January 26, 2026
Let \(\{(X_i,Y_i)\}_{i=1}^\infty\) be an independent and identically distributed (iid) sequence of pairs of random variables. Let \(F:\mathbb{R}\to[0,1]\) and \(G:\mathbb{R}\to[0,1]\) be the marginal cumulative distribution functions (cdfs) for \(X_i\) and \(Y_i\) respectively. Let \(Q:(0,1)\to\mathbb{R}\) be the quantile function (qf) for \(Y_i\); that is, \[Q(u)=\inf\{y\in\mathbb{R}:G(y)\geq u\}.\] For each \(n\in\mathbb{N}\), define the empirical cdfs \(F_n:\mathbb{R}\to[0,1]\) and \(G_n:\mathbb{R}\to[0,1]\) by \[F_n(x)=\frac{1}{n}\sum_{i=1}^n\mathbb{1}(X_i\leq x),\quad G_n(y)=\frac{1}{n}\sum_{i=1}^n\mathbb{1}(Y_i\leq y),\] and define the empirical qf \(Q_n:(0,1)\to\mathbb{R}\) by \[Q_n(u)=\inf\{y\in\mathbb{R}:G_n(y)\geq u\}.\] Define \(R:[0,1]\to[0,1]\) and \(R_n:[0,1]\to[0,1]\) by \[R(u) = F(Q(u)) \,\,\, \text{and} \,\,\, R_n(u) = F_n(Q_n(u))\,\,\, \text{for } u \in (0,1),\] \[\begin{alignedat}{3} R(0) & = \lim_{u\downarrow 0} R(u), \qquad R_n(0) & = \lim_{u\downarrow 0} R_n(u),\\ R(1) & = \lim_{u\uparrow 1} R(u), \qquad R_n(1) & = \lim_{u\uparrow 1} R_n(u). \end{alignedat}\] We refer to \(R\), \(R_n\) and \(\sqrt{n}(R_n-R)\) as the P-P curve, P-P plot and P-P process, respectively. The P-P curve is also known as the ordinal dominance curve or the receiver operating characteristic curve.
Our main result is the following theorem.
Theorem 1. \(\sqrt{n}(R_n-R)\) converges in distribution in \(L^1[0,1]\) if and only if \(R\) is absolutely continuous.
Here \(L^1[0,1]\) denotes the usual space of Lebesgue measurable and integrable real-valued functions on \([0,1]\), equipped with the norm \(\lVert\cdot\rVert_1\). Convergence in distribution (\(\rightsquigarrow\)) in a metric space is understood in the standard sense of [1]. As usual, functions in \(L^1[0,1]\) that are equal almost everywhere (a.e.) are identified.
When \(R\) is absolutely continuous, the limit in distribution of the P-P process in \(L^1[0,1]\) is a centered Gaussian process \(\mathcal{R}:[0,1]\to\mathbb{R}\) whose covariance structure depends on \(R\) and on the dependence between the paired observations. Let \(C:[0,1]^2\to[0,1]\) be a copula for \((X_i,Y_i)\); that is, a bivariate cdf on \([0,1]^2\) with uniform margins satisfying \[C(F(x),G(y))=\mathbb{P}(X_i\leq x,Y_i\leq y).\] The copula \(C\) is uniquely determined on \(\overline{\mathop{\mathrm{ran}}(F)}\times\overline{\mathop{\mathrm{ran}}(G)}\), by Sklar’s theorem ([2]; see also [3]). Let \(\mathcal{B}:[0,1]^2\to\mathbb{R}\) be a centered Gaussian process with covariance kernel \[\mathrm{Cov}(\mathcal{B}(u,v),\mathcal{B}(u',v'))=C(u\wedge u',v\wedge v')-C(u,v)C(u',v').\] The process \(\mathcal{B}\) is called a tied-down Brownian sheet with intensity measure \(C\) [4]. The marginal processes \(\mathcal{B}_1:[0,1]\to\mathbb{R}\) and \(\mathcal{B}_2:[0,1]\to\mathbb{R}\) defined by \(\mathcal{B}_1(u)=\mathcal{B}(u,1)\) and \(\mathcal{B}_2(u)=\mathcal{B}(1,u)\) are Brownian bridges. If \(R\) is absolutely continuous then it admits an a.e.derivative \(r\in L^1[0,1]\), and the distributional limit \(\mathcal{R}\) of the P-P process in \(L^1[0,1]\) is given by \[\mathcal{R}(u)=\mathcal{B}_1(R(u))-r(u)\mathcal{B}_2(u).\] The values taken by \(r\) at points where \(R\) is not differentiable are immaterial; for definiteness one may set \(r(u)=1\) on that null set.
Absolute continuity of \(R\) does not imply continuity of \(F\) and \(G\), so the copula \(C\) and the covariance kernel of \(\mathcal{B}\) need not be uniquely determined on all of \([0,1]^2\). However, \(\mathcal{R}\) depends on \(\mathcal{B}\) only through its values on \(\overline{\mathop{\mathrm{ran}}(F)}\times\overline{\mathop{\mathrm{ran}}(G)}\). Indeed, \(\mathop{\mathrm{ran}}(R)\subseteq\overline{\mathop{\mathrm{ran}}(F)}\), and \(r(u)=0\) whenever \(u\notin\overline{\mathop{\mathrm{ran}}(G)}\). Hence the distribution of \(\mathcal{R}\) is uniquely determined by the construction above, even when \(F\) or \(G\) is discontinuous.
Let \(\ell^\infty[0,1]\) be the space of bounded real-valued functions on \([0,1]\) equipped with the uniform norm. Prior work has identified sufficient conditions for the P-P process to converge in distribution in \(\ell^\infty[0,1]\). Absolute continuity of \(R\) is not sufficient for such convergence. Indeed, if \(r\) is not locally bounded on \((0,1)\) then almost every sample path of \(\mathcal{R}\) is unbounded. To circumvent this issue, prior research relying on convergence in distribution of the P-P process in \(\ell^\infty[0,1]\) has sometimes required \(R\) to have a bounded derivative on \([0,1]\). See, for instance, [5], [6], [7] and [8]. This requirement is convenient, but it excludes some basic examples. For instance, if \(F\) and \(G\) are normal with different means and equal variances, or with \(G\) having a smaller variance than \(F\), then \(R\) is continuously differentiable on \((0,1)\) but the derivative \(r\) diverges to infinity at one or both endpoints. [9] accommodate this feature of the normal case by assuming only that \(r\) is locally bounded on \((0,1)\), not necessarily bounded, and asserting convergence in distribution of the P-P process only in the truncated space \(\ell^\infty[a,b]\) where \(0<a<b<1\).
