January 01, 1970
We establish logarithmic large-deviation bounds for sums of independent nonnegative random variables with regularly varying tails. The normalization is chosen at the extreme-value scale and the speed is \(\log n\). In contrast with Cramér’s theorem, the resulting rate function is determined only by the tail index. The proof transfers a maximum large-deviation principle to sums in the one-big-jump region.
Let \((X_k)_{k\ge1}\) be independent and identically distributed real-valued random variables, and set \(S_n=X_1+\cdots+X_n\). Cramér’s theorem [1] is the classical large-deviation result for sums of independent random variables with light tails. If the logarithmic moment generating function \(\Lambda(\lambda)=\log\mathbb{E}[e^{\lambda X_1}]\) is finite in a neighbourhood of the origin, then the empirical mean \(S_n/n\) satisfies a large-deviation principle with speed \(n\) and rate function \[\Lambda^*(x)=\sup_{\lambda\in\mathbb{R}}\{\lambda x-\Lambda(\lambda)\}.\] In particular, for \(x>\mathbb{E}[X_1]\), \[\label{eq:cramer} \lim_{n\to\infty}\frac{1}{n}\log \mathbb{P}(S_n\ge nx) = -\Lambda^*(x),\tag{1}\] so upper-tail probabilities are exponentially small in \(n\). We refer to [2] for background on Cramér’s theorem and large-deviation theory.
This framework is no longer appropriate for heavy-tailed random variables. If the upper tail of \(X_1\) is regularly varying, then positive exponential moments are infinite and the Cramér transform does not describe rare upper-tail events. In this case, large deviations occur on a polynomial rather than an exponential scale. The relevant mechanism is the one-big-jump principle: a large value of the sum is typically caused by one exceptionally large summand, rather than by a collective displacement of all summands.
The purpose of this note is to formulate this heavy-tailed polynomial regime as a Cramér-type large-deviation principle. Let \(a_n\) be the upper \(1-1/n\) quantile of \(X_1\), and consider the normalized sums \[Y_n=\left(\frac{S_n}{a_n}\right)^{1/\log n}.\] This normalization converts polynomial excesses above the extreme-value scale into fixed levels: indeed, \(\{S_n\ge a_n n^x\}\) is equivalent to \(\{Y_n\ge e^x\}\).
We prove large-deviation bounds for \((Y_n)\) in the far right tail, where the one-big-jump mechanism governs the deviation. The speed is \(\log n\), and the rate function is \(I(y)=\alpha\log y\), where \(\alpha\) is the tail index. The statement is set-valued: it gives lower bounds for arbitrary open sets and upper bounds for sets that are closed in \(\mathbb{R}\), both restricted to the right-tail region \(E_\alpha\) introduced below. Thus the result does not only identify the decay of a single threshold probability, but describes the logarithmic decay of probabilities \(\mathbb{P}(Y_n\in A)\) for general right-tail deviation sets \(A\).
For the particular upper-tail event \(S_n\ge a_n n^x\), the result gives \[\lim_{n\to\infty}\frac{1}{\log n} \log\mathbb{P}(S_n\ge a_n n^x) = -\alpha x, \qquad x>\left(1-\frac{1}{\alpha}\right)^+.\] Thus the exponential speed \(n\) in Cramér’s theorem 1 is replaced by the logarithmic speed \(\log n\), the exponential scale is replaced by the natural polynomial scale of regularly varying tails, and the rate is determined only by the tail index.
The proof uses the large-deviation estimate for the normalized maximum established in [3].
Relation with the literature. The one-big-jump principle is classical in the theory of heavy-tailed and subexponential distributions. In its basic form, for a fixed number of summands \(m\), it states that \[\label{eq:Nagaev} \mathbb{P}(S_m>t)\sim m\bar F(t), \qquad t\to\infty.\tag{2}\] Thus the asymptotic parameter is the threshold \(t\), while the number of summands is fixed. More refined Nagaev-type and subexponential results allow the number of summands to vary and give uniform estimates on suitable big-jump regions; see, for instance, [4]–[6]. These results are mainly concerned with sharp or uniform tail equivalents of the form 2 .
