Suppose $X_1,X_2,..$ are i.i.d random variables, taking only the values $-1,0,1,2$, with
$P(X_1 =-1) = 0.5\\
P(X_1 =0) = 0.1\\
P(X_1 =1) = 0.2\\
P(X_1 =2) = 0.2
$
Use martingale techniques to compute $P( \exists n \geq 1~such~that~S_n=-8)$.
I understand the symmetric and asymmetric random walk questions.
A question from the martingale
I am not sure how to start this problem or how it satisfy the Optional Stopping theorem. Can anyone give me some hints to solve this problem?
Best Answer
Define $\tau_j=\inf\{n:S_n=j\}$. We want an $r \neq 1$ s.t. $E[r^{S_n}|\mathscr{F}_{n-1}]=r^{S_{n-1}}$ and $(r^{S_n})_{n \in \mathbb{N}}$ is a martingale. We then compute $$E[r^{S_n-S_{n-1}}|\mathscr{F}_{n-1}]=E[r^{X_1}]=p_{-1}r^{-1}+p_0+p_1 r+p_2r^2=1\implies r^*\approx 0.871$$ Notice that $r^{S_n}$ converges to $0$ a.s. We use optional stopping: $$1=E[r^{S_{\tau_{-b}\wedge n}}]=r^{-b}P(\tau_{-b}<n)+\underbrace{E[r^{S_n}\mathbf{1}_{\{\tau_{-b}\geq n\}}]}_{\textrm{DCT }\to 0}$$ Thus $P(\tau_{-b}<\infty)=r^{b}$ by continuity of measures. The solution has the same form suggested by @Henry in the comments.