TITLE: Is the expectation of the supremum of a random walk with negative drift finite?
QUESTION [5 upvotes]: Let $(\xi_n)_{n\in\mathbb N}$ be iid random variables with negative mean, let $S_n=\sum_{k=1}^n\xi_k$, and let $M=\sup_{n\ge 0}S_n$. Is it true that if the $\xi_n$ are nice (say, finite variance), then $M$ has finite expectation?
This question implies the result in the case where $\xi_n$ are $\{\pm1\}$-valued random variables (apply the Monotone convergence theorem to $M_n:=\sup_{0\le k\le n}S_k$), but I'm interested in the general case.
I wasn't able to find anything that looked relevant (other than the question I linked above) when Googling.

REPLY [5 votes]: By replacing $\xi_k$'s by $\xi_k \vee (-m)$ for some appropriately chosen $m > 0$ if necessary, we may assume WLOG that $\xi_k$'s are bounded below by $-m$ and we do so.
Let $M_n = \max_{0\leq k\leq n}S_k$. Also, define $W_0 = 0$ and $W_{n+1}=(W_n + \xi_{n+1})^+$, where $x^+ := x \vee 0$ is the positive part of the real number $x$.

Lemma. $M_n $ and $ W_n$ have the same distribution for all $n \geq 0$.

From the lemma and the monotone convergence theorem, we get
$$ \mathbf{E}[M] = \lim_{n\to\infty} \mathbf{E}[M_n]  = \lim_{n\to\infty} \mathbf{E}[W_n]. $$
In light of this, we show that $\mathbf{E}[W_n]$ is bounded. Assume otherwise that $\mathbf{E}[W_n] \to \infty$. Then
\begin{align*}
\mathbf{E}\left[ (W_{n+1}^2 - W_n^2) \mathbf{1}_{\{W_n \geq m\}} \right]
&= \mathbf{E}\left[ (2\xi_{n+1} W_n + \xi_{n+1}^2) \mathbf{1}_{\{W_n \geq m\}} \right] \\
&= 2 \mathbf{E}[\xi_1] \mathbf{E}\left[ W_n \mathbf{1}_{\{W_n \geq m\}} \right] + \mathbf{E}[\xi_1^2] \mathbf{P}(W_n \geq m) \\
&\to -\infty \quad \text{as } n \to \infty.
\end{align*}
On the other hand, using the fact that $W_{n+1} \leq W_n + \xi_{n+1}^+$ always holds, we get
\begin{align*}
\mathbf{E}\left[ (W_{n+1}^2 - W_n^2) \mathbf{1}_{\{W_n < m\}} \right]
&\leq \mathbf{E}\left[ ((W_n + \xi^+_{n+1})^2 - W_n^2) \mathbf{1}_{\{W_n < m\}} \right]\\
&= \mathbf{E}\left[ (2\xi_{n+1}^+ W_n + (\xi^+_{n+1})^2) \mathbf{1}_{\{W_n < m\}} \right] \\
&\leq 2m \mathbf{E}[\xi_1^+] + \mathbf{E}[(\xi_1^+)^2] \\
&< \infty.
\end{align*}
These altogether show that $\mathbf{E}\left[ W_{n+1}^2 - W_n^2 \right]$ is eventually negative.  However, this contradicts
$$ \mathbf{E}\left[ W_{n+1}^2 - W_n^2 \right] = \mathbf{E}\left[ M_{n+1}^2 - M_n^2 \right] \geq 0 \quad\text{for all } n. $$
From this, we conclude that $\mathbf{E}[W_n]$ is bounded as required.

Proof of Lemma. Define
$$ F_n(x) = (\xi_n + x)^+ . $$
Also, let $S_{p,n} = \sum_{i=1}^{n} \xi_{i+p}$ and $M_{p,n} = \max_{0 \leq k \leq n} S_{p,k}$. Then
$$ M_{n}
= \max\Bigl\{ 0, \max_{1\leq k \leq n} S_k \Bigr\}
= \Bigl(\max_{1\leq k \leq n} S_k \Bigr)^+ 
= \Bigl(\xi_1 + \max_{1\leq k \leq n} S_{1,k-1} \Bigr)^+
= F_1(M_{1,n-1}). $$
So, repeatedly applying this relation shows that $M_n = (F_1 \circ F_2 \circ \cdots \circ F_n)(0)$. However, from the recurrence relation, we know that $ W_n = (F_n \circ F_{n-1} \circ \cdots \circ F_1)(0) $. Since $(\xi_1, \xi_2, \ldots, \xi_n) $ has the same distribution as $ (\xi_n, \xi_{n-1}, \ldots, \xi_1)$, we are done. $\square$

Disclaimer. This proof is motivated by the queuing theory:

The lemma above seems to have been known at least since Lindley.

It is known that $\mathbf{E}[M^k] < \infty$ if and only if $\mathbf{E}[(\xi_1^+)^{k+1}] < \infty$; see J. Kiefer and J. Wolfowitz. In fact, the above proof is an adaptation of the proof from their paper.