My professor gave me a very unclear proof of this theorem. It was so messy and unclear, I was unable to write down all the details of the proof.

Theorem: Suppose $\tau \in T$, where $T$ is the set of all stopping times. Then $E[M_{\tau}] = E[M_{0}]$, where $M_n$ is a discrete time martingale.

What I wrote down in my lecture:

Define $N_{*} = max(k: k \in T)$. Write $\tau = \sum_{k=1}^{N_{*}}k I_{\tau=k}$. Then $$E[M_{\tau}] = E[\sum_{k=1}^{N_{*}}M_k I_{\tau=k}] = \sum_{k=1}^{N_{*}}E[M_k I_{\tau=k}]$$

That's all I wrote down. How can I finish the proof?

Earlier in the lecture, we proved that $\forall m < n$

$$E[M_n | F_m] = M_m$$

I'm thinking that if we set $m=0$ we can maybe use this:

$$E[M_n | F_0] = M_0$$

But I'm honestly not sure what to do here.

  • $\begingroup$ This is not true in general. Are you assuming that $T$ is bounded? $\endgroup$ Sep 16, 2015 at 2:56
  • $\begingroup$ @NateEldredge - Yes. $\endgroup$
    – Kerry
    Sep 16, 2015 at 3:58

1 Answer 1


Recall that $\mathbb{E}_Y [ \mathbb{E}_X[X|Y] ] = \mathbb{E}[X] $

We'll use this to inductively show that $\mathbb{E}[X_n] = \mathbb{E}[X_0]$ for a martinagale sequence.

Note that $\mathbb{E}[X_1] = \mathbb{E}[ \mathbb{E} [X_1|X_0]] = \mathbb{E}[X_0]$. Assume the hypothesis for $X_n$.

$\mathbb{E}[X_{n+1}| \mathcal{F}_n] = X_n$

Take expectation on both sides

$\mathbb{E}[\mathbb{E}[X_{n+1}| \mathcal{F}_n]] = \mathbb{E}[X_n] = \mathbb{E}[X_0]$.

The proof then follows, assuming $\tau < \infty$ a.s.:

\begin{align*} \mathbb{E}[M_\tau] &= \underset{k = 1}{\overset{\infty}{\sum}} \mathbb{E}[M_\tau | \tau = k] P(\tau = k) \\ &= \underset{k = 1}{\overset{\infty}{\sum}} \mathbb{E}[M_k] P(\tau = k) \\ &=\underset{k = 1}{\overset{\infty}{\sum}} \mathbb{E}[M_0] P(\tau = k) \\ &= \mathbb{E}[M_0]. \end{align*}


  • $\begingroup$ thanks. I suppose the proof was very simple. Not sure why my professor had to ramble for 20 mins about this. I'm a bit confused when you wrote: $E_Y$ in the first line. What does this notation mean? $\endgroup$
    – Kerry
    Sep 16, 2015 at 4:14
  • $\begingroup$ The subscript was just to clarify that the expectation was over realisations of $Y$. I wasn't sure if you'd come across that before, and seeing it written like that had helped me grasp it quicker the first time I had seen the iterated expectation. $\endgroup$ Sep 16, 2015 at 9:26
  • $\begingroup$ Yeah, it seems like a really simple idea, maybe (s)he'd been trying to address the non-finite $T$ case or something? Otherwise there's precious little meat in this. $\endgroup$ Sep 16, 2015 at 9:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .