Let $(X_n)$ be a sequence of iid random variables such that:
$$\mathbb{P}(X_k=-1)=q \\
\mathbb{P}(X_k=1)=p=1-q$$
(two points distribution)
Let $\tau$ be the first moment when number of successes ($X_k=1$) is greater than number of failures ($X_k=-1$).
Calculate $\mathbb{E}\tau$ – the first moment when the sum of $X_n$, $n=1,2\dots$ is greater than zero.
My attempt:
I think we should use Wald equality. $\tau$ obviously is a stopping time. Also $\mathbb{E}|X_i|<\infty$ so:
$$\mathbb{E}\sum_{k=1}^\tau X_k=\mathbb{E}\tau \mathbb{E}X_i$$
I think that $\mathbb{E}\sum_{k=1}^\tau X_k = 1$ Am I correct here? Then $\mathbb{E}\tau=\frac{1}{p-q}$ Is that correct? And what if $p=1/2$ ?
Best Answer
Let me sum up the comments in the question. your proof is not complete. I am not sure why you did not follow up on the comments.
As Did pointed out either $E(\tau)=\infty$ or $E(\tau)=\frac{1}{p-q}$ (note $E(\tau)<\infty$ is a condition for Wald's theorem to be satisfied. This is not clear in your case)
Note that when $p\leq 1/2$, this is negative/blows up so $E(\tau)=\infty$. This is not surprising
What else needs doing:
When $p > 1/2$, you need to show $E(\tau) <\infty$, then you can conclude $E(\tau)=\frac{1}{p-q}$, by Wald
You need to show these results about Markov Chains:
When $p>1/2$, consider 1 as an absorbing state and the mc is positive recurrent with $E(\tau) <\infty$