Singularity Analysis for Heavy-Tailed Random Variables

Abstract

We propose a novel complex-analytic method for sums of i.i.d. random variables that are heavy-tailed and integer-valued. The method combines singularity analysis, Lindelöf integrals, and bivariate saddle points. As an application, we prove three theorems on precise large and moderate deviations which provide a local variant of a result by Nagaev (Transactions of the sixth Prague conference on information theory, statistical decision functions, random processes, Academia, Prague, 1973). The theorems generalize five theorems by Nagaev (Litov Mat Sb 8:553–579, 1968) on stretched exponential laws $p(k) = c\exp ( -k^\alpha )$ and apply to logarithmic hazard functions $c\exp ( - (\log k)^\beta )$, $\beta >2$; they cover the big-jump domain as well as the small steps domain. The analytic proof is complemented by clear probabilistic heuristics. Critical sequences are determined with a non-convex variational problem.

Appendices

Appendix A. Proofs of Lemmas 2.14 and 2.15

Proof of Lemma 2.14

The proof of Assumption 2.1 is straightforward and left to the reader. The function $q(\zeta ) = \zeta ^\alpha$ is analytic in $\mathrm {Re}\,\zeta >0$ and $p(\zeta )= \exp (- \zeta ^\alpha )$ satisfies, for all $k \in \mathbb {N}$,

$$\begin{aligned} |\zeta ^k p(\zeta )| = |\zeta |^k \,\mathrm{e}^{- |\zeta |^\alpha \cos (\alpha \mathrm{arg}(\zeta ))}\, \le |\zeta |^k \,\mathrm{e}^{- |\zeta |^\alpha \cos (\alpha \pi /2)}\,. \end{aligned}$$

(A.1)

Since $\alpha \in (0,1)$, we have $\cos \frac{\alpha \pi }{2} >0$ and Eq. (A.1) shows that $|\zeta ^k p(\zeta )|$ is integrable along $\mathrm {Re}\,\zeta =1/2$ and that $p(\zeta )$ grows slower than any exponential $\exp (\varepsilon |\zeta |)$. This proves Assumption 2.2.

The equation $q''(x_n^*) = -1/(n\sigma ^2)$ can be solved explicitly. $N_n^*$ and $N_n^{**}$ are best determined with the scaling relation (2.24). They have already been determined in [12], we omit the proof. For the insensitivity scale, we notice that

$$\begin{aligned} n \sigma ^2 q'(N_n)^2 = n \sigma ^2 \alpha ^2 N_n^{2 \alpha -2} \end{aligned}$$

(A.2)

which goes to zero if and only if $N_n \gg n^{-1/(2-2\alpha )}$. $\square$

Proof of Lemma 2.15

The function $q(x) = - \log c +(\log x)^\beta$ is clearly smooth on $(1,\infty )$. Then $q'(x) = \beta (\log x)^{\beta -1}/ x$ and as $x\rightarrow \infty$,

$$\begin{aligned} q''(x) \sim - \frac{\beta (\log x)^{\beta -1}}{x^2},\quad q'''(x) \sim \frac{2 \beta (\log x)^{\beta -1}}{x^3}. \end{aligned}$$

(A.3)

Assumption 2.1 is easily checked. For Assumption 2.2 we note $q(\zeta ) = c + (\log \zeta )^\beta$ is analytic in $\mathrm {Re}\,\zeta >1$. Fix $b>1$ and write $\zeta = r \exp ( \mathrm{i}\theta )$. As $|\zeta | \rightarrow \infty$ along $\mathrm {Re}\,\zeta = b$ i.e. $\zeta = b + \mathrm{i}y$, the argument $\theta$ goes to $\pm \pi /2$ and we have

$$\begin{aligned} \mathrm {Re}\,(\log |\zeta | + \mathrm{i}\theta )^\beta&= \mathrm {Re}\,\Bigl ( \log |y| + \frac{1}{2}\log \Bigl (1+ \frac{b^2}{y^2}\Bigr ) + \mathrm{i}\theta \Bigr )^\beta \end{aligned}$$

(A.4)

$$\begin{aligned}&= (\log |y|)^\beta + o(1), \end{aligned}$$

(A.5)

conditions(i) and (ii) in Assumption 2.2 are easily checked. Assumption 2.3(iii) follows from a computation similar to (A.3). Set $y_0 = \sqrt{r^2 - b^2}$. We have for $\zeta = b + \mathrm{i}y$, $y\ge y_r$, uniformly in r,

$$\begin{aligned} \mathrm {Re}\,q(\zeta ) - \mathrm {Re}\,q(z_r)&= (\log y)^\beta - (\log y_0)^\beta + o(1)\nonumber \\&\ge \beta (\log y_0)^{\beta -1} \log \frac{y}{y_0} + o(1) \end{aligned}$$