To obtain convergence in distribution of the P-P process in \(\ell^\infty[0,1]\) when \(r\) is locally bounded on \((0,1)\) but unbounded near an endpoint, additional control of the endpoint behavior is needed. [10] impose such control through a Čibisov–O’Reilly condition, derived from earlier work of [11] and [12] on weighted approximation of empirical processes. Their subsequent discussion indicates that this type of condition is close to minimal.
Theorem 1 above contrasts with Theorem 3.1 in [10] because it includes no counterpart to the Čibisov–O’Reilly condition. The reason is that Theorem 1 is concerned with convergence in distribution in \(L^1[0,1]\), and no assumption on \(r\) is needed to ensure that \(\mathcal{R}\) has a.s.integrable sample paths. The density \(r\) is always integrable because \(R\) is nondecreasing and bounded; therefore, since \(\mathcal{B}_1\) and \(\mathcal{B}_2\) have a.s.bounded sample paths, it follows that \(\mathcal{R}\) has a.s.integrable sample paths. While convergence in distribution of the P-P process in \(L^1[0,1]\) is weaker than convergence in distribution in \(\ell^\infty[0,1]\), it may be enough to establish good behavior of statistics representable as integral-type functionals of the P-P plot. See [13] for an application of Theorem 1 involving a Wilcoxon-Mann-Whitney statistic.
Theorem 1 complements the main result of [14], which gives a necessary and sufficient condition for convergence in distribution of the quantile process \(\sqrt{n}(Q_n-Q)\) in \(L^1(0,1)\). The condition is that \(Q\) is locally absolutely continuous and satisfies \[\int_0^1\sqrt{u(1-u)}\,\mathrm{d}Q(u)<\infty.\] In [14], sufficiency is established by the delta-method. The proof given here for the P-P process also uses the delta-method, but it does not rely on the results of [14]. The argument involves two applications of the delta-method: one for the generalized inverse map from cdfs to qfs, and one for the composition of a cdf with a qf. For the generalized inverse map we use the standard Hadamard differentiability result stated in [1], which goes back to [15] and [16]. Composition requires more work. Standard results on Hadamard differentiability of composition, such as the one stated in [17], assume a uniform differentiability condition stronger than absolute continuity. We therefore prove a tailored lemma in which composition is treated as a map into \(L^1[0,1]\). This is Lemma 1.
Sections 2 and 3 establish, respectively, the sufficiency and necessity of absolute continuity for convergence in distribution. Section 4 briefly indicates the modifications needed to handle P-P plots computed from two independent samples of different sizes rather than from a single sample of pairs.
In this section we prove that absolute continuity of \(R\) is sufficient for convergence in distribution of the P-P process in \(L^1[0,1]\). We also establish a corresponding bootstrap approximation. Both conclusions follow from the same underlying argument based on the delta-method.
Let \(W_n=(W_{1,n},\ldots,W_{n,n})\) be a multinomial random vector with equal probabilities over the categories \(1,\ldots,n\), independent of \(\{(X_i,Y_i)\}_{i=1}^n\). Define the bootstrap empirical cdfs \(F^\ast_n:\mathbb{R}\to[0,1]\) and \(G_n^\ast:\mathbb{R}\to[0,1]\) by \[F_n^\ast(x)=\frac{1}{n}\sum_{i=1}^nW_{i,n}\mathbb{1}(X_i\leq x),\,\,\, G_n^\ast(y)=\frac{1}{n}\sum_{i=1}^nW_{i,n}\mathbb{1}(Y_i\leq y).\] Let \(Q_n^\ast:(0,1)\to\mathbb{R}\) be the qf corresponding to \(G_n^\ast\), and define \(R_n^\ast:[0,1]\to[0,1]\) by \[R_n^\ast(u) = F_n^\ast(Q_n^\ast(u)) \,\,\, \text{for } u \in (0,1),\] \[R_n^\ast(0) = \lim_{u\downarrow 0} R_n^\ast(u), \qquad R_n^\ast(1) = \lim_{u\uparrow 1} R_n^\ast(u).\]
Theorem 2 (Sufficiency). If \(R\) is absolutely continuous then \(\sqrt{n}(R_n-R)\rightsquigarrow\mathcal{R}\) in \(L^1[0,1]\), and \(\sqrt{n}(R_n^\ast-R_n)\rightsquigarrow\mathcal{R}\) in \(L^1[0,1]\) conditionally on \(\{(X_i,Y_i)\}_{i=1}^\infty\) in probability.
We prove this theorem by combining Donsker’s theorem with the delta-method and its bootstrap analogue; see Theorems 20.8 and 23.9 in [1]. The main technical point is to verify Hadamard differentiability of a composition map under absolute continuity alone. Standard composition results, such as Lemma 3.10.28 in [17], are not directly applicable here because they require a stronger uniform differentiability condition. We therefore begin with a tailored lemma for composition as a map into \(L^1[0,1]\).
For two normed spaces \(\mathcal{X}\) and \(\mathcal{Y}\) we denote by \(\mathcal{X}\otimes\mathcal{Y}\) the product space equipped with the max-norm. Define subsets \(\mathbb{D}_1\) and \(\mathbb{D}_2\) of \(\ell^\infty[0,1]\) by \[\begin{align} \mathbb{D}_1&=\{A\in\ell^\infty[0,1]:\mathop{\mathrm{ran}}(A)\subseteq[0,1]\\ &\quad\quad\text{ and }A\text{ is Lebesgue measurable}\}\\ \text{and}\quad\mathbb{D}_2&=\{B\in\ell^\infty[0,1]:B\text{ is Borel measurable}\}. \end{align}\] Define the composition map \(\phi:\mathbb{D}_1\times\mathbb{D}_2\to L^1[0,1]\) by \[\phi(A,B)(u)=B\circ A(u)=B(A(u)),\quad u\in[0,1].\] The Lebesgue measurability of \(A\) and Borel measurability of \(B\) together guarantee that \(B\circ A\) is Lebesgue measurable, and thus the boundedness of \(B\) guarantees that \(B\circ A\in L^1[0,1]\).
