A Sharp Abelian Theorem for the Laplace Transform

Abstract

This paper states asymptotic equivalents for the moments of the Esscher transform of a distribution on $\mathbb {R}$ with smooth density in the upper tail. As a by-product it provides a tail approximation for its moment generating function, and shows that the Esscher transforms have a Gaussian behavior for large values of the parameter.

Appendix: Proofs

The following lemma provides a simple argument for the local uniform convergence of regularly varying functions.

Lemma 1

Consider $l(t)\in RV(\alpha ), \alpha \in \mathbb {R}$. For any function $f$ such that $f(t)\underset{t\rightarrow \infty }{=}o(t)$, it holds

$$\begin{aligned} \sup _{\vert x\vert \le f(t)}\vert l(t+x)\vert \underset{t\rightarrow \infty }{\sim }\vert l(t)\vert . \end{aligned}$$

(17)

If $f(t)=at$ with $0<a<1$, then it holds

$$\begin{aligned} \sup _{\vert x\vert \le at}\vert l(t+x)\vert \underset{t\rightarrow \infty }{\sim }(1+a)^\alpha \vert l(t)\vert . \end{aligned}$$

(18)

Proof

By Theorem 1.5.2 of Bingham et al. (1987), if $l(t)\in RV(\alpha )$, then for all $I$

$$ \sup _{\lambda \in I}\left| \frac{l(\lambda t)}{l(t)}-\lambda ^\alpha \right| \underset{t\rightarrow \infty }{\longrightarrow }0, $$

with $I=[A, B]\ (0<A\le B<\infty )$ if $\alpha =0$, $I=(0, B]\ (0<B<\infty )$ if $\alpha >0$ and $I=[A, \infty ) \ (0<A<\infty )$ if $\alpha <0$.

Putting $\lambda =1+x/t$ with $f(t)\underset{t\rightarrow \infty }{=}o(t)$, we obtain

$$ \sup _{\vert x\vert \le f(t)}\left| \frac{l(t+x)}{l(t)}\right| -\left( 1+\frac{f(t)}{t}\right) ^\alpha \underset{t\rightarrow \infty }{\longrightarrow }0, $$

which implies (17).

When $f(t)=at$ with $0<a<1$, we get

$$ \sup _{\vert x\vert \le at}\left| \frac{l(t+x)}{l(t)}\right| -(1+a)^\alpha \underset{t\rightarrow \infty }{\longrightarrow }0, $$

which implies (18).   $\square$

Now we quote some simple expansions pertaining to the function $h$ under the two cases considered in the above Sect. 2.

Lemma 2

We have under Case 1,

$$\begin{aligned} h^{\prime }(x)&=\frac{h(x)}{x}[\beta +\varepsilon (x)],\\ h^{(2)}(x)&=\frac{h(x)}{x^{2}}[\beta (\beta -1)+a\varepsilon (x)+\varepsilon ^{2}(x)+x\varepsilon ^{\prime }(x)],\\ h^{(3)}(x)&=\frac{h(x)}{x^{3}}[\beta (\beta -1)(\beta -2)+b\varepsilon (x)+c\varepsilon ^{2}(x)+\varepsilon ^{3}(x)\\&\qquad \qquad \qquad +x\varepsilon ^{\prime }(x)(d+e\varepsilon (x))+x^{2}\varepsilon ^{(2)}(x)]. \end{aligned}$$

where $a,b,c,d,e$ are some real constants.

Corollary 1

We have under Case 1, $h^{\prime }(x)\underset{x\rightarrow \infty }{\sim }\beta h(x)/x$ and $|h^{(i)}(x)|\le C_{i}h(x)/x^{i},i=1,2,3$, for some constants $C_{i}$ and for large $x$.

Corollary 2

We have under Case 1, $\hat{x}(t)=\psi (t)\in RV(1/\beta )$ (see Theorem (1.5.15) of Bingham et al. (1987)) and $\hat{\sigma }^{2}(t)=\psi ^{\prime }(t)\sim \beta ^{-1}\psi (t)/t\in RV(1/\beta -1)$.

It also holds

Lemma 3

We have under Case 2,

$$ \psi ^{(2)}(t)\underset{t\rightarrow \infty }{\sim }-\frac{\psi (t)\varepsilon (t)}{t^{2}}\ and\ \psi ^{(3)}(t)\underset{t\rightarrow \infty }{\sim }2\frac{\psi (t)\varepsilon (t)}{t^{3}}. $$

Lemma 4

We have under Case 2,

$$\begin{aligned} h^{\prime }(\psi (t))&=\frac{1}{\psi ^{\prime }(t)}=\frac{t}{\psi (t)\varepsilon (t)},\\ h^{(2)}(\psi (t))&=-\frac{\psi ^{(2)}(t)}{(\psi ^{\prime }(t))^{3}}\underset{t\rightarrow \infty }{\sim }\frac{t}{\psi ^{2}(t)\varepsilon ^{2}(t)},\\ h^{(3)}(\psi (t))&=\frac{3(\psi ^{(2)}(t))^{2}-\psi ^{(3)}(t)\psi ^{\prime }(t)}{(\psi ^{\prime }(t))^{5}}\underset{t\rightarrow \infty }{\sim }\frac{t}{\psi ^{3}(t)\varepsilon ^{3}(t)}. \end{aligned}$$

Corollary 3

We have under Case 2, $\hat{x}(t)=\psi (t)\in RV(0)$ and $\hat{\sigma }^{2}(t)=\psi ^{\prime }(t)=\psi (t)\varepsilon (t)/t\in RV(-1)$. Moreover, we have $h^{(i)}(\psi (t))\in RV(1),i=1,2,3$.

Before beginning the proofs of our results we quote that the regularity conditions (9) and (12) pertaining to the function $h$ allow for the above simple expansions. Substituting the constant $c$ in (9) and (12) by functions $x\rightarrow c(x)$ which converge smoothly to some positive constant $c$ adds noticeable complexity.

We now come to the proofs of five lemmas which provide the asymptotics leading to Theorems 1 and 2.

Lemma 5

It holds

$$ \frac{\log \hat{\sigma }}{\int _{1}^{t}\psi (u)du}\underset{t\rightarrow \infty }{\longrightarrow }0. $$

Proof

By Corollaries 2 and 3, we have that $\psi (t)\in RV(1/\beta )$ in Case 1 and $\psi (t)\in RV(0)$ in Case 2. Using Theorem 1 of Feller (1971), Sect. 8.9 or Proposition 1.5.8 of Bingham et al. (1987), we obtain

$$\begin{aligned} \int _{1}^{t}\psi (u)du\underset{t\rightarrow \infty }{\sim } \left\{ \begin{array}{ll} t\psi (t)/(1+1/\beta )\in RV(1+1/\beta ) &{} \text{ if } h\in RV(\beta )\\ t\psi (t)\in RV(1) &{} \text{ if } h \text{ is } \text{ rapidly } \text{ varying } \end{array} \right. . \end{aligned}$$

(19)

Also by Corollaries 2 and 3, we have that $\hat{\sigma }^2\in RV(1/\beta -1)$ in Case 1 and $\hat{\sigma }^2\in RV(-1)$ in Case 2. Thus $t\mapsto \log \hat{\sigma }\in RV(0)$ by composition and

$$ \frac{\log \hat{\sigma }}{\int _{1}^{t}\psi (u)du}\underset{t\rightarrow \infty }{\sim } \left\{ \begin{array}{ll} \frac{\beta +1}{\beta }\times \frac{\log \hat{\sigma }}{t\psi (t)}\in RV\left( -1-\frac{1}{\beta }\right) &{} \text{ if } h\in RV(\beta )\\ \frac{\log \hat{\sigma }}{t\psi (t)}\in RV(-1) &{} \text{ if } h \text{ is } \text{ rapidly } \text{ varying } \end{array} \right. , $$

which proves the claim.    $\square$

The next steps of the proof make use of the function

$$ L(t):=\left( \log t\right) ^{3}. $$

Lemma 6

We have

$$ \sup _{|x|\le \hat{\sigma }L(t)}\left| \frac{h^{(3)}(\hat{x}+x)}{h^{(2)}(\hat{x})}\right| \hat{\sigma }L^{4}(t)\underset{t\rightarrow \infty }{\longrightarrow }0. $$