The present paper uses this one-big-jump regime for a different purpose: rather than deriving a sharp equivalent for a prescribed threshold, we organize the polynomial deviation probabilities into large-deviation bounds for a normalized sequence. The threshold is chosen as a function of the sample size, namely \(a_n n^x\), and the asymptotic parameter is \(n\to\infty\). In this sense, the result provides a logarithmic large-deviation substitute for Cramér’s theorem in a setting where the classical exponential theory cannot be applied.
Let \(X,X_1,X_2,\ldots\) be independent and identically distributed nonnegative random variables with distribution function \(F\) and survival function \(\bar F=1-F.\) Throughout the paper we assume that \(\bar F\) is regularly varying with index \(-\alpha\), where \(\alpha>0\); that is, \[\bar F(x)=x^{-\alpha}L(x), \qquad x>0,\] where \(L\) is slowly varying: \[\lim_{x\to\infty}\frac{L(xy)}{L(x)}=1, \qquad y>0.\] For background on regular variation and the facts used below, we refer to [7].
Define the upper \(1-1/n\) quantile \[a_n=F^{\leftarrow}\left(1-\frac{1}{n}\right), \qquad F^{\leftarrow}(u)=\inf\{x\in\mathbb{R}:F(x)\ge u\}.\]
Write \[S_n=X_1+\cdots+X_n,\qquad Y_n=\left(\frac{S_n}{a_n}\right)^{1/\log n}.\] Set \[c_\alpha=\left(1-\frac{1}{\alpha}\right)^+, \qquad d_\alpha=e^{c_\alpha}, \qquad E_\alpha=(d_\alpha,\infty).\]
We present the main result of the note.
Theorem 1. In the setup above, the following large-deviation bounds hold on \(E_\alpha\), with speed \(\log n\) and rate function \(I(y)=\alpha\log y\).
For every open set \(G\subset E_\alpha\), \[\liminf_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n\in G) \ge -\inf_{y\in G}\alpha\log y.\]
For every set \(F\subset E_\alpha\) that is closed in \(\mathbb{R}\), \[\limsup_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n\in F) \le -\inf_{y\in F}\alpha\log y.\]
Moreover, for every \(x>c_\alpha\) one has \[\lim_{n\to\infty} \frac{1}{\log n} \log\mathbb{P}(S_n\ge a_n n^x) = -\alpha x.\]
Suppose that \(X_1,X_2,\ldots\) are as in Section 2. Throughout this section we use the notation \[S_n=X_1+\cdots+X_n, \qquad X_{(n)}=\max\{X_1,\ldots,X_n\}.\] The corresponding normalized sum and maximum are \[Y_n=\left(\frac{S_n}{a_n}\right)^{1/\log n}, \qquad Z_n=\left(\frac{X_{(n)}}{a_n}\right)^{1/\log n}.\] For \(x>0\), set \[t_n(x)=a_n x^{\log n}.\] We also write \[c_\alpha=\left(1-\frac{1}{\alpha}\right)^+, \qquad d_\alpha=e^{c_\alpha}, \qquad E_\alpha=(d_\alpha,\infty).\]
The proof uses the following sharp large-deviation estimate for the normalized maximum, established in [3]. The version proved there assumes a weak von Mises condition and gives a stronger Borel-set statement, namely an exact logarithmic limit. In the present paper, we only need the upper-tail estimate below. For completeness, we include a short proof under the present assumptions in the Appendix.
Lemma 1. In the present setup, for every \(x>1\) one has \[\lim_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Z_n>x)=-\alpha\log x.\]
Next, we prove some auxiliary lemmas.