Let \(C[0,1]\) be the space of continuous real-valued functions on \([0,1]\) equipped with the uniform norm, a subspace of \(\ell^\infty[0,1]\). Let \(I:[0,1]\to[0,1]\) be the identity map.
Lemma 1 (Hadamard differentiability of composition). Let \(B:[0,1]\to\mathbb{R}\) be absolutely continuous with density \(b:[0,1]\to\mathbb{R}\). Then the composition map \(\phi:\mathbb{D}_1\times\mathbb{D}_2\subset\ell^\infty[0,1]\otimes\ell^\infty[0,1]\to L^1[0,1]\) is Hadamard differentiable at \((I,B)\) tangentially to \(C[0,1]\times C[0,1]\). The derivative \(\phi'_{I,B}:C[0,1]\times C[0,1]\to L^1[0,1]\) is given by \(\phi'_{I,B}(\alpha,\beta)=\beta+b\alpha\).
Proof. Let \(\{t_n\}_{n=1}^\infty\) be a sequence of positive real numbers such that \(t_n\to0\). Let \(\{\alpha_n\}_{n=1}^\infty\) be a sequence in \(\ell^\infty[0,1]\) such that \(I+t_n\alpha_n\in\mathbb{D}_1\) for all \(n\) and such that \(\alpha_n\to\alpha\in C[0,1]\). Let \(\{\beta_n\}_{n=1}^\infty\) be a sequence in \(\ell^\infty[0,1]\) such that \(B+t_n\beta_n\in\mathbb{D}_2\) for all \(n\) and such that \(\beta_n\to\beta\in C[0,1]\). It suffices to show that \(\lVert t_n^{-1}[(B+t_n\beta_n)\circ(I+t_n\alpha_n)-B]-\beta-b\alpha\rVert_1\to0\). By the triangle inequality, this norm is bounded by the following sum of three terms: \[\begin{gather} \lVert(\beta_n-\beta)\circ(I+t_n\alpha_n)\rVert_1+\lVert\beta\circ(I+t_n\alpha_n)-\beta\rVert_1\\+\lVert t_n^{-1}[B\circ(I+t_n\alpha_n)-B]-b\alpha\rVert_1. \end{gather}\] The first term converges to zero because \(\beta_n\) converges uniformly to \(\beta\). The second term converges to zero by the dominated convergence theorem because \(\beta\) is continuous. It remains to show that the third term converges to zero.
Since \(B\) is absolutely continuous with density \(b\) we have, for each \(u\in[0,1]\), \[\begin{gather} t_n^{-1}[B(u+t_n\alpha_n(u))-B(u)]\\=t_n^{-1}\int_0^1[\mathbb{1}(v\leq u+t_n\alpha_n(u))-\mathbb{1}(v\leq u)]b(v)\,\mathrm{d}v. \end{gather}\] Define the map \(\zeta_n\in\mathbb{D}_2\) by \[\zeta_n(u)=t_n^{-1}\int_0^1[\mathbb{1}(v\leq u+t_n\alpha(u))-\mathbb{1}(v\leq u)]b(v)\,\mathrm{d}v.\] Then by Fubini’s theorem we have \[\begin{gather} \lVert t_n^{-1}[B\circ(I+t_n\alpha_n)-B]-\zeta_n\rVert_1\\\leq t_n^{-1}\!\!\int_0^1\!\!\!\int_0^1\! \lvert\mathbb{1}(v\leq u+t_n\alpha_n(u))-\mathbb{1}(v\leq u+t_n\alpha(u))\rvert\lvert b(v)\rvert\mathrm{d}u\mathrm{d}v. \end{gather}\] Using the change-of-variables \(u\mapsto v-t_nw\) we may rewrite the right-hand side of the inequality as \[\begin{gather} \int_0^1\!\int_{-\infty}^\infty\mathbb{1}(t_n^{-1}(v-1)\leq w\leq t_n^{-1}v)\,\bigl|\mathbb{1}(w\leq\alpha_n(v-t_nw))-\\\mathbb{1}(w\leq\alpha(v-t_nw))\bigr|\,\lvert b(v)\rvert\,\mathrm{d}w\,\mathrm{d}v. \end{gather}\] The integrand, as a function of \(v\) and \(w\), is bounded by \(\mathbb{1}(\lvert w\rvert\leq\lVert\alpha\rVert_\infty+1)\lvert b(v)\rvert\) for all sufficiently large \(n\), and converges a.e.to zero because \(\alpha_n\) converges uniformly to \(\alpha\) and because \(\alpha\) is continuous. (Convergence of the integrand to zero need not hold on the set \(\{(v,w):w=\alpha(v)\}\), but this is a null set.) Thus the double integral over \(w\) and \(v\) converges to zero by the dominated convergence theorem. This shows that \[\lVert t_n^{-1}[B\circ(I+t_n\alpha_n)-B]-\zeta_n\rVert_1\to0.\] It therefore remains to show that \(\lVert\zeta_n-b\alpha\rVert_1\to0\).