Proof

Case 1. By Corollary 1 and by (11) we have

$$ |h^{(3)}(x)|\le C\frac{|h^{(2)}(x)|}{x}, $$

for some constant $C$ and $x$ large. Since, by Corollary 2, $\hat{x}\in RV(1/\beta )$ and $\hat{\sigma }^{2}\in RV(1/\beta -1)$, we have

$$ \frac{|x|}{\hat{x}}\le \frac{\hat{\sigma }L(t)}{\hat{x}}\in RV\left( -\frac{1}{2}-\frac{1}{2\beta }\right) $$

and $|x|/\hat{x}\underset{t\rightarrow \infty }{\longrightarrow }0$ uniformly in $\{x:|x|\le \hat{\sigma }L(t)\}$. For large $t$ and all $x$ such that $|x|\le \hat{\sigma }L(t)$, we have

$$ |h^{(3)}(\hat{x}+x)|\le C\frac{|h^{(2)}(\hat{x}+x)|}{\hat{x}+x}\le C\sup _{|x|\le \hat{\sigma }L(t)}\frac{|h^{(2)}(\hat{x}+x)|}{\hat{x}+x} $$

whence

$$ \sup _{|x|\le \hat{\sigma }L(t)}|h^{(3)}(\hat{x}+x)|\le C\sup _{|x|\le \hat{\sigma }L(t)}\frac{|h^{(2)}(\hat{x}+x)|}{\hat{x}+x} $$

where

$$ \sup _{|x|\le \hat{\sigma }L(t)}\frac{|h^{(2)}(\hat{x}+x)|}{\hat{x} +x}\underset{t\rightarrow \infty }{\sim }\frac{|h^{(2)}(\hat{x})|}{\hat{x}}, $$

using (17) for the regularly varying function $|h^{(2)}(\hat{x})|\in RV(\theta /\beta )$, with $f(t)=\hat{\sigma }L(t)\underset{t\rightarrow \infty }{=}o(\hat{x})$. Thus for $t$ large enough and for all $\delta >0$

$$ \sup _{|x|\le \hat{\sigma }L^{4}(t)}\left| \frac{h^{(3)}(\hat{x}+x)}{h^{(2)}(\hat{x})}\right| \hat{\sigma }L^{4}(t)\le C\frac{\hat{\sigma }L^{4}(t)}{\hat{x}}(1+\delta )\in RV\left( \frac{1}{2\beta }-\frac{1}{2}-\frac{1}{\beta }\right) , $$

which proves Lemma 6 in Case 1.

Case 2. By Lemma 4, we have that $h^{(3)}(\psi (t))\in RV(1)$. Using (18), we have for $0<a<1$ and $t$ large enough

$$ \sup _{|v|\le at}|h^{(3)}(\psi (t+v))|\underset{t\rightarrow \infty }{\sim }(1+a)h^{(3)}(\psi (t)). $$

In the present case $\hat{x}\in RV(0)$ and $\hat{\sigma }^{2}\in RV(-1)$. Setting $\psi (t+v)=\hat{x}+x=\psi (t)+x$, we have $x=\psi (t+v)-\psi (t)$ and $A:=\psi (t-at)-\psi (t)\le x\le \psi (t+at)-\psi (t)=:B$, since $t\mapsto \psi (t)$ is an increasing function. It follows that

$$ \sup _{|v|\le at}h^{(3)}(\psi (t+v))=\sup _{A\le x\le B}h^{(3)}(\hat{x}+x). $$

Now note that (cf. p. 127 in Bingham et al. (1987))

$$\begin{aligned} B&=\psi (t+at)-\psi (t)=\int _{t}^{t+at}\psi ^{\prime }(z)dz \\&=\int _{t}^{t+at} \frac{\psi (z)\varepsilon (z)}{z}dz\underset{t\rightarrow \infty }{\sim } \psi (t)\varepsilon (t)\log (1+a), \end{aligned}$$

since $\psi (t)\varepsilon (t)\in RV(0)$. Moreover, we have

$$ \frac{\hat{\sigma }L(t)}{\psi (t)\varepsilon (t)}\in RV(-1) \text{ and } \frac{\hat{\sigma }L(t)}{\psi (t)\varepsilon (t)}\underset{t\rightarrow \infty }{\longrightarrow }0. $$

It follows that $\hat{\sigma }L(t)\underset{t\rightarrow \infty }{=}o(B)$ and in a similar way, we have $\hat{\sigma }L(t)\underset{t\rightarrow \infty }{=}o(A)$. Using Lemma 4 and since $\hat{\sigma }L^{4}(t)\in RV(-1/2)$, it follows that for $t$ large enough and for all $\delta >0$

$$\begin{aligned} \sup _{|x|\le \hat{\sigma }L(t)}\frac{|h^{(3)}(\psi (t+v))|}{|h^{(2)}(\psi (t))|}\hat{\sigma }L^{4}(t)&\le \sup _{A\le x\le B}\frac{|h^{(3)} (\psi (t+v))|}{|h^{(2)}(\psi (t))|}\hat{\sigma }L^{4}(t)\\&\le (1+a)\frac{\hat{\sigma }L^{4}(t)}{\psi (t)\varepsilon (t)}(1+\delta )\in RV\left( -\frac{1}{2}\right) , \end{aligned}$$

which concludes the proof of Lemma 6 in Case 2.    $\square$

Lemma 7

We have

$$\begin{aligned}&|h^{(2)}(\hat{x})|\hat{\sigma }^{4}\underset{t\rightarrow \infty }{\longrightarrow }0,\\&|h^{(2)}(\hat{x})|\hat{\sigma }^{3}L(t)\underset{t\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

Proof

Case 1. By Corollaries 1 and 2, we have

$$ |h^{(2)}(\hat{x})|\hat{\sigma }^{4}\le \frac{C_{2}}{\beta ^{2}t}\in RV(-1) $$

and

$$ |h^{(2)}(\hat{x})|\hat{\sigma }^{3}L(t)\le \frac{C_{2}}{\beta ^{3/2}}\frac{L(t)}{\sqrt{t\psi (t)}}\in RV\left( -\frac{1}{2\beta }-\frac{1}{2}\right) , $$

proving the claim.

Case 2. We have by Lemma 4 and Corollary 3

$$ h^{(2)}(\hat{x})\hat{\sigma }^{4}\underset{t\rightarrow \infty }{\sim }\frac{1}{t}\in RV(-1) $$

and

$$ h^{(2)}(\hat{x})\hat{\sigma }^{3}L(t)\underset{t\rightarrow \infty }{\sim } \frac{L(t)}{\sqrt{t\psi (t)\varepsilon (t)}}\in RV\left( -\frac{1}{2}\right) , $$

which concludes the proof of Lemma 7.    $\square$

We now define some functions to be used in the sequel. A Taylor–Lagrange expansion of $K(x,t)$ in a neighborhood of $\hat{x}$ yields

$$\begin{aligned} K(x,t)=K(\hat{x},t)-\frac{1}{2}h^{\prime }(\hat{x})(x-\hat{x})^{2}-\frac{1}{6}h^{(2)}(\hat{x})(x-\hat{x})^{3}+\varepsilon (x,t), \end{aligned}$$

(20)

where, for some $\theta \in (0,1)$,

$$\begin{aligned} \varepsilon (x,t)=-\frac{1}{24}h^{(3)}(\hat{x}+\theta (x-\hat{x}))(x-\hat{x})^{4}.\end{aligned}$$

(21)

Lemma 8

We have

$$ \sup _{y\in [-L(t),L(t)]}\frac{|\xi (\hat{\sigma }y+\hat{x},t)|}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\underset{t\rightarrow \infty }{\longrightarrow }0, $$

where $\xi (x,t)=\varepsilon (x,t)+q(x)$ and $\varepsilon (x,t)$ is defined in (21).