(A.6)

hence

$$\begin{aligned} \int _{y_r}^\infty \,\mathrm{e}^{- \mathrm {Re}\,q(b + \mathrm{i}y)}\, \mathrm {d}y&\le \,\mathrm{e}^{-\mathrm {Re}\,q(b+\mathrm{i}y_0) + o(1)}\, \int _1^\infty \,\mathrm{e}^{- (\log y_0)^{\beta -1} \log s}\, y_0 \mathrm {d}s \nonumber \\&\sim \,\mathrm{e}^{-\mathrm {Re}\,q(b+\mathrm{i}y_0)}\, \frac{y_0}{(\log y_0)^{\beta -1}}= \,\mathrm{e}^{-\mathrm {Re}\,q(b+\mathrm{i}y_0) + O(\log r)}\,, \end{aligned}$$

(A.7)

which proves Assumption2.3(i). Next let $\zeta \in \mathbb {C}$ with $\mathrm {Re}\,\zeta >1$, write $\zeta = r \exp (\mathrm{i}\theta )$, then

$$\begin{aligned} \zeta q'(\zeta ) = \beta (\log \zeta )^{\beta -1} = \beta (\log r + \mathrm{i}\theta )^{\beta -1} = q'(r) \Bigl (1 + \frac{\mathrm{i}\theta }{\log r }\Bigr )^{\beta -1} \end{aligned}$$

(A.8)

and for large r and $\theta \in (0,\pi /2)$,

$$\begin{aligned} \frac{\mathrm {Im}\,\zeta q'(\zeta )}{\mathrm {Im}\,\zeta q'(r)} \sim \frac{\beta -1}{r \log r} \frac{\theta }{\sin \theta } \rightarrow 0 \end{aligned}$$

(A.9)

and so Assumption 2.3(ii) holds.

We now turn to the asymptotic behavior of the sequences $x_n^*$, $N_n^*$ and $N_n^{**}$. Since $q''(x_n) = -\frac{1}{n\sigma ^2}$, it is clear that $x_n \rightarrow \infty$ as $n \rightarrow \infty$. The equation is

$$\begin{aligned} \frac{1}{n\sigma ^2} \sim \frac{\beta (\log x_n^*)^{\beta -1}}{(x_n^*)^2}. \end{aligned}$$

(A.10)

Consequently,

$$\begin{aligned} (x_n^*)^2\sim & {} \beta n \sigma ^2 \bigl ( \tfrac{1}{2} \log (x_n^*)^2 \bigr )^{\beta -1}\nonumber \\\sim & {} \beta n \sigma ^2 \left( \tfrac{1}{2} \log n\right) ^{\beta -1} \left( 1 + \frac{\log \beta \sigma ^2 + (\beta -1) \log \log x_n^*}{\log n} \right) ^{\beta -1}. \end{aligned}$$

(A.11)

The last bracket is asymptotically equal to 1 and we get the expression for $x_n^*$. Next, we have from (2.10)

$$\begin{aligned} N_n^* = x_n^* + n \sigma ^2 \frac{\beta (\log x_n^*)^{\beta -1}}{x_n^*} \sim 2 x_n^*. \end{aligned}$$

(A.12)

The last asymptotics follows from (A.10).

We now turn to $N_n^{**}$. It is asymptotically given by the solution of the equations

$$\begin{aligned} \frac{N_n^2}{2n\sigma ^2}&= q(x_n) + \frac{(N_n-x_n)^2}{2n\sigma ^2}, \end{aligned}$$

(A.13)

$$\begin{aligned} q'(x_n)&= \frac{N_n-x_n}{n\sigma ^2}. \end{aligned}$$

(A.14)

Equation (A.14) is equivalent to

$$\begin{aligned} x_n^2 - N_n x_n + \beta n \sigma ^2 (\log x_n)^{\beta -1} = 0. \end{aligned}$$

(A.15)

The relevant solution is

$$\begin{aligned} x_n = \tfrac{1}{2} \Bigl ( N_n + \sqrt{ N_n^2 - 4 \beta n \sigma ^2 (\log x_n)^{\beta -1}} \Bigr ) = N_n \Bigl ( 1 - \frac{\beta n \sigma ^2}{N_n^2} (\log x_n)^{\beta -1} (1 + o(1)) \Bigr ). \end{aligned}$$

(A.16)

It follows that $N_n - x_n \sim \frac{\beta n \sigma ^2}{N_n} (\log x_n)^{\beta -1}$. We insert this in (A.13); using $\log x_n \sim \log N_n$, we get

$$\begin{aligned} \frac{N_n^4}{2n\sigma ^2} - N_n^2 (\log N_n)^\beta - \tfrac{1}{2} \beta ^2 n \sigma ^2 (\log N_n)^{2\beta -2} = o(1). \end{aligned}$$

(A.17)

The relevant solution is

$$\begin{aligned} \begin{aligned} N_n^2&\sim n \sigma ^2 \Bigl [ (\log N_n)^\beta + \sqrt{(\log N_n)^{2\beta } + \beta ^2 (\log N_n)^{2\beta -2}} \Bigr ] \\&\sim 2n \sigma ^2 (\tfrac{1}{2} \log N_n^2)^\beta \sim 2^{1-\beta } n \sigma ^2 \bigl ( \log n + \log 2\sigma ^2 + \beta \log \log N_n \bigr )^\beta . \end{aligned} \end{aligned}$$

(A.18)

Only the term $\log n$ matters in the last bracket and the result follows.