Lemma 2. Fix \(x\in E_\alpha\) and \(h>0\). Set \[b_n(x,h)=\frac{t_n(x)}{2h\log n}.\] Then \[\mathbb{P}\left( \sum_{k=1}^n X_k\mathbf{1}_{\{X_k\le b_n(x,h)\}} \ge t_n(x) \right) \le n^{-h}\] for all sufficiently large \(n\).
Proof. Write \(t_n=t_n(x)\) and \(b_n=b_n(x,h)\), and put \(q=\log x+1/\alpha\). Since \(x>e^{c_\alpha}\), we have \(q>1/(\alpha\wedge1)\). Choose \(p\) such that \(q^{-1}<p<\alpha\wedge1\). Then \(p<1\), \(p<\alpha\), and \(pq>1\).
Let \(T_k=X_k\mathbf{1}_{\{X_k\le b_n\}}\). Since \(0\le T_k\le b_n\) and \(p<1\), for \(r=1,2\), \[T_k^r\le b_n^{r-p}X_k^p.\] Thus, taking expectations, \[\mathbb{E}[T_k^r]\le b_n^{r-p}\mathbb{E}[X_k^p].\] By Lemma 4, \(\mathbb{E}[X_1^p]<\infty\), and by Lemma 5(ii), \(t_n=n^{q+o(1)}\). Hence, for \(r=1,2\), \[\begin{align} \frac{n\mathbb{E}[T_1^r]}{b_n^{r-1}t_n} &\le \mathbb{E}[X_1^p]\frac{n b_n^{1-p}}{t_n} \nonumber\\ &= \mathbb{E}[X_1^p]\frac{n t_n^{-p}}{(2h\log n)^{1-p}} \nonumber\\ &= \mathbb{E}[X_1^p]\frac{n^{1-pq+o(1)}}{(2h\log n)^{1-p}}. \end{align}\] Since \(pq>1\) and \(p<1\), the last term converges to \(0\). Consequently, for all sufficiently large \(n\), \[n\mathbb{E}[T_1]\le \frac{t_n}{2}, \qquad 8n\mathbb{E}[T_1^2]\le \frac{2}{3}b_n t_n.\]
Then, for \(n\) large enough, \[\left\{\sum_{k=1}^n T_k\ge t_n\right\} \subset \left\{\sum_{k=1}^n\bigl(T_k-\mathbb{E}[T_1]\bigr)\ge \frac{t_n}{2}\right\}.\] By the one-sided Bernstein inequality for bounded independent random variables; see [8], \[\mathbb{P}\left(\sum_{k=1}^n T_k\ge t_n\right) \le \exp\left( -\frac{t_n^2}{8n\mathbb{E}[T_1^2]+(4/3)b_n t_n} \right).\] The denominator is at most \(2b_n t_n\), and therefore \[\mathbb{P}\left(\sum_{k=1}^n T_k\ge t_n\right) \le \exp\left(-\frac{t_n}{2b_n}\right) = n^{-h}.\] The proof is complete. ◻
Lemma 3. For every \(x\in E_\alpha\), \[\lim_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n>x) = -\alpha\log x.\]
Proof. Since \(Y_n\ge Z_n\), applying Lemma 1 we get the lower bound \[\liminf_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n>x)\ge \liminf_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Z_n>x) = -\alpha\log x.\]
We now prove the upper bound. Fix \(x\in E_\alpha\) and \(h>0\). We have \[\begin{align} \{Y_n>x\} &= \{S_n>t_n(x)\}\\ &\subset \{X_{(n)}>b_n(x,h)\} \cup \left\{ \sum_{k=1}^n X_k\mathbf{1}_{\{X_k\le b_n(x,h)\}}>t_n(x) \right\}. \end{align}\] By Lemma 2, the second event has probability at most \(n^{-h}\) for all sufficiently large \(n\).