Proof

For $y\in [-L(t),L(t)]$, by 21, it holds

$$ \left| \frac{\varepsilon (\hat{\sigma }y+\hat{x},t)}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right| \le \left| \frac{h^{(3)}(\hat{x}+\theta \hat{\sigma }y)(\hat{\sigma }y)^{4}}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right| \le \left| \frac{h^{(3)}(\hat{x}+\theta \hat{\sigma }y)\hat{\sigma }L^{4}(t)}{h^{(2)}(\hat{x})}\right| , $$

with $\theta \in (0,1)$. Let $x=\theta \hat{\sigma }y$. It then holds $|x|\le \hat{\sigma }L(t)$. Therefore by Lemma 6

$$ \sup _{y\in [-L(t),L(t)]}\left| \frac{\varepsilon (\hat{\sigma }y+\hat{x},t)|}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right| \le \sup _{|x|\le \hat{\sigma }L(t)}\left| \frac{h^{(3)}(\hat{x}+x)}{h^{(2)}(\hat{x})}\hat{\sigma }L^{4}(t)\right| \underset{t\rightarrow \infty }{\longrightarrow }0. $$

It remains to prove that

$$\begin{aligned} \sup _{y\in [-L(t),L(t)]}\left| \frac{q(\hat{\sigma }y+\hat{x})}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right| \underset{t\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

(22)

Case 1. By (11) and by composition, $|h^{(2)}(\hat{x})|\in RV(\theta /\beta )$. Using Corollary 1 we obtain

$$ |h^{(2)}(\hat{x})\hat{\sigma }^{3}|\underset{t\rightarrow \infty }{\sim }\frac{|h^{(2)}(\hat{x})|\psi ^{3/2}(t)}{\beta ^{3/2}t^{3/2}}\in RV\left( \frac{\theta }{\beta }+\frac{3}{2\beta }-\frac{3}{2}\right) . $$

Since, by (14), $|q(\hat{x})|\in RV(\eta /\beta )$, for $\eta <\theta -3\beta /2+3/2$ and putting $x=\hat{\sigma }y$, we obtain

$$\begin{aligned} \sup _{y\in [-L(t),L(t)]}\left| \frac{q(\hat{\sigma }y+\hat{x} )}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right|&=\sup _{|x|\le \hat{\sigma }L(t)}\left| \frac{q(\hat{x}+x)}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right| \\&\underset{t\rightarrow \infty }{\sim }\frac{|q(\hat{x})|}{|h^{(2)}(\hat{x})\hat{\sigma }^{3}|}\in RV\left( \frac{\eta -\theta }{\beta }-\frac{3}{2\beta }+\frac{3}{2}\right) , \end{aligned}$$

which proves (22).

Case 2. By Lemma 4 and Corollary 3, we have

$$ |h^{(2)}(\hat{x})\hat{\sigma }^{3}|\underset{t\rightarrow \infty }{\sim }\frac{1}{\sqrt{t\psi (t)\varepsilon (t)}}\in RV\left( -\frac{1}{2}\right) . $$

As in Lemma 6, since by (15), $q(\psi (t))\in RV(\eta )$, then we obtain, with $\eta <-1/2$

$$\begin{aligned} \sup _{y\in [-L(t),L(t)]}\left| \frac{q(\hat{\sigma }y+\hat{x} )}{h^{(2)}(\hat{x})\hat{\sigma }^{3}}\right|&=\sup _{|x|\le \hat{\sigma }L(t)}\left| \frac{q(\hat{x}+x)}{h^{(2)}(\hat{x})\hat{\sigma }^{3} }\right| \\&\le \sup _{|v|\le at}\left| \frac{q(\psi (t+v))}{h^{(2)}(\hat{x} )\hat{\sigma }^{3}}\right| \\&\le (1+a)^{\eta }q(\psi (t))\sqrt{t\psi (t)\varepsilon (t)}(1+\delta )\in RV\left( \eta +\frac{1}{2}\right) , \end{aligned}$$

for all $\delta >0$, with $a<1$, $t$ large enough and $\eta +1/2<0$. This proves (22).    $\square$

Lemma 9

For $\alpha \in \mathbb {N}$, denote

$$ \varPsi (t,\alpha ):=\int _{0}^{\infty }(x-\hat{x})^{\alpha }e^{tx}p(x)dx. $$

Then

$$ \varPsi (t,\alpha )\underset{t\rightarrow \infty }{=}\hat{\sigma }^{\alpha +1}e^{K(\hat{x},t)}T_{1}(t,\alpha )(1+o(1)), $$

where

$$\begin{aligned} T_{1}(t,\alpha )=\int _{-\frac{L^{1/3}(t)}{\sqrt{2}}}^{\frac{L^{1/3}(t)}{\sqrt{2}}}y^{\alpha }\exp \left( -\frac{y^{2}}{2}\right) dy-\frac{h^{(2)}(\hat{x} )\hat{\sigma }^{3}}{6}\int _{-\frac{L^{1/3}(t)}{\sqrt{2}}}^{\frac{L^{1/3} (t)}{\sqrt{2}}}y^{3+\alpha }\exp \left( -\frac{y^{2}}{2}\right) dy. \end{aligned}$$

(23)

Proof

We define the interval $I_{t}$ as follows

$$ I_{t}:=\left( -\frac{L^{\frac{1}{3}}(t)\hat{\sigma }}{\sqrt{2}},\frac{L^{\frac{1}{3}}(t)\hat{\sigma }}{\sqrt{2}}\right) . $$

For large enough $\tau$, when $t\rightarrow \infty$ we can partition $\mathbb {R}_{+}$ into

$$ \mathbb {R}_{+}=\{x:0<x<\tau \}\cup \{x:x\in \hat{x}+I_{t}\}\cup \{x:x\ge \tau ,x\not \in \hat{x}+I_{t}\}, $$

where for $x>\tau$, $q(x)<\log 2$. Thus we have

$$\begin{aligned} p(x)<2e^{-g(x)}. \end{aligned}$$

(24)

For fixed $\tau$, $\{x:0<x<\tau \}\cap \{x:x\in \hat{x}+I_{t} \}=\emptyset$. Therefore $\tau <\hat{x}-\frac{L^{\frac{1}{3}}(t)\hat{\sigma }}{\sqrt{2}}\le \hat{x}$ for $t$ large enough. Hence it holds

$$\begin{aligned} \varPsi (t,\alpha )=:\varPsi _{1}(t,\alpha )+\varPsi _{2}(t,\alpha )+\varPsi _{3}(t,\alpha ), \end{aligned}$$

(25)

where

$$\begin{aligned} \varPsi _{1}(t,\alpha )&=\int _{0}^{\tau }(x-\hat{x})^{\alpha }e^{tx}p(x)dx,\\ \varPsi _{2}(t,\alpha )&=\int _{x\in \hat{x}+I_{t}}(x-\hat{x})^{\alpha } e^{tx}p(x)dx,\\ \varPsi _{3}(t,\alpha )&=\int _{x\not \in \hat{x}+I_{t},x\ge \tau }(x-\hat{x})^{\alpha }e^{tx}p(x)dx. \end{aligned}$$

We estimate $\varPsi _{1}(t,\alpha )$, $\varPsi _{2}(t,\alpha )$ and $\varPsi _{3} (t,\alpha )$ in Step 1, Step 2 and Step 3.

