Stopping time proof with discrete martingale

Question

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.

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*}

Q.E.D.