First, we recall that
\({\hat{Q}}_{n}{\;\buildrel {d}\over = \;}M_{\beta ,n}\) for all
\(n\in {\mathbb {N}}\), so that
\({\mathbb {P}}({\hat{Q}}_{n} = 0) = {\mathbb {P}}(M_{\beta ,n}=0)\),
\({\mathbb {E}}[{\hat{Q}}_{n}]={\mathbb {E}}[M_{\beta ,n}]\) and
\(\mathrm{Var}\,\,{\hat{Q}}_{n}=\mathrm{Var}\,\,M_{\beta ,n}\) as defined in (
52). Our starting point is Spitzer’s identity, see [
3, p. 230],
$$\begin{aligned} {\mathbb {E}}[\mathrm{e}^{it M_{\beta ,n}}] = \exp \left( \sum _{k=1}^\infty \frac{1}{k} ({\mathbb {E}}[\mathrm{e}^{itS_{k,n}^+}]-1)\right) , \end{aligned}$$
(57)
with
\(S_{k,n}\) as in (
52), and
\(M_{\beta ,n}\) the all-time maximum of the associated random walk. Simple manipulations of (
57) give
$$\begin{aligned} \mathrm{ln}\,{\mathbb {P}}(M_{\beta ,n} = 0)&= -\sum _{k=1}^\infty \frac{1}{k}\,{\mathbb {P}}(S_{k,n} > 0), \end{aligned}$$
(58)
$$\begin{aligned} {\mathbb {E}}[M_{\beta ,n}]&= \sum _{k=1}^\infty \frac{1}{k} {\mathbb {E}}[S^+_{k,n}] = \sum _{k=1}^\infty \frac{1}{k}\int _0^\infty {\mathbb {P}}(S_{k,n} > x) \mathrm{d}x, \end{aligned}$$
(59)
$$\begin{aligned} \mathrm{Var}\,M_{\beta ,n}&= \sum _{k=1}^\infty \frac{1}{k} {\mathbb {E}}[(S^{+}_{k,n})^2] =\sum _{k=1}^\infty \frac{1}{k}\int _0^\infty {\mathbb {P}}(S_{k,n} > \sqrt{x}) \mathrm{d}x. \end{aligned}$$
(60)
By Lemma
1, we know
$$\begin{aligned} {\mathbb {P}}(S_{k,n}> y) = {\mathbb {P}}\left( {\sum _{i=1}^k} Y_{i,n}> y \right) \rightarrow {\mathbb {P}}\left( \sum _{i=1}^k Z_i> y\right) , \end{aligned}$$
for
\(n\rightarrow \infty \), where the
\(Z_i\) are independent and identically normally distributed with mean
\(-\beta \) and variance 1. Because equivalent expressions to (
58)–(
60) apply to the limiting Gaussian random walk, it is sufficient to show that the sums converge uniformly in
n, so that we can apply dominated convergence to prove the result.
We start with the empty-queue probability. To justify interchangeability of the infinite sum and limit, note
$$\begin{aligned} {\mathbb {P}}(S_{k,n}> 0) \le {\mathbb {P}}(|S_{k,n}+k\beta | > k\beta )\le \frac{k}{\beta ^2k^2} = \frac{1}{\beta ^2k}, \end{aligned}$$
where we used that
\({\mathbb {E}}[ S_{k,n}] = k{\mathbb {E}}[Y_{1,n}] = -k\beta \) and
\(\mathrm{Var}\,S_{k,n} = k\), and apply Chebyshev’s inequality, so that
$$\begin{aligned} \sum _{k=1}^\infty \frac{1}{k}{\mathbb {P}}(S_{k,n} > 0) \le \sum _{k=1}^\infty \frac{1}{\beta ^2 k^2} < \infty , \quad \forall n\in {\mathbb {N}}. \end{aligned}$$
Hence,
$$\begin{aligned} \lim _{n\rightarrow \infty } \mathrm{ln}\,{\mathbb {P}}({\hat{Q}}_{n}= 0)&= \lim _{n\rightarrow \infty } - \sum _{k=1}^\infty \frac{1}{k}{\mathbb {P}}(S_{k,n}> 0) = -\sum _{k=1}^\infty \frac{1}{k} \lim _{n\rightarrow \infty }{\mathbb {P}}(S_{k,n}> 0)\\&= -\sum _{k=1}^\infty \frac{1}{k} {\mathbb {P}}\left( {\sum _{i=1}^k} Z_i > 0\right) = \mathrm{ln}\,{\mathbb {P}}(M_\beta = 0). \end{aligned}$$
Finding a suitable upper bound on
\(\frac{1}{k}\int _0^\infty {\mathbb {P}}({\hat{Q}}_{n}>x) \mathrm{d}x\) and
\(\frac{1}{k}\int _0^\infty {\mathbb {P}}({\hat{Q}}_{n}>\sqrt{x}) \mathrm{d}x\) requires a bit more work. We initially focus on the former, and the latter follows easily. The following inequality from [
32] proves to be very useful:
$$\begin{aligned} {\mathbb {P}}({\bar{S}}_k>y) \le C_r\,\left( \frac{k\,\sigma ^2}{y^2}\right) ^r + k\,{\mathbb {P}}(X>y/r), \end{aligned}$$
(61)
where
\({\bar{S}}_k\) is the sum of
k i.i.d. random variables distributed as
X, with
\({\mathbb {E}}[X] = 0\) and
\(\mathrm{Var}\,\, X=\sigma ^2\),
\(y > 0\),
\(r>0\) and
\(C_r\) a constant only depending on
r. We take
\(r=3\) for brevity in the remainder of the proof, although any
\(r>2\) will suffice. We have, from (
61) with
\(X={\hat{A}}_n\), so that
\({\mathbb {E}}[X]=0,\ \mathrm{Var}\,\, X = 1\), and
\(r=3\),
\(y=x+k\beta \),
$$\begin{aligned} {\mathbb {P}}(S_{k,n}> x)= & {} {\mathbb {P}}\left( \sum _{i=1}^k {\hat{A}}_{i,n}> x+k\beta \right) \nonumber \\&\le C_3 \left( \frac{k}{(x+k\beta )^2}\right) ^3 +\, k\,{\mathbb {P}}\left( {\hat{A}}_{1,n}>\frac{x+k\beta }{3}\right) . \end{aligned}$$
(62)
The quantity
\((k/(x+k\beta )^2)^3\) is independent of
n, and we have
$$\begin{aligned} \sum _{k=1}^\infty \frac{1}{k}\int _0^\infty \left( \frac{k}{(x+k\beta )^2}\right) ^3\, \mathrm{d}x&= \sum _{k=1}^\infty k^2 \int _0^\infty \frac{\mathrm{d}x}{(x+k\beta )^6} \nonumber \\&= \sum _{k=1}^\infty \frac{k^2}{5(k\beta )^5} = \frac{1}{5\beta ^5} \sum _{k=1}^\infty \frac{1}{k^3} < \infty . \end{aligned}$$
(63)
Next, by assumption, there is an
\(M_3>0\) such that
$$\begin{aligned} {\mathbb {E}}[(\max \{{\hat{A}}_{1,n},0\})^3] = \int _0^\infty t^3 \, \mathrm{d}P_{{\hat{A}}_{1,n}}(t) \le M_3 \end{aligned}$$
(64)
for all
\(n=1,2,\ldots \). It follows that for all
\(x>0\),
\(k=1,2,\ldots \), and all
\(n=1,2,\ldots \),
$$\begin{aligned} {\mathbb {P}}\left( {\hat{A}}_{1,n} > \frac{x+k\beta }{3}\right)&= \int _{\frac{x+k\beta }{3}}^\infty \mathrm{d}P_{{\hat{A}}_{1,n}}(t) \le \int _{\frac{x+k\beta }{3}}^\infty \frac{t^3}{\left( \frac{x+k\beta }{3}\right) ^3} \, \mathrm{d}P_{{\hat{A}}_{1,n}}(t) \nonumber \\&\le \frac{27}{(x+k\beta )^3} \, {\mathbb {E}}[(\max \{{\hat{A}}_{1,n},0\})^3]\le \frac{27M_3}{(x+k\beta )^3}. \end{aligned}$$
(65)
The quantity
\(1/(x+k\beta )^3\) is independent of
n, and we have
$$\begin{aligned} \sum _{k=1}^\infty \int _0^{\infty } \frac{\mathrm{d}x}{(x+k\beta )^3} = \sum _{k=1}^\infty \frac{1}{2k^2\beta ^2} < \infty . \end{aligned}$$
(66)
Thus we see that for all
\(x>0\),
\(k=1,2,\ldots \), and all
\(n=1,2,\ldots \),