Define the map \(\chi_n:[0,1]^2\to\mathbb{R}\) by \[\chi_n(u,v)=t_n^{-1}[\mathbb{1}(v\leq u+t_n\alpha(u))-\mathbb{1}(v\leq u)],\] so that \[\zeta_n(u)=\int_0^1\chi_n(u,v)b(v)\,\mathrm{d}v\] for each \(u\in[0,1]\). Then \[\begin{align} \lVert\zeta_n-b\alpha\rVert_1&\leq \int_0^1\left\vert b(u)\int_0^1\chi_n(u,v)\,\mathrm{d}v-b(u)\alpha(u)\right\vert\mathrm{d}u\notag\\ &\quad\quad+\int_0^1\left\vert\int_0^1\chi_n(u,v)(b(v)-b(u))\,\mathrm{d}v\right\vert\mathrm{d}u.\label{eq:comp2terms} \end{align}\tag{1}\] We now show that the first term on the right-hand side converges to zero. Use the change-of-variables \(v\mapsto u+t_nw\) to obtain, for each \(u\in[0,1]\), \[\begin{gather} \int_0^1\chi_n(u,v)\,\mathrm{d}v=\int_{-u/t_n}^{(1-u)/t_n}\chi_n(u,u+t_nw)\,t_n\,\mathrm{d}w\\ =\int_{-u/t_n}^{(1-u)/t_n}[\mathbb{1}(w\leq\alpha(u))-\mathbb{1}(w\leq0)]\,\mathrm{d}w.\label{eq:compterm1} \end{gather}\tag{2}\] The final integral converges to \(\alpha(u)\) for each \(u\in(0,1)\) and is bounded by \(\lVert\alpha\rVert_\infty\). Consequently \[b(u)\int_0^1\chi_n(u,v)\,\mathrm{d}v-b(u)\alpha(u)\] converges to zero for each \(u\in(0,1)\) and is bounded by \(2\lVert\alpha\rVert_\infty\lvert b(u)\rvert\), an integrable function of \(u\). Thus an application of the dominated convergence theorem shows that the first term on the right-hand side of 1 converges to zero. It remains to show that the second term converges to zero.
Fix \(\epsilon>0\). The continuous maps from \([0,1]\) to \(\mathbb{R}\) are dense in \(L^1[0,1]\), so there exists a continuous \(b_\epsilon:[0,1]\to\mathbb{R}\) such that \(\lVert b_\epsilon-b\rVert_1\leq\epsilon\). Use Fubini’s theorem to bound the second term on the right-hand side of 1 by \[\begin{gather} \int_0^1\left(\int_0^1\lvert\chi_n(u,v)\rvert\,\mathrm{d}v\right)\lvert b_\epsilon(u)-b(u)\rvert\,\mathrm{d}u\\+\int_0^1\left(\int_0^1\lvert\chi_n(u,v)\rvert\,\mathrm{d}u\right)\lvert b_\epsilon(v)-b(v)\rvert\,\mathrm{d}v\\+\int_0^1\int_0^1\lvert\chi_n(u,v)\rvert\,\lvert b_\epsilon(u)-b_\epsilon(v)\rvert\,\mathrm{d}v\,\mathrm{d}u.\label{eq:comptriangle} \end{gather}\tag{3}\] The first term is bounded by \(\lVert\alpha\rVert_\infty\epsilon\) because we have \(\smallint_0^1\lvert\chi_n(u,v)\rvert\,\mathrm{d}v\leq\lVert\alpha\rVert_\infty\) by applying the change-of-variables \(v\mapsto u+t_nw\) as in 2 . To bound the second term we use the change-of-variables \(u\mapsto v-t_nw\) to obtain \[\begin{align} \int_0^1\lvert\chi_n(u,v)\rvert\,\mathrm{d}u&=\int_{(v-1)/t_n}^{v/t_n}\lvert\chi_n(v-t_nw,v)\rvert\,t_n\,\mathrm{d}w\\&\leq\int_{-\infty}^\infty\lvert\mathbb{1}(w\leq\alpha(v-t_nw))-\mathbb{1}(w\leq0)\rvert\,\mathrm{d}w. \end{align}\] The final integrand can be nonzero only if \(\lvert w\rvert\leq\lVert\alpha\rVert_\infty\). Since \(\alpha\) is uniformly continuous there exists an integer \(N_{\epsilon,1}\) such that \(\lvert\alpha(v-t_nw)-\alpha(v)\rvert\leq\epsilon\) for all \(n\geq N_{\epsilon,1}\), all \(v\in[0,1]\) and all \(w\in\mathbb{R}\) such that \(\lvert w\rvert\leq\lVert\alpha\rVert_\infty\). Consequently \[\begin{align} &\int_{-\infty}^\infty\lvert\mathbb{1}(w\leq\alpha(v-t_nw))-\mathbb{1}(w\leq0)\rvert\,\mathrm{d}w\\&\quad\leq\int_{-\infty}^\infty\mathbb{1}\big([0\wedge(\alpha(v)-\epsilon)]\leq w\leq[0\vee(\alpha(v)+\epsilon)]\big)\,\mathrm{d}w\\&\quad\leq\lvert\alpha(v)\rvert+2\epsilon\leq\lVert\alpha\rVert_\infty+2\epsilon \end{align}\] for all \(n\geq N_{\epsilon,1}\) and all \(v\in[0,1]\). The second term in 3 is therefore bounded by \(\lVert\alpha\rVert_\infty\epsilon+2\epsilon^2\) for all \(n\geq N_{\epsilon,1}\). To bound the third term, observe that since \(b_\epsilon\) is uniformly continuous there exists a real number \(\delta_{\epsilon}>0\) such that \(\lvert b_\epsilon(u)-b_\epsilon(v)\rvert\leq\epsilon\) whenever \(\lvert u-v\rvert\leq\delta_{\epsilon}\). Further observe that \(\chi_n(u,v)\) can be nonzero only if \(\lvert u-v\rvert\leq t_n\lVert\alpha\rVert_\infty\). Therefore there exists an integer \(N_{\epsilon,2}\) such that, for all \(n\geq N_{\epsilon,2}\), the third term in 3 is bounded by \(\epsilon\smallint_0^1\smallint_0^1\lvert\chi_n(u,v)\rvert\,\mathrm{d}v\,\mathrm{d}u\). Thus, since \(\smallint_0^1\lvert\chi_n(u,v)\rvert\,\mathrm{d}v\leq\lVert\alpha\rVert_\infty\), the third term is bounded by \(\lVert\alpha\rVert_\infty\epsilon\) for all \(n\geq N_{\epsilon,2}\).
Combining the preceding bounds shows that the second term on the right-hand side of 1 is at most \(3\lVert\alpha\rVert_\infty\epsilon+2\epsilon^2\) for all sufficiently large \(n\). Therefore, since \(\epsilon\) may be chosen arbitrarily small, the second term converges to zero. ◻
Lemma 1 supplies the only nonstandard differentiability argument needed for the proof of Theorem 2. The remainder of the proof combines Donsker’s theorem with the delta-method and its bootstrap analogue.