Step 1: Since $q$ is bounded, we consider

$$ \log d=\sup _{x\in (0, \tau )}q(x) $$

and for $t$ large enough, we have

$$ \left| \varPsi _{1}(t,\alpha )\right| \le \int _{0}^{\tau }\left| x-\hat{x}\right| ^{\alpha }e^{tx}p(x)dx\le d\int _{0}^{\tau }\hat{x}^{\alpha }e^{tx}dx, $$

since when $0<x<\tau <\hat{x}$ then $\left| x-\hat{x}\right| =\hat{x}-x<\hat{x}$ for $t$ large enough and $g$ is positive.Since, for $t$ large enough, we have

$$ \int _{0}^{\tau }\hat{x}^{\alpha }e^{tx}dx=\hat{x}^{\alpha }\frac{e^{t\tau }}{t}-\frac{\hat{x}^{\alpha }}{t}\le \hat{x}^{\alpha }\frac{e^{t\tau }}{t}, $$

we obtain

$$\begin{aligned} \left| \varPsi _{1}(t,\alpha )\right| \le d\hat{x}^{\alpha }\frac{e^{t\tau }}{t}. \end{aligned}$$

(26)

We now show that for $h\in \fancyscript{R}$, it holds

$$\begin{aligned} \hat{x}^{\alpha }\frac{e^{t\tau }}{t}\underset{t\rightarrow \infty }{=}o(\vert \hat{\sigma }^{\alpha +1}\vert e^{K(\hat{x},t)}\vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert ), \end{aligned}$$

(27)

with $K(x,t)$ defined as in (16). This is equivalent to

$$ \frac{\hat{x}^{\alpha }e^{t\tau }}{t\vert \hat{\sigma }^{\alpha +4}h^{(2)}(\hat{x})\vert }\underset{t\rightarrow \infty }{=}o(e^{K(\hat{x},t)}), $$

which is implied by

$$\begin{aligned} -(\alpha +4)\log \vert \hat{\sigma }\vert -\log t+\alpha \log \hat{x}+\tau t-\log \vert h^{(2)}(\hat{x})\vert \underset{t\rightarrow \infty }{=}o(K(\hat{x},t)), \end{aligned}$$

(28)

if $K(\hat{x},t)\underset{t\rightarrow \infty }{\longrightarrow }\infty$.

Setting $u=h(v)$ in $\int _{1}^{t}\psi (u)du$, we have

$$ \int _{1}^{t}\psi (u)du=t\psi (t)-\psi (1)-g(\psi (t))+g(\psi (1)). $$

Since $K(\hat{x},t)=t\psi (t)-g(\psi (t))$, we obtain

$$\begin{aligned} K(\hat{x},t)=\int _{1}^{t}\psi (u)du+\psi (1)-g(\psi (1))\underset{t\rightarrow \infty }{\sim }\int _{1}^{t}\psi (u)du. \end{aligned}$$

(29)

Let us denote (19) by

$$\begin{aligned} K(\hat{x},t)\underset{t\rightarrow \infty }{\sim }at\psi (t), \end{aligned}$$

(30)

with

$$ a=\left\{ \begin{array} [c]{ll} \frac{\beta }{\beta +1} &{} \text{ if } h\in RV(\beta )\\ 1 &{} \text{ if } h \text{ is } \text{ rapidly } \text{ varying } \end{array} \right. . $$

We have to show that each term in (28) is $o(K(\hat{x},t))$.

• 1. By Lemma 5, $\log \hat{\sigma }\underset{t\rightarrow \infty }{=}o(\int _{1}^{t}\psi (u)du)$. Hence $\log \hat{\sigma }\underset{t\rightarrow \infty }{=}o(K(\hat{x},t))$.

• 2. By Corollaries 2 and 3, we have

  $$ \frac{t}{K(\hat{x},t)}\underset{t\rightarrow \infty }{\sim }\frac{1}{a\psi (t)}\underset{t\rightarrow \infty }{\longrightarrow }0. $$

  Thus $t\underset{t\rightarrow \infty }{=}o(K(\hat{x},t))$.

• 3. Since $\hat{x}=\psi (t)\underset{t\rightarrow \infty }{\longrightarrow }\infty$, it holds

  $$ \left| \frac{\log \hat{x}}{K(\hat{x},t)}\right| \le C\frac{\psi (t)}{K(\hat{x},t)}, $$

  for some positive constant $C$ and $t$ large enough. Moreover by (30), we have

  $$ \frac{\psi (t)}{K(\hat{x},t)}\underset{t\rightarrow \infty }{\sim }\frac{1}{at}\underset{t\rightarrow \infty }{\longrightarrow }0. $$

  Hence $\log \hat{x}\underset{t\rightarrow \infty }{=}o(K(\hat{x},t))$.

• 4. Using (30), $\log \vert h^{(2)}(\hat{x})\vert \in RV(0)$ and $\log \vert h^{(2)}(\hat{x})\vert \underset{t\rightarrow \infty }{=}o(K(\hat{x},t))$.

• 5. Since $\log t\underset{t\rightarrow \infty }{=}o(t)$ and $t\underset{t\rightarrow \infty }{=}o(K(\hat{x},t))$, we obtain $\log t\underset{t\rightarrow \infty }{=}o(K(\hat{x},t))$.

Since (28) holds and $K(\hat{x},t)\underset{t\rightarrow \infty }{\longrightarrow }\infty$ by (29) and (30), we then get (27).

Eqs. (26) and (27) yield together

$$\begin{aligned} \left| \varPsi _{1}(t,\alpha )\right| \underset{t\rightarrow \infty }{=}o(\vert \hat{\sigma }^{\alpha +1}\vert e^{K(\hat{x},t)}\vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert ). \end{aligned}$$

(31)

When $\alpha$ is even,

$$\begin{aligned} T_{1}(t,\alpha )=\int _{-\frac{t^{1/3}}{\sqrt{2}}}^{\frac{t^{1/3}}{\sqrt{2}}}y^{\alpha }\exp \left( -\frac{y^{2}}{2}\right) dy\underset{t\rightarrow \infty }{\sim }\sqrt{2\pi }M_{\alpha }, \end{aligned}$$

(32)

where $M_{\alpha }$ is the moment of order $\alpha$ of a standard normal distribution. Thus by Lemma 7 we have

$$\begin{aligned} \frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{T_{1}(t,\alpha )}\underset{t\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

(33)

When $\alpha$ is odd,

$$\begin{aligned} T_{1}(t,\alpha )=-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\int _{-\frac{l^{\frac{1}{3}}}{\sqrt{2}}}^{\frac{l^{\frac{1}{3}}}{\sqrt{2}}}y^{3+\alpha }\exp \left( -\frac{y^{2}}{2}\right) dy\underset{t\rightarrow \infty }{\sim }-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\sqrt{2\pi }M_{\alpha +3}, \end{aligned}$$

(34)

where $M_{\alpha +3}$ is the moment of order $\alpha +3$ of a standard normal distribution. Thus we have

$$\begin{aligned} \frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{T_{1}(t,\alpha )}\underset{t\rightarrow \infty }{\sim }-\frac{6}{\sqrt{2\pi }M_{\alpha +3}}. \end{aligned}$$

(35)

