Martingale and finite stopping time [closed]

Solution 1:

A nice example to know. The original "martingale": keep doubling your bet until you win.

I bet on a sequence of independent tosses of a fair coin. At time $n$ I bet $2^n$, so that when the coin is H, my fortune increases by $2^n$ but when the coin is T, my fortune decreases by $2^n$. [I have unlimited funds to bet, and the casino never closes.] This is the martincale $X_n$. So: $$ X_n = \sum_{k=0}^n \epsilon_k 2^k $$ where the random signs $\epsilon_k$ are $+1$ or $-1$ independently with probability $1/2$ for each. And of course $\mathcal F_n = \sigma\{\epsilon_0,\epsilon_1,\dots,\epsilon_n\}$ has $2^{n+1}$ atoms of probability $2^{-n-1}$.

Then let $$ \tau = \inf\{n : X_n > 0\} = \inf\{n : \epsilon_n = +1\} . $$ Things to check:

$\tau$ is a stopping time,
$[\tau = n\} = \{\epsilon_k=-1\text{ for }0 \le k \le n-1\text{ and }\epsilon_n = 1\}$,
$\tau < \infty$ a.s.,
$X_\tau = 1$ a.s.,
$\{\tau \ge n\} = \{\epsilon_k=-1\text{ for }0 \le k \le n-1\}$,
$$ \mathbb P[\mathbf1_{\tau\ge n} X_n = 0] = 1-2^{-n} \\ \mathbb P[\mathbf1_{\tau\ge n} X_n = 1] = 2^{-n-1} \\ \mathbb P[\mathbf1_{\tau\ge n} X_n = -2^{n+1}-1] = 2^{-n-1} $$ So that $$ \mathbb E\big[\mathbf1_{\tau\ge n} |X_n|\big] = 1\cdot2^{-n-1} +(2^{n+1}-1)) \cdot2^{-n-1} = 1 . $$

Solution 2:

Another example is the simple symmetric random walk with $X_0=0$ and $\tau=\inf\{n: X_n=1\}$. Because $X$ stopped at time $\tau$ is still a martingale, you have $$ 0=\Bbb E[X_{n\wedge\tau}]=\Bbb E[X_n; n<\tau]+\Bbb P[n\ge \tau]. $$ And because $X_n\le 0$ for $n<\tau$, this leads to $$ \Bbb P[\tau\le n]=\Bbb E[|X_n|; n<\tau]. $$ The left side converges to $1$ as $n\to\infty$.