Proof of Theorem 2. Let \(\{(U_i,V_i)\}_{i=1}^\infty\) be an iid sequence of pairs of uniform random variables such that the bivariate distribution of each pair \((U_i,V_i)\) is given by the copula \(C\). Let \(C_n:[0,1]^2\to[0,1]\) be the bivariate empirical cdf for \(\{(U_i,V_i)\}_{i=1}^n\). Let \(\tilde{F}_n:[0,1]\to[0,1]\) and \(\tilde{G}_n:[0,1]\to[0,1]\) be the empirical cdfs for \(\{U_i\}_{i=1}^n\) and \(\{V_i\}_{i=1}^n\), respectively. Let \(\tilde{Q}_n:(0,1)\to[0,1]\) be the empirical qf for \(\{V_i\}_{i=1}^n\). The classical Donsker theorem establishes that \[\sqrt{n}(C_n-C)\rightsquigarrow\mathcal{B}\,\,\,\text{in }\ell^\infty([0,1]^2).\] An application of the continuous mapping theorem using the map \(h(u,v)\mapsto(h(1,v),h(R(u),1))\) from \(\ell^\infty([0,1]^2)\) into \(\ell^\infty[0,1]\otimes\ell^\infty[0,1]\) therefore shows that \[\label{eq:donskerjoint} \sqrt{n}((\tilde{G}_n,\tilde{F}_n\circ R)-(I,R))\rightsquigarrow(\mathcal{B}_2,\mathcal{B}_1\circ R)\,\,\,\text{in }\ell^\infty[0,1]\otimes\ell^\infty[0,1],\tag{4}\] where \(\mathcal{B}_1(\cdot)=\mathcal{B}(\cdot,1)\) and \(\mathcal{B}_2(\cdot)=\mathcal{B}(1,\cdot)\). Now an application of the delta-method using the generalized inverse map, suitable Hadamard differentiability of which is supplied by Lemma 21.4(ii) in [1], shows that \[\sqrt{n}((\tilde{Q}_n,\tilde{F}_n\circ R)-(I,R))\rightsquigarrow(-\mathcal{B}_2,\mathcal{B}_1\circ R)\,\,\,\text{in }\ell^\infty[0,1]\otimes\ell^\infty[0,1].\] Apply the delta-method again using the composition map \(\phi\) defined above. Lemma 1 supplies the requisite Hadamard differentiability. This yields \[\label{eq:conv1} \sqrt{n}(\tilde{F}_n\circ R\circ\tilde{Q}_n-R)\rightsquigarrow\mathcal{B}_1\circ R-r\mathcal{B}_2=\mathcal{R}\quad\text{in }L^1[0,1].\tag{5}\]
It remains to replace \(\tilde{F}_n\circ R\circ\tilde{Q}_n\) with \(R_n\) in the preceding display. Let \(P:(0,1)\to\mathbb{R}\) be the qf for \(X_i\). The function \(\tilde{F}_n\circ F\) is the empirical cdf for \(\{P(U_i)\}_{i=1}^n\) because \[\tilde{F}_n(F(x))=\frac{1}{n}\sum_{i=1}^n\mathbb{1}(U_i\leq F(x))=\frac{1}{n}\sum_{i=1}^n\mathbb{1}(P(U_i)\leq x).\] The function \(Q\circ \tilde{Q}_n\) is the empirical qf for \(\{Q(V_i)\}_{i=1}^n\) because applying \(Q\) to the order statistics of \(\{V_i\}_{i=1}^n\) produces the order statistics of \(\{Q(V_i)\}_{i=1}^n\). We have \[\begin{gather} \label{eq:eqd} \big((P(U_1),Q(V_1)),\ldots,(P(U_n),Q(V_n))\big)\\ \mathrel{\mathop{=}\limits^{ \vbox to0.5ex{\kern 0.2\ex@ \scriptstyle\mathrm{\scriptscriptstyle{D}\,}\vss}}}\big((X_1,Y_1),\ldots,(X_n,Y_n)\big) \end{gather}\tag{6}\] because the distribution of each pair \((P(U_i),Q(V_i))\) has margins \(F\) and \(G\) and admits the copula \(C\). Therefore \[(\tilde{F}_n\circ F)\circ(Q\circ\tilde{Q}_n) \mathrel{\mathop{=}\limits^{ \vbox to0.5ex{\kern 0.2\ex@ \scriptstyle\mathrm{\scriptscriptstyle{D}\,}\vss}}}F_n\circ Q_n\] as random elements of \(L^1[0,1]\). Since \[(\tilde{F}_n\circ F)\circ(Q\circ\tilde{Q}_n)=\tilde{F}_n\circ R\circ\tilde{Q}_n\quad\text{and}\quad F_n\circ Q_n=R_n\] on \((0,1)\), we deduce that \(\tilde{F}_n\circ R\circ\tilde{Q}_n \mathrel{\mathop{=}\limits^{ \vbox to0.5ex{\kern 0.2\ex@ \scriptstyle\mathrm{\scriptscriptstyle{D}\,}\vss}}}R_n\) as random elements of \(L^1[0,1]\), which justifies replacing \(\tilde{F}_n\circ R\circ\tilde{Q}_n\) with \(R_n\) in 5 . This proves the first assertion of the theorem.