As for the first event, choose \(\eta>0\) so small that \(xe^{-\eta}> 1\). Note that \[b_n(x,h) = \frac{t_n(x)}{2h\log n} = \frac{a_n x^{\log n}}{2h\log n} = \frac{t_n(xe^{-\eta}) n^\eta}{2h\log n}.\] Since \(n^\eta/\log n\to\infty\), we have \(b_n(x,h)\ge t_n(xe^{-\eta})\) for all sufficiently large \(n\). Hence \[\{X_{(n)}>b_n(x,h)\} \subset \{X_{(n)}>t_n(xe^{-\eta})\} = \{Z_n>xe^{-\eta}\}.\] By Lemma 1, applied to \(xe^{-\eta}>1\), \[\limsup_{n\to\infty} \frac{1}{\log n} \log\mathbb{P}(X_{(n)}>b_n(x,h)) \le -\alpha\log(xe^{-\eta}) = -\alpha\log x+\alpha\eta.\] Combining the bounds obtained above, we obtain \[\begin{align} \limsup_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n>x) &\le \bigl(-\alpha\log x+\alpha\eta\bigr)\vee(-h). \end{align}\] Letting first \(h\to\infty\) and then \(\eta\downarrow0\) gives \[\limsup_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n>x) \le -\alpha\log x.\] This completes the proof. ◻
We now turn to the proof of the main result.
Proof of Theorem 1. Let \(G\subset E_\alpha\) be open.
Given a fixed \(x_0\in G\), we choose \(a,b\in E_\alpha\) such that \(a<x_0<b\) and \((a,b]\subset G\).
The function \(x\mapsto I(x)=\alpha\log x\) is strictly increasing and \(a<b\), hence \(I(a)<I(b)\). Fix \(\varepsilon>0\) such that \[2\varepsilon<I(b)-I(a).\] By Lemma 3, for all sufficiently large \(n\), \[n^{-I(a)-\varepsilon}\le \mathbb{P}(Y_n>a),\qquad \mathbb{P}(Y_n>b)\le n^{-I(b)+\varepsilon}.\] Therefore \[\begin{align} \mathbb{P}(a<Y_n\le b) &= \mathbb{P}(Y_n>a)-\mathbb{P}(Y_n>b) \\ &\ge n^{-I(a)-\varepsilon}-n^{-I(b)+\varepsilon} \\ &= n^{-I(a)-\varepsilon} \left(1-n^{I(a)-I(b)+2\varepsilon}\right). \end{align}\] Since \(I(a)-I(b)+2\varepsilon<0\), the term in parentheses converges to \(1\). Consequently, \[\liminf_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(a<Y_n\le b) \ge -I(a)-\varepsilon.\] Letting first \(\varepsilon\downarrow0\) and then \(a\uparrow x_0\) give \[\liminf_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n\in G) \ge -I(x_0).\] Since \(x_0\in G\) was arbitrary, \[\liminf_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n\in G) \ge -\inf_{x\in G}I(x).\]
Now let \(F\subset E_\alpha\) be closed in \(\mathbb{R}\). If \(F=\emptyset\), there is nothing to prove. Since \(F\) is closed in \(\mathbb{R}\) and \(F\subset E_\alpha=(d_\alpha,\infty)\), we have \[a_F:=\inf F\in F \qquad\text{and}\qquad a_F>d_\alpha.\] For every \(\delta\in(0,a_F-d_\alpha)\), \[F\subset(a_F-\delta,\infty).\] Hence \[\mathbb{P}(Y_n\in F) \le \mathbb{P}(Y_n>a_F-\delta).\] By Lemma 3, \[\limsup_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n\in F) \le -I(a_F-\delta).\] Letting \(\delta\downarrow0\) and using the continuity of \(I\), we obtain \[\limsup_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Y_n\in F) \le -I(a_F)=-\inf_{x\in F}I(x),\] where we have applied in the last equality that \(I\) is increasing. This completes the proof of the upper bound.
Finally, we note that the final assertion of Theorem 1 follows from Lemma 3 above. The proof is complete. ◻
This appendix is devoted to some auxiliary results.
