of \(\lambda\) (which depends on the estimated covariance) and the estimated value itself. As a result,  is also unbiased in this setting. However, in the non-Gaussian case,  is biased, while the use of sample-splitting retains the unbiasedness of , as we will see in 4.3↩︎

  4. Notably, the condition for  uses \(n/2\), the size of a single fold, while the condition for  uses \(n\), which illustrates a cost of sample splitting in the Gaussian setting.↩︎

  5. It is generally known that  is biased in finite samples [27], [28], but we do not know of other results giving the exact form, so we produce it here. While [27] show that the bias of a similar estimator is \(O(1/n)\), they do not give the exact form.↩︎