The proof of the second assertion is similar. Define \(C_n^\ast:[0,1]^2\to[0,1]\), \(\tilde{F}_n^\ast:[0,1]\to[0,1]\) and \(\tilde{G}_n^\ast:[0,1]\to[0,1]\) by \[C_n^\ast(u,v)=\frac{1}{n}\sum_{i=1}^nW_{i,n}\mathbb{1}(U_i\leq u,V_i\leq v),\] \[\tilde{F}_n^\ast(u)=C_n^\ast(u,1),\quad\tilde{G}_n^\ast(v)=C_n^\ast(1,v).\] Let \(\tilde{Q}_n^\ast\) be the qf corresponding to \(\tilde{G}_n^\ast\). The bootstrap version of Donsker’s theorem yields \[\begin{gather} \sqrt{n}(C_n^\ast-C_n)\rightsquigarrow\mathcal{B}\quad\text{in }\ell^\infty([0,1]^2)\\\text{conditionally on } \{(U_i,V_i)\}_{i=1}^\infty\text{ in probability.} \end{gather}\] Applying the continuous mapping theorem and then the delta-method twice as above yields \[\begin{gather} \label{eq:conv2} \sqrt{n}(\tilde{F}_n^\ast\circ R\circ\tilde{Q}_n^\ast-\tilde{F}_n\circ R\circ\tilde{Q}_n)\rightsquigarrow\mathcal{R}\quad\text{in }L^1[0,1]\\\text{conditionally on } \{(U_i,V_i)\}_{i=1}^\infty\text{ in probability.} \end{gather}\tag{7}\] Fix an arbitrary bounded Lipschitz map \(h:L^1[0,1]\to\mathbb{R}\). For \(z=((x_1,y_1),\ldots,(x_n,y_n))\in(\mathbb{R}^2)^n\), let \(T_n(z)\in L^1[0,1]\) denote the P-P plot based on the sample \(z\), and for a bootstrap weight vector \(w\in\mathbb{R}^n\) let \(T_n^\ast(z,w)\in L^1[0,1]\) denote the corresponding bootstrap P-P plot. Define \[\Gamma_n(z)=\mathbb{E}\big[h\big(\!\sqrt n(T_n^\ast(z,W_n)-T_n(z))\big)\big].\] Then \[\begin{gather} \Gamma_n\big((X_1,Y_1),\ldots,(X_n,Y_n)\big)\\=\mathbb{E}\big[h\big(\!\sqrt n(R_n^\ast-R_n)\big)\,\big|\,\{(X_i,Y_i)\}_{i=1}^\infty\big] \end{gather}\] and \[\begin{gather} \Gamma_n\big((P(U_1),Q(V_1)),\ldots,(P(U_n),Q(V_n))\big)\\=\mathbb{E}\big[h\big(\!\sqrt{n}(\tilde{F}_n^\ast\circ R\circ\tilde{Q}_n^\ast-\tilde{F}_n\circ R\circ\tilde{Q}_n)\big)\,\big|\,\{(U_i,V_i)\}_{i=1}^\infty\big] \end{gather}\] a.s. By 7 , the final conditional expectation converges in probability to \(\mathbb{E}h(\mathcal{R})\). In view of 6 , it follows that \[\mathbb{E}\big[h\big(\!\sqrt n(R_n^\ast-R_n)\big)\,\big|\,\{(X_i,Y_i)\}_{i=1}^\infty\big]\to\mathbb{E}h(\mathcal{R})\] in probability. Since \(h\) was arbitrary, this proves the second assertion. ◻
We now show that convergence in distribution in \(L^1[0,1]\) cannot hold unless \(R\) is absolutely continuous. For clarity we state this as a separate theorem.
Theorem 3 (Necessity). If \(R\) is not absolutely continuous then \(\sqrt{n}(R_n-R)\) does not converge in distribution in \(L^1[0,1]\).
The proof is similar in spirit to the necessity argument in [14], but the P-P setting requires a separate treatment because the process involves composition of the empirical cdf with the empirical quantile function. We begin with two lemmas. Let \(\bar R:(0,1)\to[0,1]\) be the left-continuous version of \(R\) defined by \(\bar R(u)=\sup_{v\in(0,u)}R(v)\). This construction makes \(\bar R\) a valid qf. In fact \(\bar R\) is the qf for each \(F(Y_i)\).
Lemma 2. \(\bar{R}\) is the qf for each \(F(Y_i)\) and \(F\circ Q_n\) is the empirical qf for \(\{F(Y_i)\}_{i=1}^n\).
Proof. Let \(U\) be a random variable distributed uniformly on \((0,1)\). Then \(\bar{R}(U) \mathrel{\mathop{=}\limits^{ \vbox to0.5ex{\kern 0.2\ex@ \scriptstyle\mathrm{\scriptscriptstyle{a.s.}\,}\vss}}}R(U)=F(Q(U)) \mathrel{\mathop{=}\limits^{ \vbox to0.5ex{\kern 0.2\ex@ \scriptstyle\mathrm{\scriptscriptstyle{D}\,}\vss}}}F(Y_i)\). This shows that \(\bar{R}\) is the qf for each \(F(Y_i)\). The second assertion of the lemma is true because applying \(F\) to the order statistics of \(\{Y_i\}_{i=1}^n\) produces the order statistics of \(\{F(Y_i)\}_{i=1}^n\). ◻
Lemma 3. \(\{\!\sqrt{n}\lVert R_n-R\rVert_1\}_{n=1}^\infty\) is uniformly integrable.
