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Given some probability space $(\Omega, \mathcal{F},\mathbb{P})$, $F_{n}$ some filtration, and $X_{n}$ some martingale with $\tau$ stopping time. We know some major things such as $\tau$ is finite almost surely, $\mathbb{E}|X_{\tau}|$ is finite, and $\lim_{n\to\infty}\int_{\tau>n}|X_{n}|d\mathbb{P}=0$. The goal being to show the big result of the Stopping Theorem, $\mathbb{E}X_{\tau}=\mathbb{E}X_{0}$

I have already proven that $\lim_{n\to\infty}\mathbb{E}(X_{\tau\wedge n})=\mathbb{E}(X_{\tau})$

I am stuck trying to show that $\mathbb{E}(X_{\tau \wedge n})=\mathbb{E}(X_{0})$.

A suggestion I found was the show that $\lim_{n\to\infty}\mathbb{E}|X_{\tau \wedge n}-X_{\tau}|\to 0$, but not really exactly sure what the punchline is?

Any suggestions?

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  • $\begingroup$ Not a full answer, but my intuition about it. The process underlying the martingale will always have the same expectation while it progresses because for any value that is not stopped the expectation remains the same. The matter/problem is whether the expectation of the stopping time is finite. For instance, a random walk starting at $0$ with an absorbing boundary at $a$ will have $E[X_\tau]=a\neq 0$. That is because there's a tiny amount of probability that $\tau$ takes infinitely long and why $E[X_\tau]$ can deviate from 0. At any time it's still $E[X_t]=0$ and the non-stopped paths balance. $\endgroup$ Commented Apr 1, 2023 at 21:10
  • $\begingroup$ What do you mean by "$\tau \land n$"? $\endgroup$ Commented Apr 1, 2023 at 21:13
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    $\begingroup$ @SextusEmpiricus From the context presumably $\tau \land n$ (logical and, aka meet) is supposed to mean $\min(\tau, n)$ $\endgroup$
    – Henry
    Commented Apr 1, 2023 at 21:23

1 Answer 1

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To prove $E(X_{\tau \wedge n}) = E(X_0)$, you may show that $\{(X_n^*, \mathscr{F}_n): n = 1, 2, \ldots\}$ is a martingale. Here $X_n^* = X_{\tau \wedge n}$.

First, $X_n^*$ is measurable $\mathscr{F}_n$: since $[\tau > n] = \Omega - [\tau \leq n] \in \mathscr{F}_n$, for all $H \in \mathscr{R}^1$, it follows that \begin{align*} [X_n^* \in H] = \bigcup_{k = 0}^n[\tau = k, X_k \in H] \cup [\tau > n, X_n \in H] \in \mathscr{F}_n. \end{align*} Second, $E[|X_n^*|] < \infty$ for every $n$. This is because \begin{align*} E[|X_n^*|] = \sum_{k = 0}^{n - 1}\int_{[\tau = k]}|X_k|dP + \int_{[\tau \geq n]}|X_n|dP \leq \sum_{k = 0}^nE[|X_k|] < \infty. \end{align*}

Third, for every $A \in \mathscr{F}_n$, \begin{align*} & \int_A X_n^* dP = \int_{A \cap [\tau > n]}X_n dP + \int_{A \cap [\tau \leq n]}X_\tau dP, \tag{1} \\ & \int_A X_{n + 1}^* dP = \int_{A \cap [\tau > n]}X_{n + 1}dP + \int_{A \cap [\tau \leq n]}X_\tau dP. \tag{2} \end{align*} Because $\{(X_n, \mathscr{F_n})\}$ is a martingale and $A \cap [\tau > n] \in \mathscr{F}_n$, the right-hand sides of $(1)$ and $(2)$ coincide, which shows that $E[X_{n + 1}^*|\mathscr{F}_n] = X_n^*$.

The above three items showed that $\{(X_n^*, \mathscr{F}_n): n = 1, 2, \ldots\}$ is a martingale, it then follows that $E[X_n^*] = E[X_0^*] = E[X_0]$. This completes the proof.

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