# Understanding proof of discrete optimal sampling theorem

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?

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.

• 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. Jan 18 at 22:40
• @edamondo I edited the answer to include the proof. Jan 18 at 23:22
• 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? Jan 19 at 11:40
• @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? Jan 19 at 15:49
• 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 $(n=0,1,2,...)$ Jan 19 at 17:10