Combined with (31), (33) and (35) imply for $\alpha \in \mathbb {N}$

$$\begin{aligned} \left| \varPsi _{1}(t,\alpha )\right| \underset{t\rightarrow \infty }{=}o(\hat{\sigma }^{\alpha +1}e^{K(\hat{x},t)}T_{1}(t,\alpha )). \end{aligned}$$

(36)

Step 2: By (6) and (20)

$$\begin{aligned} \varPsi _{2}(t,\alpha )&=\int _{x\in \hat{x}+I_{t}}(x-\hat{x})^{\alpha }e^{K(x,t)+q(x)}dx\\&=\int _{x\in \hat{x}+I_{t}}(x-\hat{x})^{\alpha }e^{K(\hat{x},t)-\frac{1}{2}h^{\prime }(\hat{x})(x-\hat{x})^{2}-\frac{1}{6}h^{(2)}(\hat{x})(x-\hat{x})^{3}+\xi (x,t)}dx, \end{aligned}$$

where $\xi (x,t)=\varepsilon (x,t)+q(x)$. Making the substitution $y=(x-\hat{x})/\hat{\sigma }$, it holds

$$\begin{aligned} \varPsi _{2}(t,\alpha )=\hat{\sigma }^{\alpha +1}e^{K(\hat{x},t)} \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{\alpha }\exp \left( -\frac{y^{2}}{2}-\frac{\hat{\sigma }^{3}y^{3}}{6}h^{(2)}(\hat{x})+\xi (\hat{\sigma }y+\hat{x},t)\right) dy, \end{aligned}$$

(37)

since $h^{\prime }(\hat{x})=1/\hat{\sigma }^{2}$.

On $\left\{ y: y\in \left( -L^{\frac{1}{3}}(t)/\sqrt{2},L^{\frac{1}{3}}(t)/\sqrt{2}\right) \right\}$, by Lemma 7, we have

$$ \left| h^{(2)}(\hat{x})\hat{\sigma }^{3}y^{3}\right| \le \left| h^{(2)}(\hat{x})\hat{\sigma }^{3}L(t)\right| /2^{\frac{3}{2}} \underset{t\rightarrow \infty }{\longrightarrow }0. $$

Perform the first-order Taylor expansion

$$\begin{aligned} \exp \left( -\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}y^{3}+\xi (\hat{\sigma }y+\hat{x},t)\right) \underset{t\rightarrow \infty }{=} 1-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}y^{3}+\xi (\hat{\sigma }y+\hat{x},t)+o_{1}(t,y), \end{aligned}$$

where

$$\begin{aligned} o_{1}(t,y)=o\left( -\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}y^{3}+\xi (\hat{\sigma }y+\hat{x},t)\right) . \end{aligned}$$

(38)

We obtain

$$ \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{\alpha }\exp \left( -\frac{y^{2}}{2}-\frac{\hat{\sigma }^{3}y^{3}}{6}h^{(2)}(\hat{x})+\xi (\hat{\sigma }y+\hat{x},t)\right) dy =: T_{1}(t,\alpha )+T_{2}(t,\alpha ), $$

where $T_{1}(t,\alpha )$ is defined in (23) and

$$\begin{aligned} T_{2}(t,\alpha ):=\int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left( \xi (\hat{\sigma }y+\hat{x} ,t)+o_{1}(t,y)\right) y^{\alpha }e^{-\frac{y^{2}}{2}}dy. \end{aligned}$$

(39)

Using (38) we have for $t$ large enough

$$\begin{aligned} \left| T_{2}(t,\alpha )\right|&\le \sup _{y\in [-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}},\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}]}\left| \xi (\hat{\sigma }y+\hat{x},t)\right| \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left| y\right| ^{\alpha }e^{-\frac{y^{2}}{2}}dy\\&\qquad +\int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left( \left| o\left( \frac{h^{(2)} (\hat{x})\hat{\sigma }^{3}}{6}y^{3}\right) \right| +\left| o(\xi (\hat{\sigma }y+\hat{x},t))\right| \right) |y|^{\alpha }e^{-\frac{y^{2}}{2}}dy, \end{aligned}$$

where $\sup _{y\in [-L^{\frac{1}{3}}(t)/\sqrt{2},L^{\frac{1}{3}}(t)/\sqrt{2}]}\left| \xi (\hat{\sigma }y+\hat{x},t)\right| \le \sup _{y\in [-L(t),L(t)]}\left| \xi (\hat{\sigma }y+\hat{x},t)\right|$ since $L^{\frac{1}{3}}(t)/\sqrt{2}\le L(t)$ holds for $t$ large enough. Thus

$$\begin{aligned} \left| T_{2}(t,\alpha )\right|&\le 2\sup _{y\in [-L(t),L(t)]}\left| \xi (\hat{\sigma }y+\hat{x},t)\right| \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left| y\right| ^{\alpha }e^{-\frac{y^{2}}{2}}dy\\&\qquad \qquad +\left| o\left( \frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\right) \right| \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}} ^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left| y\right| ^{3+\alpha }e^{-\frac{y^{2}}{2}}dy\\&\underset{t\rightarrow \infty }{=}\left| o\left( \frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\right) \right| \left( \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left| y\right| ^{\alpha }e^{-\frac{y^{2}}{2}}dy+\int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\left| y\right| ^{3+\alpha }e^{-\frac{y^{2}}{2}}dy\right) , \end{aligned}$$

where the last equality holds from Lemma 8. Since the integrals in the last equality are both bounded, it holds

$$\begin{aligned} T_{2}(t,\alpha )\underset{t\rightarrow \infty }{=}o(h^{(2)}(\hat{x})\hat{\sigma }^{3}). \end{aligned}$$

(40)

When $\alpha$ is even, using (32) and Lemma 7

$$\begin{aligned} \left| \frac{T_{2}(t,\alpha )}{T_{1}(t,\alpha )}\right| \le \frac{\vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert }{\sqrt{2\pi }M_{\alpha }} \underset{t\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

(41)

When $\alpha$ is odd, using (34), we get

$$\begin{aligned} \frac{T_{2}(t,\alpha )}{T_{1}(t,\alpha )}\underset{t\rightarrow \infty }{=} -\frac{6}{\sqrt{2\pi }M_{\alpha +3}}o(1)\underset{t\rightarrow \infty }{\longrightarrow }0. \end{aligned}$$

(42)

Now with $\alpha \in \mathbb {N}$, by (41) and (42)

$$ T_{2}(t,\alpha )\underset{t\rightarrow \infty }{=}o(T_{1}(t,\alpha )), $$

which, combined with (37), yields

$$\begin{aligned} \varPsi _{2}(t,\alpha )=c\hat{\sigma }^{\alpha +1}e^{K(\hat{x},t)}T_{1}(t,\alpha )(1+o(1)). \end{aligned}$$

(43)

Step 3: The Three Chords Lemma implies, for $x\mapsto K(x,t)$ concave and $(x,y,z)\in \mathbb {R}_{+}^{3}$ such that $x<y<z$

$$\begin{aligned} \frac{K(y,t)-K(z,t)}{y-z}\le \frac{K(x,t)-K(z,t)}{x-z}\le \frac{K(x,t)-K(y,t)}{x-y}. \end{aligned}$$

(44)

Since $x\mapsto K(x,t)$ is concave and attains its maximum in $\hat{x}$, we consider two cases: $x<\hat{x}$ and $x\ge \hat{x}$. After some calculus using (44) in each case, we get

$$\begin{aligned} K(x,t)-K(\hat{x},t)\le \frac{K(\hat{x}+sgn(x-\hat{x})\frac{L^{1/3} (t)\hat{\sigma }}{\sqrt{2}})-K(\hat{x},t)}{sgn(x-\hat{x})\frac{L^{1/3} (t)\hat{\sigma }}{\sqrt{2}}}(x-\hat{x}), \end{aligned}$$

