Let $X = \{X_n\}_{n=0}^{\infty}$ be a closable submartingale. Then, for any stopping time $τ, X_τ$ is integrable and, for another stopping time $σ$, $E[X_\tau |\mathcal{F}_\sigma ]\ge X_{\sigma\wedge \tau}$ , P-a.s.

The proof start by showing $E\left[X_{\infty}|\mathcal{F}_{\sigma}\right]\ge X_{\sigma}$ where $X_{\infty}$ is the closing element, and then says it suffices to show that the stopped process $X^{\tau}=\{X_{n\wedge\tau}\}_{n=0}^{\infty}$ is closable. But how does this help us to conclude?


1 Answer 1


After you've shown $X^\tau$ is closable, applying the fact that $E[X_{\infty}|\mathcal F_{\sigma}] \ge X_{\sigma}$ for any closed submartingale $X$ to $X^\tau$ we have \begin{align*} E[X_{\tau}|\mathcal F_{\sigma}] &= E[X_{\infty}^{\tau}|\mathcal F_{\sigma}] \\ &\ge X_{\sigma}^\tau \\ &= X_{\sigma \wedge \tau}. \end{align*}

To show $X_\tau = X_\infty^\tau$, note that on the event $\{\tau < \infty \}$, we have $$X_{\infty}^\tau = \lim_{n \rightarrow \infty} X_n^\tau = \lim_{n \rightarrow \infty} X_{n \wedge \tau} = X_\tau$$ because $n \wedge \tau = \tau$ for all $n$ sufficiently large (how large is "sufficiently large" depends on $\omega$, but we're sending $n \rightarrow \infty$ so that doesn't matter here). On the event $\{\tau = \infty\}$, $X_\tau = X_\infty$ by definition, and $$X_{\infty}^\tau = \lim_{n \rightarrow \infty} X_n^\tau = \lim_{n \rightarrow \infty} X_{n \wedge \tau} = \lim_{n \rightarrow \infty} X_{n} = X_\infty = X_\tau.$$

Lemma: Let $X_n$ be a closable submartingale with closing element $Z$, i.e. $X_n \le \mathbb{E}[Z|\mathcal F_n]$ for all $n$. Then there exists an integrable $X_\infty$ with $(X_n) \rightarrow X_\infty$ a.s. and $X_n \le \mathbb{E}[X_\infty|\mathcal F_n]$.

Proof: By monotonicity of $x \mapsto x^+$ and the conditional Jensen inequality, we have $(X_n)^+ \le (\mathbb{E}[Z|\mathcal F_n])^+ \le \mathbb{E}[Z^+|\mathcal F_n]$ so $$\sup_n \mathbb{E}[(X_n)^+] \le \sup_n \mathbb{E}[\mathbb{E}[Z^+|\mathcal F_n]] = \mathbb{E}[Z^+] < \infty.$$ Therefore by Doob's submartingale convergence theorem, there exists an integrable random variable $X_\infty$ with $(X_n) \rightarrow X_\infty$ a.s.

Now to show $X_m \le \mathbb{E}[X_\infty|\mathcal F_m]$, first note that $0 \le \mathbb{E}[Z|\mathcal F_n] - X_n$, so by the conditional Fatou lemma we have \begin{align*} \mathbb{E}\left[\left.\liminf_{n \rightarrow \infty}(\mathbb{E}[Z|\mathcal F_n] - X_n)\right| \mathcal F_m\right] &\le \liminf_{n \rightarrow \infty} \mathbb{E}[\mathbb{E}[Z|\mathcal F_n]-X_n|\mathcal F_m] \\ &= \liminf_{n \rightarrow \infty} \mathbb{E}[Z|\mathcal F_m] - \liminf_{n \rightarrow \infty} \mathbb{E}[X_n | \mathcal F_m]. \end{align*} There is a theorem saying that since $Z$ is $\mathcal F_{\infty}$ measurable, $\lim_{n \rightarrow \infty}\mathbb{E}[Z|\mathcal F_n] = Z$ (see Williams Probability with martingales). Since $(X_n) \rightarrow X_\infty$, the left side of the inequality is therefore $\mathbb{E}[Z-X_\infty|\mathcal F_m]$. Rewriting, we've shown \begin{align*} \liminf_{n \rightarrow \infty} \mathbb{E}[X_n | \mathcal F_m] &\le \mathbb{E}[X_\infty|\mathcal F_m]. \end{align*} Since $X_m \le \mathbb{E}[X_n|\mathcal F_m]$ for all $n \ge m$, we therefore conclude $X_m \le \liminf_{n \rightarrow \infty} \mathbb{E}[X_n | \mathcal F_m] \le \mathbb{E}[X_\infty|\mathcal F_m]$ as desired.

  • $\begingroup$ How did you show the first equality? Formally, I understand it as $X^{\tau}_{\infty}=X_{\tau\wedge\infty}=X_{\tau}$ but this is just formally. $\endgroup$
    – edamondo
    Jan 18 at 22:40
  • $\begingroup$ @edamondo I edited the answer to include the proof. $\endgroup$ Jan 18 at 23:22
  • $\begingroup$ I have still one doubt. $X_{\infty}$ is by definition the closing element but you treat it as if it is the limit of the $X_n$'s. I know that this is equivalent for martingales, but I am not sure about submartingales. Is being a closing element equivalent to being the limit of the submartingale? $\endgroup$
    – edamondo
    Jan 19 at 11:40
  • $\begingroup$ @edamondo Right, sorry, I was thinking that was the definition of a closing element. What is the definition you are using for a closing element of a submartingale? $\endgroup$ Jan 19 at 15:49
  • $\begingroup$ My definition is : A submartingale $\{X_n\}_{n=0}^{\infty}$ for which there exists an $F_{\infty}:=\sigma(\cup_{n=0}^{\infty}\mathcal{F}_n)$-measurable and integrable random variable $X_{\infty}$ satisfying $X_n\le E[X_{\infty}|\mathcal{F}_n]$ $P$-as $(n=0,1,2,...)$ $\endgroup$
    – edamondo
    Jan 19 at 17:10

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