# Proof of Doob's Maximal Inequality for Positive Supermartingales

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.

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*}
• Thank you for your quick answer. I understand that OST guarantees that $\mathbb E (X_{T \wedge n}) \leq \mathbb E (X_0)$ but why is it true that we can apply this inequality to the $\liminf$? Feb 23 at 13:07
• @FD_bfa This is just a standard property of sequences: if $(a_n)$ is a sequence of real numbers satisfying $a_n \le x$ for all $n$, then $\liminf (a_n) \le x$, and in fact every subsequential limit is less than or equal to $x$ Feb 23 at 15:22