(45)

where

$$\begin{aligned} sgn(x-\hat{x})= \left\{ \begin{array}{ll} 1 &{} \text{ if } x\ge \hat{x}\\ -1 &{} \text{ if } x<\hat{x} \end{array} \right. . \end{aligned}$$

Using Lemma 7, a third-order Taylor expansion in the numerator of (45) gives

$$ K(\hat{x}+sgn(x-\hat{x})\frac{L^{1/3}(t)\hat{\sigma }}{\sqrt{2}})-K(\hat{x},t)\le -\frac{1}{4}h^{\prime }(\hat{x})L^{2/3}(t)\hat{\sigma }^{2}=-\frac{1}{4}L^{2/3}(t), $$

which yields

$$ K(x,t)-K(\hat{x},t)\le -\frac{\sqrt{2}}{4}\frac{L^{1/3}(t)}{\hat{\sigma }}\vert x-\hat{x}\vert . $$

Using (24), we obtain for large enough fixed $\tau$

$$\begin{aligned} \vert \varPsi _{3}(t,\alpha )\vert&\le 2\int _{x\not \in \hat{x}+I_{t},x>\tau }\vert x-\hat{x}\vert ^{\alpha }e^{K(x,t)}dx\\&\le 2e^{K(\hat{x},t)}\int _{\vert x-\hat{x}\vert >\frac{L^{1/3} (t)\hat{\sigma }}{\sqrt{2}},x>\tau }\vert x-\hat{x}\vert ^{\alpha }\exp \left( -\frac{\sqrt{2}}{4}\frac{L^{1/3}(t)}{\hat{\sigma }}\vert x-\hat{x}\vert \right) dx\\&=2e^{K(\hat{x},t)}\hat{\sigma }^{\alpha +1}\left[ \int _{\frac{L^{1/3} (t)}{\sqrt{2}}}^{+\infty }y^{\alpha }e^{-\frac{\sqrt{2}}{4}L^{1/3}(t)y} dy \right. \\&\qquad \qquad \qquad \qquad \qquad \left. +\int _{\frac{\tau -\hat{x}}{\hat{\sigma }}}^{-\frac{L^{1/3}(t)}{\sqrt{2}} }(-y)^{\alpha }e^{\frac{\sqrt{2}}{4}L^{1/3}(t)y}dy\right] \\&:= 2e^{K(\hat{x},t)}\hat{\sigma }^{\alpha +1}(I_{\alpha }+J_{\alpha }). \end{aligned}$$

It is easy but a bit tedious to show by recursion that

$$\begin{aligned} I_{\alpha }&=\int _{\frac{L^{1/3}(t)}{\sqrt{2}}}^{+\infty }y^{\alpha } \exp \left( -\frac{\sqrt{2}}{4}L^{1/3}(t)y\right) dy\\&=\exp \left( -\frac{1}{4}L^{2/3}(t)\right) \sum _{i=0}^{\alpha }2^{\frac{4i+3-\alpha }{2}}L^{\frac{\alpha -(2i+1)}{3}}(t)\frac{\alpha !}{(\alpha -i)!}\\&\underset{t\rightarrow \infty }{\sim }2^{\frac{3-\alpha }{2}}\exp \left( -\frac{1}{4}L^{2/3}(t)\right) L^{\frac{\alpha -1}{3}}(t) \end{aligned}$$

and

$$\begin{aligned} J_{\alpha }&=\int _{\frac{\tau -\hat{x}}{\hat{\sigma }}}^{-\frac{L^{1/3} (t)}{\sqrt{2}}}(-y)^{\alpha }\exp \left( \frac{\sqrt{2}}{4}L^{1/3}(t)y\right) dy\\&=I_{\alpha }-\exp \left( \frac{\sqrt{2}}{4}L^{1/3}(t)\frac{\tau -\hat{x} }{\hat{\sigma }}\right) \sum _{i=0}^{\alpha }\left( \frac{\hat{x}-\tau }{\hat{\sigma }}\right) ^{\alpha -i}L^{-\frac{i+1}{3}}(t)2^{\frac{3i+3}{2}} \frac{\alpha !}{(\alpha -i)!}\\&=I_{\alpha }+M(t), \end{aligned}$$

with $\hat{x}/\hat{\sigma }\in RV((1+1/\beta )/2)$ when $h\in RV(\beta )$ and $\hat{x}/\hat{\sigma }\in RV(1/2)$ when $h$ is rapidly varying. Moreover, $\tau -\hat{x}<0$, thus $M(t)\underset{t\rightarrow \infty }{\longrightarrow }0$ and we have for some positive constant $Q_1$

$$ \vert \varPsi _{3}(t,\alpha )\vert \le Q_1e^{K(\hat{x},t)}\hat{\sigma }^{\alpha +1} \exp \left( -\frac{1}{4}L^{2/3}(t)\right) L^{\frac{\alpha -1}{3}}(t). $$

With (43), we obtain for some positive constant $Q_2$

$$ \left| \frac{\varPsi _{3}(t,\alpha )}{\varPsi _{2}(t,\alpha )}\right| \le \frac{Q_2\exp (-\frac{1}{4}L^{2/3}(t))L^{\frac{\alpha -1}{3}}(t)}{\vert T_{1}(t,\alpha )\vert }. $$

In Step 1, we saw that $T_{1}(t,\alpha )\underset{t\rightarrow \infty }{\sim }\sqrt{2\pi }M_{\alpha }$, for $\alpha$ even and $T_{1}(t,\alpha )\underset{t\rightarrow \infty }{\sim }-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\sqrt{2\pi }M_{\alpha +3}$, for $\alpha$ odd. Hence for $\alpha$ even and $t$ large enough

$$\begin{aligned} \left| \frac{\varPsi _{3}(t,\alpha )}{\varPsi _{2}(t,\alpha )}\right| \le Q_3\frac{\exp (-\frac{1}{4}L^{2/3}(t))L^{\frac{\alpha -1}{3}}(t)}{\sqrt{2\pi }M_{\alpha }}\underset{t\rightarrow \infty }{\longrightarrow }0, \end{aligned}$$

(46)

and for $\alpha$ odd and $t$ large enough

$$ \left| \frac{\varPsi _{3}(t,\alpha )}{\varPsi _{2}(t,\alpha )}\right| \le Q_4\frac{\exp (-\frac{1}{4}L^{2/3}(t))L^{\frac{\alpha -1}{3}}(t)}{\frac{\vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert }{6}\sqrt{2\pi }M_{\alpha +3}}, $$

for positive constants $Q_3$ and $Q_4$.

As in Lemma 7, we have

$$ \vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert \in RV\left( \frac{\theta }{\beta }+\frac{3}{2\beta }-\frac{3}{2}\right) \text{ if } h\in RV(\beta ) $$

and

$$ \vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert \in RV\left( -\frac{1}{2}\right) \text{ if } h \text{ is } \text{ rapidly } \text{ varying }. $$

Let us denote

$$ \vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert =t^{\rho }L_{1}(t), $$

for some slowly varying function $L_{1}$ and $\rho <0$ defined as

$$ \rho = \left\{ \begin{array}{ll} \frac{\theta }{\beta }+\frac{3}{2\beta }-\frac{3}{2} &{} \text{ if } h\in RV(\beta )\\ -\frac{1}{2} &{} \text{ if } h \text{ is } \text{ rapidly } \text{ varying } \end{array} \right. . $$

We have for some positive constant $C$

$$ \left| \frac{\varPsi _{3}(t,\alpha )}{\varPsi _{2}(t,\alpha )}\right| \le C\exp \left( -\frac{1}{4}L^{2/3}(t)-\rho \log t-\log L_{1}(t)\right) L^{\frac{\alpha -1}{3}}(t)\underset{t\rightarrow \infty }{\longrightarrow }0, $$

since $-(\log t)^{2}/4-\rho \log t-\log L_{1}(t)\underset{t\rightarrow \infty }{\sim }-(\log t)^{2}/4\underset{t\rightarrow \infty }{\longrightarrow }-\infty$.