$$\begin{aligned} \frac{1}{k}\,{\mathbb {P}}(S_{k,n} > x ) \le \frac{C_3 k^2}{(x+k\beta )^6} + \frac{27M_3}{(x+k\beta )^3}, \end{aligned}$$
(67)
with
$$\begin{aligned} \sum _{k=1}^\infty \int _0^\infty \left( \frac{C_3 k^2}{(x+k\beta )^6} + \frac{27 M_3}{(x+k\beta )^3}\right) \, \mathrm{d}x < \infty . \end{aligned}$$
(68)
By Lebesgue’s theorem on dominated convergence, it then follows that
$$\begin{aligned} \lim _{n\rightarrow \infty } {\mathbb {E}}[{\hat{Q}}_n]&= \lim _{n\rightarrow \infty } \sum _{k=1}^\infty \frac{1}{k}\int _0^\infty {\mathbb {P}}(S_{k,n}>x)\, \mathrm{d}x \nonumber \\&= \sum _{k=1}^\infty \int _0^\infty {\mathbb {P}}\left( \sum _{i=1}^k Z_i> x\right) \, \mathrm{d}x = {\mathbb {E}}[M_\beta ]. \end{aligned}$$
(69)
In a similar fashion, by using (
61) with
\(y = \sqrt{x}+k\beta \), we have
$$\begin{aligned} {\mathbb {P}}(S_{k,n}> \sqrt{x} ) \le C_3 \left( \frac{k}{(\sqrt{x}+k\beta )^2} \right) ^3 + k\,{\mathbb {P}}\left( {\hat{A}}_{1,n} > \frac{\sqrt{x}+k\beta }{3}\right) . \end{aligned}$$
(70)
Now we have
$$\begin{aligned} \sum _{k=1}^\infty \frac{1}{k}\int _0^\infty \left( \frac{k}{(\sqrt{x}+k\beta )^2} \right) ^3 \mathrm{d}x = \frac{1}{10\beta ^4} \sum _{k=1}^\infty \frac{1}{k^2} < \infty , \end{aligned}$$
(71)
where we have used that, by partial integration,
$$\begin{aligned} \int _0^\infty \frac{\mathrm{d}x}{(\sqrt{x}+k\beta )^6} = \int _0^\infty \frac{2u\, \mathrm{d}u}{(u+k\beta )^6} = \frac{1}{10(k\beta )^4}. \end{aligned}$$
(72)
Next, by assumption, there in an
\(M_4>0\) such that
$$\begin{aligned} {\mathbb {E}}[(\max \{{\hat{A}}_{1,n},0\})^4] = \int _0^\infty t^4 \, \mathrm{d}P_{{\hat{A}}_{1,n}}(t) \le M_4 \end{aligned}$$
(73)
for all
\(n=1,2,\ldots \). Then, as in (
65), we get that for all
\(x>0\),
\(k=1,2,\ldots \), and all
\(n=1,2,\ldots \),
$$\begin{aligned} {\mathbb {P}}\left( {\hat{A}}_{1,n}> \frac{\sqrt{x}+k\beta }{3}\right) \le \frac{81 M_4}{(\sqrt{x}+k\beta )^4}, \end{aligned}$$
(74)
while, as in (
66) and (
72),
$$\begin{aligned} \sum _{k=1}^\infty \int _0^\infty \frac{\mathrm{d}x}{(\sqrt{x}+k\beta )^4} = \sum _{k=1}^\infty \frac{1}{3k^2\beta ^2} < \infty . \end{aligned}$$
(75)
Therefore, for all
\(x>0\),
\(k=1,2,\ldots \), and all
\(n=1,2,\ldots \),
$$\begin{aligned} \frac{1}{k}\,{\mathbb {P}}\left( S_{k,n}> \sqrt{x}\right) \le \frac{C_3k^2}{(\sqrt{x}+k\beta )^6} + \frac{81 M_4}{(\sqrt{x}+k\beta )^4}, \end{aligned}$$
(76)
with
$$\begin{aligned} \sum _{k=1}^\infty \int _0^\infty \left( \frac{C_3k^2}{(\sqrt{x}+k\beta )^6} + \frac{81 M_4}{(\sqrt{x}+k\beta )^4}\right) \, \mathrm{d}x < \infty . \end{aligned}$$
(77)
Hence, we conclude, as in (
69), that
\(\lim _{n\rightarrow \infty } \mathrm{Var}\,{\hat{Q}}_n = \mathrm{Var}\,M_\beta \). This completes the proof.
\(\square \)