I'm trying to prove the following generalized version of Doob's optional sampling theorem:

Let $X$ be a square integrable martingale with respect to a filtration $\mathbb F = \left\{ \mathcal F_n \right\}_{n \in \mathbb N}$ with square variation process $\langle X \rangle$. Let $\tau$ be a finite stopping time, and suppose $\mathbb E\left[\langle X \rangle_\tau \right] < \infty$. Then $$\mathbb E\left[ \left(X_\tau - X_0 \right)^2 \right] = \mathbb E\left[\langle X \rangle_\tau \right] \quad \textrm{and} \quad \mathbb E\left[ X_\tau \right] = \mathbb E\left[X_0 \right]$$

I know that $\langle X \rangle$ is the unique predictable process for which $\left(X_n^2 - \langle X \rangle_n\right)_{n \in \mathbb N}$ is a martingale, and it can be expressed in equations as $$\langle X \rangle = \sum_{i=1}^n \mathbb E\left[ \left(X_i - X_{i-1}\right)^2 \bigg| \mathcal F_{i-1}\right] \quad \textrm{and} \quad \mathbb E\left[\langle X\rangle_n\right] = \mathbf{Var}\left[X_n - X_0 \right].$$

And I know that if $\tau$ is bounded, then $\mathbb E\left[X_\tau\right] = \mathbb E\left[X_0\right]$. So I know in particular that $\mathbb E\left[X_{\tau \wedge T}\right] = \mathbb E\left[X_0\right]$ for every $T \in \mathbb N$. Clearly $X_{\tau \wedge T} \mathbb 1_{\{\tau = N\}} \to X_N \mathbb 1_{\{\tau = N\}}$ as $T \to \infty$, and $$\left|X_{\tau \wedge T} \mathbb 1_{\{\tau = N\}}\right| \leq \max_{1 \leq t \leq N} |X_t|,$$ so by dominated convergence, $$ \mathbb E\left[ X_{\tau \wedge T} \mathbb 1_{\{\tau = N\}}\right] \xrightarrow{T \to \infty} \mathbb E\left[ X_N \mathbb 1_{\{\tau = N\}}\right] $$ for each $N \in \mathbb N$. I think, therefore, it follows that $$ \mathbb E\left[X_0\right] = \lim_{T \to \infty} \mathbb E\left[X_{\tau \wedge T}\right] = \lim_{T \to \infty} \sum_{N = 0}^\infty\mathbb E\left[X_{\tau \wedge T} \mathbb 1_{\{\tau = N\}} \right] = \sum_{N = 0}^\infty\mathbb E\left[X_{\tau} \mathbb 1_{\{\tau = N\}} \right] = \mathbb E\left[X_\tau\right]. $$ But I'm not sure if passing this limit through the sum is valid. It requires dominated convergence, i.e. that the terms $\mathbb E\left[X_{\tau \wedge T} \mathbb 1_{\{\tau = N\}} \right]$ are uniformly bounded by some function, but I don't know what that function is. Is there a better approach?

  • $\begingroup$ By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_{\tau \wedge n}$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well. $\endgroup$ – Shalop Mar 23 '19 at 4:13
  • $\begingroup$ Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind. $\endgroup$ – Shalop Mar 23 '19 at 4:24
  • $\begingroup$ Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text. $\endgroup$ – D Ford Mar 23 '19 at 4:26
  • 1
    $\begingroup$ Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_{\tau \wedge n}$ again. From the given hypothesis $E[\langle X \rangle_{\tau}]<\infty$ you can basically see (using orthogonality of the increments of $M$) that $\sum_n E[(M_{n+1}-M_n)^2]<\infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $\sum \|a_n\|^2<\infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results. $\endgroup$ – Shalop Mar 23 '19 at 5:01
  • $\begingroup$ By orthogonality of increments of $M$, do you mean $\mathbb E\left[\left(M_{n+1} - M_n\right)\left(M_n - M_{n-1}\right)\right] = 0$? I'm having a hard time seeing how this with the hypothesis $\mathbb E\left[\langle X\rangle_\tau\right]<\infty$ implies $\sum_n \mathbb E\left[\left(M_{n+1}-M_n\right)^2\right] < \infty$. Also, if $\sum||a_n||^2 < \infty$, would it not be the case that $a_n \to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point. $\endgroup$ – D Ford Mar 23 '19 at 15:05


  1. Show that for any square-integrable martingale $(M_n,\mathcal{F}_n)_{n \geq 1}$ it holds that $$\mathbb{E}(M_n M_m \mid \mathcal{F}_m) = M_m^2, \qquad m \leq n.$$ Conclude that $$\mathbb{E}((M_n-M_m)^2) = \mathbb{E}(M_n^2-M_m^2) = \mathbb{E}(\langle M \rangle_n)- \mathbb{E}(\langle M \rangle_m), \qquad m \leq n.$$
  2. Using Step 1 for the martingale $M_n := X_{n \wedge \tau}$ gives $$\mathbb{E}((X_{n \wedge \tau}-X_{m \wedge \tau})^2) = \mathbb{E}(\langle X \rangle_{\tau \wedge n} - \langle X \rangle_{\tau \wedge m}), \qquad m \leq n.$$ Deduce from the monotonicity of $\langle X \rangle$ and the fact that $\mathbb{E}(\langle X \rangle_{\tau})< \infty$ that $$\mathbb{E}((X_{n \wedge \tau}-X_{m \wedge \tau})^2) \xrightarrow[]{m,n \to \infty} 0.$$
  3. By the completeness of $L^2(\mathbb{P})$ it follows that $Z := \lim_{n \to \infty} X_{n \wedge \tau}$ exists in $L^2(\mathbb{P})$. Show that $Z=X_{\tau}$.
  4. Combine the identities $$\mathbb{E}((X_{n \wedge \tau}-X_0)^2) = \mathbb{E}(\langle X \rangle_{\tau}) \quad \text{and} \quad \mathbb{E}(X_{\tau \wedge n}) = \mathbb{E}(X_0)$$ with Step 3 to prove the assertion.

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