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I'm trying to understand a pretty standard proof about the possible events for Martingales with bounded increments. Specifically, assume

1) $X_1, X_2, \ldots$ is a martingale

2) $|X_{n+1} - X_n| \le M$

3) $C = \{ \omega : \lim X_{n}(\omega) \text{ exists and is finite} \}$

4) $D = \{ \omega : \lim\sup X_{n}(\omega) = \infty, \lim \inf X_n(\omega) = - \infty \}$

Then $P(C \cup D) = 1$

In particular, the proof in Durrett, page 204-205, they mention that if I fix a constant $K$ and let $N$ be the first time $X_n \le -K$, then I can use the fact that a non-negative (super)martingale has a limit almost surely on $ \{ \omega : N(\omega) = \infty \}$. It's precisely this last statement, "on ${ \omega : N(\omega) = \infty }$" that I don't understand. Can someone clarify why this doesn't apply for $ \{ \omega : N(\omega) < \infty \}$?


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$X_{N \wedge n}$ is a martingale bounded below and as such has a limit (call it $X$) almost surely. On $\{\omega : N(\omega) = \infty\}$, $X_n$ converges to $X$ as well. Certainly $X_n$ need not converge on the set $\{ N < \infty\}$; consider simple random walk. – Nate Eldredge Nov 25 '11 at 4:33
Then on $\{ \omega : N(\omega) = \infty \}$, isn't $X_{n \land N} = X_{n}$? Can you elaborate how $\{ \omega : N(\omega) < \infty \}$ relates to the random walk martingale example? – duckworthd Nov 25 '11 at 4:40
I added an answer expanding on this. – Nate Eldredge Nov 25 '11 at 12:26
up vote 5 down vote accepted

I'll elaborate on my comment.

To make the notation more explicit, I'll write $N_K = \min \{ n : X_n \le -K \}$, and abbreviate $\{\omega : N_K(\omega) = \infty\}$ as $\{N_K = \infty\}$. Since $X_{n \wedge N_K} \ge -K-M$ for all $n$, $X_{n \wedge N_K}$ is a martingale which is bounded below, thus it converges a.s. But on the set $\{N_K = \infty\}$, $X_n = X_{n \wedge N_K}$ for all $n$, so $X_n$ converges a.s. on $\{N_K = \infty\}$.

$X_n$ need not converge on the set $\{N_K < \infty\}$. If $X_n$ is simple random walk, then $P(N_K < \infty) = 1$, and $X_n$ converges with probability 0. (This is why the theorem read $P(C \cup D) = 1$ instead of $P(C) = 1$.)

To complete the proof, since $K$ was arbitrary, we have that $X_n$ converges a.s. on $\bigcup_K \{N_K = \infty\}$. This latter set is precisely $\{\liminf X_n > -\infty\}$. Replacing $X_n$ by $-X_n$ we can also show that $X_n$ converges a.s. on $\{\limsup X_n < \infty\}$. Together, these are equivalent to the statement $P(C \cup D) = 1$.

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Thanks! It all makes sense now. – duckworthd Nov 25 '11 at 19:42
Very useful explanation... – d.k.o. Jan 9 '15 at 7:56

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