Hence we obtain

$$\begin{aligned} \varPsi _{3}(t,\alpha )\underset{t\rightarrow \infty }{=}o(\varPsi _{2}(t,\alpha )). \end{aligned}$$

(47)

The proof is completed by combining (25), (36), (43), and (47).    $\square$

Proof

(Proof of Theorem 1) By Lemma 9, if $\alpha =0$, it holds

$$ T_{1}(t,0)\underset{t\rightarrow \infty }{\longrightarrow }\sqrt{2\pi }, $$

since $L(t)\underset{t\rightarrow \infty }{\longrightarrow }\infty$. Approximate the moment generating function of $X$

$$\begin{aligned} \varPhi (t) =\varPsi (t,0)\underset{t\rightarrow \infty }{=}\hat{\sigma }e^{K(\hat{x},t)}T_{1}(t,0)(1+o(1))\underset{t\rightarrow \infty }{=}\sqrt{2\pi }\hat{\sigma } e^{K(\hat{x},t)}(1+o(1)). \end{aligned}$$

(48)

If $\alpha =1$, it holds

$$ T_{1}(t,1)\underset{t\rightarrow \infty }{=}-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}M_{4}\sqrt{2\pi }(1+o(1)), $$

where $M_{4}=3$ denotes the fourth-order moment of the standard normal distribution. Consequently, we obtain

$$\begin{aligned} \varPsi (t,1)\underset{t\rightarrow \infty }{=}-\sqrt{2\pi }\hat{\sigma } ^{2}e^{K(\hat{x},t)}\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{2} (1+o(1))\underset{t\rightarrow \infty }{=}-\varPhi (t)\frac{h^{(2)}(\hat{x} )\hat{\sigma }^{4}}{2}(1+o(1)), \end{aligned}$$

(49)

which, together with the definition of $\varPsi (t,\alpha )$, yields

$$ \int _{0}^{\infty }xe^{tx}p(x)dx=\varPsi (t,1)+\hat{x}\varPhi (t)\underset{t\rightarrow \infty }{=}\left( \hat{x}-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{4}}{2}(1+o(1))\right) \varPhi (t). $$

Hence we get

$$\begin{aligned} m(t)=\frac{\int _{0}^{\infty }xe^{tx}p(x)dx}{\varPhi (t)}=\hat{x}-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{4}}{2}(1+o(1)). \end{aligned}$$

(50)

By Lemma 7, we obtain

$$\begin{aligned} m(t)\underset{t\rightarrow \infty }{\sim }\hat{x}=\psi (t). \end{aligned}$$

(51)

If $\alpha =2$, it follows:

$$ T_{1}(t,2)\underset{t\rightarrow \infty }{=}\sqrt{2\pi }(1+o(1)). $$

Thus we have

$$\begin{aligned} \varPsi (t,2)\underset{t\rightarrow \infty }{=}\hat{\sigma }^{2}\varPhi (t)(1+o(1)). \end{aligned}$$

(52)

Using (49), (50) and (52), it follows:

$$\begin{aligned}&\int _{0}^{\infty }(x-m(t))^{2}e^{tx}p(x)dx=\int _{0}^{\infty }(x-\hat{x} +\hat{x}-m(t))^{2}e^{tx}p(x)dx\\&=\varPsi (t,2)+2(\hat{x}-m(t))\varPsi (t,1)+(\hat{x}-m(t))^{2}\varPhi (t)\\&\underset{t\rightarrow \infty }{=}\hat{\sigma }^{2}\varPhi (t)(1+o(1))-\hat{\sigma }^{2}\varPhi (t)\frac{(h^{(2)}(\hat{x})\hat{\sigma }^{3})^{2}}{4}(1+o(1))\underset{t\rightarrow \infty }{=}\hat{\sigma }^{2}\varPhi (t)(1+o(1)), \end{aligned}$$

where the last equality holds since $\vert h^{(2)}(\hat{x})\hat{\sigma }^{3}\vert \underset{t\rightarrow \infty }{\longrightarrow }0$ by Lemma 7.

Hence we obtain

$$\begin{aligned} s^{2}(t)=\frac{\int _{0}^{\infty }(x-m(t))^{2}e^{tx}p(x)dx}{\varPhi (t)}\underset{t\rightarrow \infty }{\sim }\hat{\sigma }^{2}=\psi ^{\prime }(t). \end{aligned}$$

(53)

If $\alpha =3$, it holds

$$ T_{1}(t,3)=-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{6}e^{-\frac{y^{2}}{2}}dy. $$

Thus we have

$$\begin{aligned} \varPsi (t,3)&=-\sqrt{2\pi }\hat{\sigma }^{4}e^{K(\hat{x},t)}\frac{h^{(2)} (\hat{x})\hat{\sigma }^{3}}{6}\int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}} }^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}\frac{1}{\sqrt{2\pi }}y^{6} e^{-\frac{y^{2}}{2}}dy\\&\underset{t\rightarrow \infty }{=}-M_{6}\frac{h^{(2)}(\hat{x})\hat{\sigma }^{6}}{6}\varPhi (t)(1+o(1)),\nonumber \end{aligned}$$

(54)

where $M_{6}=15$ denotes the sixth-order moment of standard normal distribution. Using (49), (50), (52) and (54), we have

$$\begin{aligned}&\int _{0}^{\infty }(x-m(t))^{3}e^{tx}p(x)dx=\int _{0}^{\infty }(x-\hat{x}+\hat{x}-m(t))^{3}e^{tx}p(x)dx\\&=\varPsi (t,3)+3(\hat{x}-m(t))\varPsi (t,2)+3(\hat{x}-m(t))^{2}\varPsi (t,1)+(\hat{x}-m(t))^{3}\varPhi (t)\\&\underset{t\rightarrow \infty }{=}-h^{(2)}(\hat{x})\hat{\sigma }^{6}\varPhi (t)(1+o(1))-h^{(2)}(\hat{x})\hat{\sigma }^{6}\varPhi (t)\frac{(h^{(2)}(\hat{x})\hat{\sigma }^{3})^{2}}{4}(1+o(1))\\&\underset{t\rightarrow \infty }{=}-h^{(2)}(\hat{x})\hat{\sigma }^{6}\varPhi (t)(1+o(1)), \end{aligned}$$

where the last equality holds since $|h^{(2)}(\hat{x})\hat{\sigma }^{3}|\underset{t\rightarrow \infty }{\longrightarrow }0$ by Lemma 7. Hence we get

$$\begin{aligned} \mu _{3}(t)&=\frac{\int _{0}^{\infty }(x-m(t))^{3}e^{tx}p(x)dx}{\varPhi (t)}\underset{t\rightarrow \infty }{\sim }-h^{(2)}(\hat{x})\hat{\sigma }^{6} \nonumber \\&=\frac{\psi ^{(2)}(t)}{(\psi ^{\prime }(t))^{3}}(\psi ^{\prime }(t))^{3}=\psi ^{(2)}(t). \end{aligned}$$

(55)