Throughout, we assume that \[\bar F(x)=x^{-\alpha}L(x), \qquad \alpha>0,\] where \(L\) is slowly varying. We use standard facts on regularly varying functions, including Potter bounds, quantile asymptotics, and polynomial tail bounds; see [7]. Recall that \[a_n=F^{\leftarrow}\left(1-\frac{1}{n}\right), \qquad t_n(x)=a_n x^{\log n}.\]
Lemma 4. If \(\bar F(x)=x^{-\alpha}L(x)\) with \(\alpha>0\), then \[\mathbb{E}[X^p]<\infty\] for every \(p\in(0,\alpha)\).
Proof. For a nonnegative random variable \(X\), \[\mathbb{E}[X^p] = p\int_0^\infty t^{p-1}\mathbb{P}(X>t)\,\mathrm dt.\] Choose \(\varepsilon>0\) such that \(p+\varepsilon<\alpha\). By regular variation, there exists \(t_0>0\) such that \[\bar F(t)\le t^{-\alpha+\varepsilon}, \qquad t\ge t_0.\] Therefore \[\int_{t_0}^\infty t^{p-1}\bar F(t)\,\mathrm dt \le \int_{t_0}^\infty t^{p-1-\alpha+\varepsilon}\,\mathrm dt <\infty.\] This proves the claim. ◻
Lemma 5. The following estimates hold.
The high quantile satisfies \[\frac{\log a_n}{\log n}\longrightarrow \frac{1}{\alpha}.\] Moreover, \[\frac{\log\bar F(a_n)}{\log n}\longrightarrow -1.\]
For every \(M>1\), \[\frac{\log t_n(x)}{\log n} \longrightarrow \frac{1}{\alpha}+\log x\] uniformly for \(x\in[1,M]\).
For every \(M>1\), \[\bar F(t_n(x)) = n^{-1-\alpha\log x+o(1)}\] uniformly for \(x\in[1,M]\).
If \((b_n)\) is a positive sequence such that \(\log b_n=o(\log n)\), then, for every \(M>1\), \[\bar F(b_n t_n(x)) = n^{-1-\alpha\log x+o(1)}\] uniformly for \(x\in[1,M]\).
Proof. By regular variation, \[\frac{\log \bar F(t)}{\log t}\longrightarrow -\alpha, \qquad t\to\infty;\] see [7]. In particular, for every \(\varepsilon\in(0,\alpha)\) and all sufficiently large \(t\), \[t^{-(\alpha+\varepsilon)} \le \bar F(t) \le t^{-(\alpha-\varepsilon)}.\]
We first prove the quantile estimate. We have \(a_n\to\infty\). Indeed, if \((a_n)\) were bounded, then \(\bar F\) would vanish eventually, contradicting regular variation with index \(-\alpha\).
Fix \(\varepsilon\in(0,\alpha)\) and let \(\rho>0\) be fixed. For all sufficiently large \(n\), both \(a_n\) and \(a_n-\rho\) are large enough for the preceding polynomial bounds to apply. By the definition of the generalized inverse, \[\bar F(a_n)\le \frac{1}{n} \qquad\text{and}\qquad \frac{1}{n}\le \bar F(a_n-\rho).\] Therefore, for all sufficiently large \(n\), \[a_n^{-(\alpha+\varepsilon)} \le \bar F(a_n) \le \frac{1}{n} \le \bar F(a_n-\rho) \le (a_n-\rho)^{-(\alpha-\varepsilon)}.\] It follows that \[n^{1/(\alpha+\varepsilon)} \le a_n \le \rho+n^{1/(\alpha-\varepsilon)}.\] Hence \[\frac{1}{\alpha+\varepsilon} \le \liminf_{n\to\infty}\frac{\log a_n}{\log n} \le \limsup_{n\to\infty}\frac{\log a_n}{\log n} \le \frac{1}{\alpha-\varepsilon}.\] Letting \(\varepsilon\downarrow0\) proves \[\frac{\log a_n}{\log n}\longrightarrow \frac{1}{\alpha}.\]
Moreover, \[\frac{\log\bar F(a_n)}{\log n} = \frac{\log\bar F(a_n)}{\log a_n} \frac{\log a_n}{\log n}.\] The first factor converges to \(-\alpha\) and the second one to \(1/\alpha\). Thus \[\frac{\log\bar F(a_n)}{\log n}\longrightarrow -1.\] This proves (i).