Proof. Let \(H:[0,1]\to[0,1]\) be the cdf for each \(F(Y_i)\) and let \(H_n:[0,1]\to[0,1]\) be the empirical cdf for \(\{F(Y_i)\}_{i=1}^n\). The \(L^1\)-distance between any two cdfs is equal to the \(L^1\)-distance between the corresponding two qfs. It therefore follows from Lemma 2 that \(\lVert H_n-H\rVert_1=\lVert F\circ Q_n-\bar{R}\rVert_1\). Observe that \[\begin{align} \sqrt{n}\rVert R_n-R\rVert_1&\leq\sqrt{n}\lVert F_n\circ Q_n-F\circ Q_n\rVert_1+\sqrt{n}\lVert F\circ Q_n-\bar{R}\rVert_1\\&\quad\quad+\sqrt{n}\lVert \bar{R}-R\rVert_1. \end{align}\] The first term is bounded by \(\sqrt{n}\lVert F_n-F\rVert_\infty\), the second term is equal to \(\sqrt{n}\lVert H_n-H\rVert_1\) and thus bounded by \(\sqrt{n}\lVert H_n-H\rVert_\infty\), and the third term is zero. Therefore it suffices that \[\{\!\sqrt{n}\lVert F_n-F\rVert_\infty\}_{n=1}^\infty\quad\text{ and }\quad\{\!\sqrt{n}\lVert H_n-H\rVert_\infty\}_{n=1}^\infty\] are uniformly integrable. This is a well-known consequence of the Dvoretzky–Kiefer–Wolfowitz inequality [1]. ◻
Proof of Theorem 3. We prove the contrapositive. Let \(\mathcal{S}\) be a random element of \(L^1[0,1]\) and assume that \(\sqrt{n}(R_n-R)\rightsquigarrow\mathcal{S}\) in \(L^1[0,1]\). Let \(\mathcal{B}_2:[0,1]\to\mathbb{R}\) be a Brownian bridge. We will establish the bound \[\begin{gather} \label{eq:openineqR} \mathbb{E}\int_{[a,b)}\lvert \mathcal{B}_2(u)\rvert\,\mathrm{d}R(u)\leq\mathbb{E}\int_a^b\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u+\sqrt{\frac{\pi}{2}}(b-a)\\\text{for all continuity points }a,b\text{ of }R\text{ with }0<a<b<1. \end{gather}\tag{8}\] The first integral is defined in the Lebesgue-Stieltjes sense, wherein a Borel measure \(\mu_R\) on \((0,1)\) is generated by assigning measure \(\bar{R}(b)-\bar{R}(a)\) to each interval \([a,b)\). Observe that \[\begin{gather} \mathbb{E}\int_a^b\lvert F(Q_n(u))-R(u)\rvert\,\mathrm{d}u\leq\mathbb{E}\int_a^b\lvert R_n(u)-R(u)\rvert\,\mathrm{d}u\\+\mathbb{E}\int_a^b\lvert F_n(Q_n(u))-F(Q_n(u))\rvert\,\mathrm{d}u. \end{gather}\] Consequently, to establish the inequality in 8 it suffices to verify that \[\lim_{n\to\infty}\mathbb{E}\sqrt{n}\int_a^b\lvert R_n(u)-R(u)\rvert\,\mathrm{d}u=\mathbb{E}\int_a^b\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u,\label{eq:Rverify1}\tag{9}\] that \[\limsup_{n\to\infty}\mathbb{E}\sqrt{n}\int_a^b\lvert F_n(Q_n(u))-F(Q_n(u))\rvert\,\mathrm{d}u\leq\sqrt{\frac{\pi}{2}}(b-a),\label{eq:Rverify2}\tag{10}\] and that \[\begin{gather} \liminf_{n\to\infty}\mathbb{E}\sqrt{n}\int_a^b\lvert F(Q_n(u))-R(u)\rvert\,\mathrm{d}u\\\geq\mathbb{E}\int_{[a,b)}\lvert \mathcal{B}_2(u)\rvert\,\mathrm{d}R(u).\label{eq:Rverify3} \end{gather}\tag{11}\] An application of the continuous mapping theorem shows that \(\sqrt{n}\smallint_a^b\lvert R_n(u)-R(u)\rvert\,\mathrm{d}u\rightsquigarrow\smallint_a^b\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u\). Convergence in distribution implies convergence in mean due to the uniform integrability established in Lemma 3. This verifies 9 and also shows that \(\mathbb{E}\smallint_0^1\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u<\infty\). Observe that \[\int_a^b\lvert F_n(Q_n(u))-F(Q_n(u))\rvert\,\mathrm{d}u\leq\lVert F_n-F\rVert_\infty(b-a).\] We have \(\mathbb{E}\sqrt{n}\lVert F_n-F\rVert_\infty\leq\sqrt{\pi/2}\) by the Dvoretzky–Kiefer–Wolfowitz inequality. This verifies 10 . Lemma 3.4 in [14], applied to the sample \(\{F(Y_i)\}_{i=1}^n\) and combined with Lemma 2 above, yields 11 whenever \(a\) and \(b\) are continuity points of \(R\).
Next we strengthen 8 by showing that \[\begin{gather} \label{eq:openineqR2} \mathbb{E}\int_{[a,b]}\lvert \mathcal{B}_2(u)\rvert\,\mathrm{d}R(u)\leq\mathbb{E}\int_a^b\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u+\sqrt{\frac{\pi}{2}}(b-a)\\\text{for all }a,b\in(0,1)\text{ such that }a<b. \end{gather}\tag{12}\] Let \(\{a_n\}_{n=1}^\infty\) and \(\{b_n\}_{n=1}^\infty\) be sequences of continuity points of \(R\) such that \(0<a_n<b_n<1\) for each \(n\) and such that \(a_n\uparrow a\) and \(b_n\downarrow b\). Then Fatou’s lemma and 8 respectively justify the inequalities \[\begin{gather} \mathbb{E}\int_{[a,b]}\lvert \mathcal{B}_2(u)\rvert\,\mathrm{d}R(u)\leq\liminf_{n\to\infty}\mathbb{E}\int_{[a_n,b_n)}\lvert \mathcal{B}_2(u)\rvert\,\mathrm{d}R(u)\\\leq\liminf_{n\to\infty}\left(\mathbb{E}\int_{a_n}^{b_n}\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u+\sqrt{\frac{\pi}{2}}(b_n-a_n)\right). \end{gather}\] Since \(\mathbb{E}\smallint_0^1\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u<\infty\), an application of the dominated convergence theorem yields 12 .