The last part of the lemma on insensitivity sequence is shown in [3, Section 8.3], the proof is therefore omitted. $\square$

Appendix B. Bivariate Hessian

As explained in Step 4 of the proof outline, the Hessian at $(t_n,\zeta _n)$ has determinant $-1 + o(1)$ and is a saddle point of $\Phi _n(t,\zeta )$, considered as a function of two real variables $t,\zeta >0$. In order to get rid of off-diagonal elements in the Hessian and to give all eigenvalues the same sign, we take complex $\zeta$ and reparametrize, as sketched in Step 5.

Lemma B.1

Let $\zeta (t)$ be the unique solution of $(\partial _\zeta \Phi _n)(t,\zeta ) =t - q'(\zeta ) = 0$. Set

$$\begin{aligned} F_n: (0,\infty ) \times \mathbb {R}\rightarrow \mathbb {C},\quad F_n(t,s) = \Phi _n (t, \zeta (t) + \mathrm{i}s). \end{aligned}$$

Then $(\nabla F_n)(t_n,0) =0$,

$$\begin{aligned} \mathrm{Hess}\, F_n(t,0) =\begin{pmatrix} \beta (t) &{} 0 \\ 0 &{} q''(\zeta (t)) \end{pmatrix}, \quad \beta (t) = - \frac{\det (\mathrm{Hess}\, \Phi _n)(t,\zeta (t))}{q''(\zeta (t))}. \end{aligned}$$

Note that $\zeta _n = \zeta (t_n)$, so as $n\rightarrow \infty$

$$\begin{aligned} \det \mathrm{Hess}\, F_n(t_n,0) = \beta (t_n) q''(\zeta _n)=- \det (\mathrm{Hess}\, \Phi _n)(t_n,\zeta _n) = 1+o(1). \end{aligned}$$

(B.1)

Proof

We have

$$\begin{aligned} \partial _t F_n(t,s)&= (\partial _t \Phi _n)(t,\zeta (t) + \mathrm{i}s) + (\partial _\zeta \Phi _n)(t,\zeta (t) + \mathrm{i}s) \zeta '(t),\nonumber \\ \partial _s F_n(t,s)&= \mathrm{i}\partial _\zeta \Phi _n(t,\zeta (t) + \mathrm{i}s). \end{aligned}$$

(B.2)

At $t=t_n$, $s=0$, we have $\zeta (t) = \zeta _n$ and $(\nabla F_n)(t_n,0) = \nabla \Phi _n(t_n,\zeta _n) =0$. For the Hessian, we compute

$$\begin{aligned} \partial _s^2 F_n(t,0)&= - \partial _\zeta ^2 \Phi _n(t, \zeta (t))= q''(\zeta (t)), \nonumber \\ \partial _t \partial _s F_n(t,0)&= \mathrm{i} (\partial _t \partial _\zeta \Phi _n)(t,\zeta (t) ) + \mathrm{i} \partial _\zeta ^2 \Phi _n(t,\zeta (t)) \nonumber \\ \partial _t^2 F_n(t,0)&= \frac{\mathrm {d}^2}{\mathrm {d}t^2} \Phi _n(t,\zeta (t)) = \beta (t). \end{aligned}$$

(B.3)

By definition of $\zeta (t)$,

$$\begin{aligned} 0= \frac{\mathrm {d}}{\mathrm {d}t} \partial _\zeta \Phi _n (t,\zeta (t)) = (\partial _t\partial _\zeta \Phi _n) (t,\zeta (t)) + \partial _\zeta ^2 \Phi _n(t,\zeta (t)) \zeta '(t). \end{aligned}$$

(B.4)

It follows that $\partial _t \partial _s F_n(t,0) =0$, and

$$\begin{aligned} \beta (t)&= \frac{\mathrm {d}^2}{\mathrm {d}t^2} \Phi _n(t,\zeta (t)) = \frac{\mathrm {d}}{\mathrm {d}t} (\partial _t \Phi _n)(t,\zeta (t)) \nonumber \\&= (\partial _t^2 \Phi _n)(t,\zeta (t)) + (\partial _t \partial _\zeta \Phi _n) (t,\zeta (t)) \zeta '(t). \end{aligned}$$

(B.5)

We solve for $\zeta '(t)$ in Eq. (B.4), insert into Eq. (B.5), and obtain the formula for $\beta (t)$. $\square$