I am working through a course on Stochastic Processes and am looking for a proof verification for an alternative to the one that I have been presented. My proof, as noted below, currently contains a gap. However, it is unclear to me whether or not this is an easily resolvable gap or if I have to refer to an alternative proof.
Let $(X_n)_{n \ge 0}$ be a positive supermartingale. Then Doob's Maximal Inequality states that for any $\lambda \ge 0$, we have the following: $$ \lambda \mathbb{P} \big{(} \max _{k \ge 0} X_k \ge \lambda \big{)} \leq \mathbb E(X_0)$$
My proof attempt uses the fact that by Doob's Optional Stopping Theorem, if we can find a stopping time $T$, then for any martingale, $\mathbb{E}(X_T) \leq \mathbb E (X_0)$ under the assumption that one of the following holds:
- $T$ is bounded
- $X$ is bounded and $T$ is finite
- $\mathbb E (T) < \infty$ and for some $K > 0$, we have that $|X_n – X_{(n-1)} | \leq K$
The gap in my proof is that I can't see how any of these conditions apply in this case. However, I suspect that the fact that we assume that $X$ is not just a supermartingale, but a positive one may allow us to relax these assumptions.
After that, inspired by the inequality in question, I let our stopping time $T$ be defined: $$T := \inf \{ n \ge 0 : X_n \ge \lambda \}$$
After this we can perform some fairly routine calculations to show the desired result:
$$ \mathbb E (X_T) = \mathbb E (X_T | T < \infty ) \mathbb{P}(T \leq \infty) + \mathbb E (X_T | T = \infty ) \mathbb{P} (T = \infty) $$
$$ \ge \mathbb E (X_T | T < \infty ) \mathbb{P}\big{(}T \leq \infty)$$
$$ \ge \lambda \mathbb{P}(T \leq \infty)$$
$$ = \lambda \mathbb{P}\big{(}\max _{k \ge 0} X_k \ge \lambda \big{)}$$
$$ \text{So: } \space \lambda \mathbb{P}\big{(}\max _{k \ge 0} X_k \ge \lambda \big{)} \leq \mathbb E (X_T) \leq \mathbb E (X_0) \space \text{ (as required) }\quad \square$$
This intuitively feels close to a complete proof, however, I am aware that this could be misleading and I may need to take a completely different avenue given the fact that I cannot currently invoke the Optional Stopping Theorem in the way that I have done. I would be grateful if this could be resolved for me.
Best Answer
To show that $\mathbb{E}[X_T] \le \mathbb{E}[X_0]$, note that we have $\mathbb{E}[X_{T \wedge n}] \le \mathbb{E}[X_0]$ for all $n$ because $T \wedge n$ is a bounded stopping time. Then, since $X_{T \wedge n} \ge 0$, Fatou's lemma gives \begin{align*} \mathbb{E}[X_T] &= \mathbb{E}[\lim X_{T \wedge n}] \\ &\le \liminf \mathbb{E}[X_{T \wedge n}] \\ &\le \mathbb{E}[X_0]. \end{align*}