Finally we use 12 to show that \(R\) is absolutely continuous on \([0,1]\). It suffices to show absolute continuity on \((0,1)\) because \(R\) is continuous at zero and one by construction. Let \(\lambda\) be the Lebesgue measure on \((0,1)\), let \(\mu_R\) be the Lebesgue-Stieltjes measure on \((0,1)\) generated by \(R\), and let \(\mathcal{J}\) be the collection of all finite unions of closed intervals with endpoints in \((0,1)\). By definition, \(R\) is absolutely continuous on \((0,1)\) if for every \(\epsilon>0\) there exists \(\delta_\epsilon>0\) such that \(\mu_R(A)\leq\epsilon\) for every \(A\in\mathcal{J}\) such that \(\lambda(A)\leq\delta_\epsilon\). For every \(A\in\mathcal{J}\) and every \(\eta\in(0,1/2)\) we have \[\begin{align} \mu_R(A)&=\int_A\mathbb{1}(\eta\leq u\leq1-\eta)\,\mathrm{d}R(u)+\int_A\mathbb{1}(u<\eta)\,\mathrm{d}R(u)\\ &\quad\quad+\int_A\mathbb{1}(u>1-\eta)\,\mathrm{d}R(u)\\ &\leq\frac{1}{\sqrt{\eta(1-\eta)}}\int_A\sqrt{u(1-u)}\,\mathrm{d}R(u)\\ &\quad\quad+\left(\bar{R}(\eta)-\lim_{u\downarrow0}\bar{R}(u)\right)+\left(\lim_{u\uparrow1}\bar{R}(u)-\bar{R}(1-\eta)\right). \end{align}\] Fix \(\epsilon>0\). Since \(\lim_{u\downarrow0}\bar{R}(u)\) and \(\lim_{u\uparrow1}\bar{R}(u)\) are finite there exists \(\eta_\epsilon\in(0,1/2)\) such that, for every \(A\in\mathcal{J}\), \[\mu_R(A)\leq\frac{1}{\sqrt{\eta_\epsilon(1-\eta_\epsilon)}}\int_A\sqrt{u(1-u)}\,\mathrm{d}R(u)+\frac{\epsilon}{3}.\] We have \(\mathbb{E}\lvert\mathcal{B}_2(u)\rvert=\sqrt{(2/\pi)u(1-u)}\) because \(\mathcal{B}_2(u)\) is normally distributed with mean zero and variance \(u(1-u)\). Therefore an application of Fubini’s theorem shows that \[\mu_R(A)\leq\sqrt{\frac{\pi}{2\eta_\epsilon(1-\eta_\epsilon)}}\,\mathbb{E}\int_A\lvert\mathcal{B}_2(u)\rvert\,\mathrm{d}R(u)+\frac{\epsilon}{3}.\] Since \(A\) is a finite union of closed intervals with endpoints in \((0,1)\), an application of the bound in 12 shows that \[\mu_R(A)\leq\sqrt{\frac{\pi}{2\eta_\epsilon(1-\eta_\epsilon)}}\left(\mathbb{E}\int_A\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u+\sqrt{\frac{\pi}{2}}\lambda(A)\right)+\frac{\epsilon}{3}.\] For each \(n\in\mathbb{N}\) we have \[\begin{align} \mathbb{E}\int_A\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u&=\mathbb{E}\int_A\mathbb{1}\big(\lvert\mathcal{S}(u)\rvert\leq n\big)\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u\\ &\quad\quad+\mathbb{E}\int_A\mathbb{1}\big(\lvert\mathcal{S}(u)\rvert>n\big)\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u\\ &\leq n\lambda(A)+\mathbb{E}\int_0^1\mathbb{1}\big(\lvert\mathcal{S}(u)\rvert>n\big)\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u. \end{align}\] Since \(\mathbb{E}\smallint_0^1\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u<\infty\) the dominated convergence theorem shows that \(\mathbb{E}\smallint_0^1\mathbb{1}\big(\lvert\mathcal{S}(u)\rvert>n\big)\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u\to0\). Therefore there exists an integer \(N_\epsilon\) not depending on \(A\) such that \[\mathbb{E}\int_A\lvert\mathcal{S}(u)\rvert\,\mathrm{d}u\leq N_\epsilon\lambda(A)+\sqrt{\frac{2\eta_\epsilon(1-\eta_\epsilon)}{\pi}}\cdot\frac{\epsilon}{3}.\] Consequently \[\mu_R(A)\leq\sqrt{\frac{\pi}{2\eta_\epsilon(1-\eta_\epsilon)}}\left(N_\epsilon\lambda(A)+\sqrt{\frac{\pi}{2}}\lambda(A)\right)+\frac{2\epsilon}{3}.\] Now if we set \[\delta_\epsilon=\frac{\sqrt{2\eta_\epsilon(1-\eta_\epsilon)}}{3\sqrt{\pi}(N_\epsilon+\sqrt{\pi/2})}\,\epsilon\] then we have \(\mu_R(A)\leq\epsilon\) for every \(A\in\mathcal{J}\) such that \(\lambda(A)\leq\delta_\epsilon\). This establishes that \(R\) is absolutely continuous. ◻
The foregoing analysis extends with only minor modifications to the case in which the P-P process is constructed from two independent samples of different sizes. To do this we introduce a nondecreasing function \(m:\mathbb{N}\to\mathbb{N}\). Instead of defining \(F_n\) to be the empirical cdf for \(\{X_i\}_{i=1}^n\), define it to be the empirical cdf for \(\{X_i\}_{i=1}^{m(n)}\). Assume that \(C\) is the product copula. In this setting \(R_n\) is the P-P plot based on two independent samples of sizes \(m(n)\) and \(n\). If we also assume that \[m(n)\to\infty\quad\text{and}\quad\frac{n}{m(n)+n}\to\rho\quad\text{for some }\rho\in[0,1)\] then Theorem 1 remains true in this modified setting. The proof of Theorem 3 requires only minor adjustments to keep track of the constant \(\rho\). The first assertion of Theorem 2 remains true provided that \(\mathcal{R}\) is defined to be \[\mathcal{R}(u)=\sqrt{\frac{\rho}{1-\rho}}\mathcal{B}_1(R(u))-r(u)\mathcal{B}_2(u),\] where \(\mathcal{B}_1\) and \(\mathcal{B}_2\) are independent Brownian bridges. In the proof of the first assertion we should define \(\tilde{F}_n\) to be the empirical cdf for \(\{U_i\}_{i=1}^{m(n)}\), so that \[\sqrt{n}\big((\tilde{F}_n,\tilde{G}_n)-(I,I)\big)\rightsquigarrow\left(\sqrt{\frac{\rho}{1-\rho}}\mathcal{B}_1,\mathcal{B}_2\right)\] in \(\ell^\infty[0,1]\otimes\ell^\infty[0,1]\) by Donsker’s theorem. An application of the continuous mapping theorem then yields 4 with \(\mathcal{B}_1\) replaced by \(\sqrt{\rho/(1-\rho)}\mathcal{B}_1\). The remainder of the argument is unchanged.
With independent samples the implementation of the bootstrap is naturally modified so that independent copies of the bootstrap weight vector \(W_n\) are separately used to construct \(F_n^\ast\) and \(G_n^\ast\). The second assertion of Theorem 2 remains true with this implementation of the bootstrap and with the modified definition of \(\mathcal{R}\). The proof extends in the same way as the proof of the first assertion.