For (ii), note that \[\frac{\log t_n(x)}{\log n} = \frac{\log a_n}{\log n}+\log x.\] The convergence is uniform for \(x\in[1,M]\) by (i).
We next prove (iii). Since \(t_n(x)\ge a_n\to\infty\) uniformly for \(x\in[1,M]\), the convergence \[\frac{\log \bar F(t)}{\log t}\longrightarrow -\alpha\] may be applied uniformly along \(t=t_n(x)\). Hence, uniformly for \(x\in[1,M]\), \[\begin{align} \frac{\log\bar F(t_n(x))}{\log n} &= \frac{\log\bar F(t_n(x))}{\log t_n(x)} \frac{\log t_n(x)}{\log n} \\ &\longrightarrow -\alpha\left(\frac{1}{\alpha}+\log x\right) = -1-\alpha\log x. \end{align}\] This is equivalent to \[\bar F(t_n(x)) = n^{-1-\alpha\log x+o(1)}\] uniformly for \(x\in[1,M]\).
Finally, let \((b_n)\) be a positive sequence with \(\log b_n=o(\log n)\). Then, uniformly for \(x\in[1,M]\), \[\frac{\log(b_n t_n(x))}{\log n} = \frac{\log b_n}{\log n} + \frac{\log t_n(x)}{\log n} = \frac{1}{\alpha}+\log x+o(1).\] In particular, \(b_n t_n(x)\to\infty\) uniformly for \(x\in[1,M]\). Applying again the convergence \[\frac{\log \bar F(t)}{\log t}\longrightarrow -\alpha\] along \(t=b_n t_n(x)\) gives \[\frac{\log\bar F(b_n t_n(x))}{\log n} \longrightarrow -1-\alpha\log x\] uniformly for \(x\in[1,M]\). This proves (iv). ◻
Proof of Lemma 1. Let \(x>1\) and put \[q_n(x)=\bar F(t_n(x)).\] By Lemma 5(iii), \[q_n(x)=n^{-1-\alpha\log x+o(1)}.\] Since \(x>1\), we have \[nq_n(x)=n^{-\alpha\log x+o(1)}\longrightarrow0.\] Moreover, \[\mathbb{P}(Z_n>x) = \mathbb{P}(X_{(n)}>t_n(x)) = 1-\bigl(1-q_n(x)\bigr)^n.\] Using the identity \[1-u^n=(1-u)(1+u+\cdots+u^{n-1}), \qquad 0\le u\le1,\] we obtain \[n(1-u)u^{n-1}\le 1-u^n\le n(1-u).\] Using Bernoulli’s inequality \(u^{n-1}\ge 1-(n-1)(1-u)\), we get \[n(1-u)[1-(n-1)(1-u)]\le 1-u^n\le n(1-u).\] Applying this with \(u=1-q_n(x)\), we obtain \[n q_n(x)(1-(n-1)q_n(x)) \le \mathbb{P}(Z_n>x) \le nq_n(x).\] Since \(nq_n(x)\to0\), we have \[1-(n-1)q_n(x)\longrightarrow 1.\] Thus \[\mathbb{P}(Z_n>x) = nq_n(x)(1+o(1)).\] Using again \[q_n(x)=n^{-1-\alpha\log x+o(1)},\] we obtain \[\mathbb{P}(Z_n>x) = n^{-\alpha\log x+o(1)}.\] Hence \[\lim_{n\to\infty} \frac{1}{\log n}\log\mathbb{P}(Z_n>x) = -\alpha\log x.\] The proof is complete. ◻