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.


1 Answer 1


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*}

  • $\begingroup$ 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 $\liminf$? $\endgroup$
    – FD_bfa
    Feb 23 at 13:07
  • 1
    $\begingroup$ @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 $x$ $\endgroup$ Feb 23 at 15:22

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