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. \(\square \)