Is the martingale stopping theorem applicable?

Question

Let $Z_n=\prod_{i=1}^nX_i$, where $X_i,i\ge1$ are independent random variables with

$$P\{X_i=2\}=P\{X_i=0\}=1/2.$$

Let $N=\min\{n:Z_n=0\}$. Is the martingale stopping theorem applicable?

Here is my solution:

$$E[N]=\sum_{k=1}^\infty k(1/2)^k=2,$$

and

$$E[|Z_{n+1} - Z_n| ~ |Z_1,\dots,Z_n] = E[|Z_n(X_{n+1}-1)|~~|Z_1,\dots,Z_n] = |Z_n|E[|X_{n+1}-1|] = 0.$$

Thus, the martingale stopping theorem is applicable.

However, it is obvious that $E[X_N]=0$ according to the definition $N$, and $E[X_1]=1$ which leads $E[X_N]=E[X_1]=1$ according to the martingale stopping theorem.

Where am I wrong?

Answer 1

If your stopping time is in $L^1$ then you must ensure that $(Z_n -Z_{n-1})$ is bounded. But in your case this fact doesn't happens once $|Z_n -Z_{n-1}|=\prod_{i=1}^{n-1}X_i|X_n-1|$.

As already mentioned, $E|X_n-1| = 1$ instead of 0.

Answer 2

There is something wrong in your calculation.

$$ E[|Z_{n+1} - Z_n| ~ |Z_1,\dots,Z_n] = E[|Z_n(X_{n+1}-1)|~~|Z_1,\dots,Z_n] = |Z_n|E[|X_{n+1}-1|] $$

should be equal to $|Z_n|$ instead of $0$.

$Z_n$ is unbounded then the martingale stopping theorem is not applicable.