We now consider $\alpha =j>3$ for even $j$. Using (50) and Lemma 9, we have

$$\begin{aligned}&\int _{0}^{\infty }(x-m(t))^{j}e^{tx}p(x)dx=\int _{0}^{\infty }(x-\hat{x}+\hat{x}-m(t))^{j}e^{tx}p(x)dx\\&=\sum _{i=0}^{j}\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( \frac{h^{(2)}(\hat{x})\hat{\sigma }^{4}}{2}\right) ^{i}\hat{\sigma }^{j-i+1}e^{K(\hat{x},t)}T_{1}(t,j-i)(1+o(1)),\nonumber \end{aligned}$$

(56)

with

$$\begin{aligned} T_{1}(t,j-i)&=\left\{ \begin{array}{ll} \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{j-i}e^{-\frac{y^{2}}{2}}dy &{} \text{ for } \text{ even } i\\ -\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\int _{-\frac{L^{\frac{1}{3}} (t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{3+j-i}e^{-\frac{y^{2}}{2}}dy &{} \text{ for } \text{ odd } i \end{array} \right. \\&\underset{t\rightarrow \infty }{=}\left\{ \begin{array}{ll} \sqrt{2\pi }M_{j-i}(1+o(1)) &{} \text{ if } i \text{ is } \text{ even }\\ -\sqrt{2\pi }\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}M_{3+j-i} &{}\text{ if } i \text{ is } \text{ odd } \end{array} \right. . \end{aligned}$$

Using (48), we obtain

$$\begin{aligned}&\int _{0}^{\infty }(x-m(t))^{j}e^{tx}p(x)dx\\&\underset{t\rightarrow \infty }{=}\sum _{i=0}^{j}\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( \frac{h^{(2)}(\hat{x})\hat{\sigma }^{4}}{2}\right) ^{i}\varPhi (t)\times \\&\quad \left[ \hat{\sigma }^{j-i}M_{j-i}(1+o(1))\mathbb {I} _{even~i}-\frac{h^{(2)}(\hat{x})\hat{\sigma }^{4}}{2}\sigma ^{j-i-1} \frac{M_{3+j-i}}{3}(1+o(1))\mathbb {I}_{odd~i}\right] \\&\underset{t\rightarrow \infty }{=}\sum _{k=0}^{j/2}\left( {\begin{array}{c}j\\ 2k\end{array}}\right) \left( \frac{h^{(2)}(\hat{x})\hat{\sigma }^{4}}{2}\right) ^{2k}\varPhi (t)\hat{\sigma }^{j-2k}M_{j-2k}(1+o(1))\\&\quad -\sum _{k=0}^{j/2-1}\left( {\begin{array}{c}j\\ 2k+1\end{array}}\right) \left( \frac{h^{(2)} (\hat{x})\hat{\sigma }^{4}}{2}\right) ^{2(k+1)}\varPhi (t)\hat{\sigma } ^{j-2k-2}\frac{M_{3+j-2k-1}}{3}(1+o(1))\\&\underset{t\rightarrow \infty }{\sim }\hat{\sigma }^{j}\varPhi (t)\times \\&\left( M_{j}+\sum _{k=1}^{j/2}\left( {\begin{array}{c}j\\ 2k\end{array}}\right) (h^{(2)}(\hat{x})\hat{\sigma }^{3})^{2k}\frac{M_{j-2k}}{2^{2k}} \right. \\&\quad \left. -\sum _{k=0}^{j/2-1}\left( {\begin{array}{c}j\\ 2k+1\end{array}}\right) (h^{(2)}(\hat{x})\hat{\sigma }^{3})^{2(k+1)}\frac{M_{3+j-2k-1}}{3\times 2^{2(k+1)}}\right) \\&\underset{t\rightarrow \infty }{=}M_{j}\hat{\sigma }^{j}\varPhi (t)(1+o(1)), \end{aligned}$$

since $|h^{(2)}(\hat{x})\hat{\sigma }^{3}|\underset{t\rightarrow \infty }{\longrightarrow }0$ by Lemma 7. Hence we get for even $j$

$$\begin{aligned} \mu _{j}(t) =\frac{\int _{0}^{\infty }(x-m(t))^{j}e^{tx}p(x)dx}{\varPhi (t)}\underset{t\rightarrow \infty }{\sim }M_{j}\hat{\sigma }^{j}\underset{t\rightarrow \infty }{\sim }M_{j}s^{j}(t), \end{aligned}$$

(57)

by (53).

To conclude, we consider $\alpha =j>3$ for odd $j$. (56) holds true with

$$\begin{aligned} T_{1}(t,j-i)&=\left\{ \begin{array}{ll} \int _{-\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{j-i}e^{-\frac{y^{2}}{2}}dy &{} \text{ for } \text{ odd } i\\ -\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}\int _{-\frac{L^{\frac{1}{3}} (t)}{\sqrt{2}}}^{\frac{L^{\frac{1}{3}}(t)}{\sqrt{2}}}y^{3+j-i}e^{-\frac{y^{2}}{2}}dy &{} \text{ for } \text{ even } i \end{array} \right. \\&\underset{t\rightarrow \infty }{=}\left\{ \begin{array}{ll} \sqrt{2\pi }M_{j-i}(1+o(1)) &{} \text{ if } i \text{ is } \text{ odd }\\ -\sqrt{2\pi }\frac{h^{(2)}(\hat{x})\hat{\sigma }^{3}}{6}M_{3+j-i} &{}\text{ if } i \text{ is } \text{ even } \end{array} \right. . \end{aligned}$$

Thus, with the same tools as above, some calculus and making use of (57),

$$ \int _{0}^{\infty }(x-m(t))^{j}e^{tx}p(x)dx\underset{t\rightarrow \infty }{=} \frac{M_{j+3}-3jM_{j-1}}{6}\times (-h^{(2)}(\hat{x})\hat{\sigma }^{j+3})\varPhi (t). $$

Hence we get for odd $j$

$$\begin{aligned} \mu _{j}(t)&=\frac{\int _{0}^{\infty }(x-m(t))^{j}e^{tx}p(x)dx}{\varPhi (t)}\underset{t\rightarrow \infty }{\sim }\frac{M_{j+3}-3jM_{j-1}}{6}\times (-h^{(2)}(\hat{x})\hat{\sigma }^{j+3})\\&\underset{t\rightarrow \infty }{\sim }\frac{M_{j+3}-3jM_{j-1}}{6}\mu _{3}(t)s^{j-3}(t),\nonumber \end{aligned}$$

(58)

by (53) and (55).

The proof is complete by considering (51), (53), (55), (57) and (58).    $\square$

Proof

(Proof of Theorem 2) It is proved incidentally in (48).    $\square$

Proof

(Proof of Theorem 3) Consider the moment generating function of the random variable

$$ Y_{t}:=\frac{\fancyscript{X}_{t}-m(t)}{s(t)}. $$

It holds for any $\lambda$

$$\begin{aligned} \log E\exp \lambda Y_{t}&=-\lambda \frac{m(t)}{s(t)}+\log \frac{\varPhi \left( t+\frac{\lambda }{s(t)}\right) }{\varPhi (t)}\\&=\frac{\lambda ^2}{2}\frac{s^{2}\left( t+\theta \frac{\lambda }{s(t)}\right) }{s^{2}(t)}=\frac{\lambda ^2}{2}\frac{\psi ^{\prime }\left( t+\theta \frac{\lambda \left( 1+o(1)\right) }{\sqrt{\psi ^{\prime }(t)}}\right) }{\psi ^{\prime }(t)}\left( 1+o(1)\right) \end{aligned}$$

as $t\rightarrow \infty ,$ for some $\theta \in \left( 0,1\right)$ depending on $t$, where we used Theorem 1. Now making use of Corollaries 2 and 3 it follows that

$$ \lim _{t\rightarrow \infty }\log E\exp \lambda Y_{t}=\frac{\lambda ^2}{2}, $$

which proves the claim.    $\square$