Let {$S_n$} be a modified random walk on the positive integers, where $P(S_n=k+1|F_n)$ is no longer $\frac{1}{2}$ but instead is $\frac{1}{2} - \frac{1}{4k}$ Use the harmonic function $f(n) = n^2$ to compute the probability that the walk, if started from 50 and absorbed when it hits 0, will ever reach 100.
I'm very confused about this problem. I mostly want a solution and explanation to the problem itself.
However, if it's not too much trouble, I would also like to know why this question discusses $P(S_n=k+1|F_n)$ and not $P(S_n=k+1|F_{n-1})$? Isn't {$S_n$} adapted to {$F_n$}, which would make $P(S_n=k+1|F_n)$ = $E(1_{S_n=k+1}|F_n) = 1_{S_n=k+1}$? What am I missing?
Also, what exactly is k?
Okay: So writing it up cleanly. We claim that $f(S_n)$ is a martingale. Note that $$ \mathbb{E}(f(S_{n+1})-f(S_n)|F_n)=1_{(S_n>0)}\left(\left(\frac{1}{2}-\frac{1}{4 S_n}\right)\left(1+2 S_n\right)+\left(\frac{1}{2}+\frac{1}{4 S_n}\right)\left(1-2 S_n\right)\right)=0 $$ Then, if $\tau$ is the first hitting time of $\{0,100\}$, we see that $f(S_{n\wedge \tau})$ is a uniformly bounded martingale by optional sampling. Applying dominated convergence, we get
$$ 10^4\mathbb{P}(S_{\tau}=100)=\mathbb{E}(f(S_{\tau}))=\lim_{n\to\infty} \mathbb{E}(f(S_{n\wedge \tau}))=\lim_{n\to\infty} \mathbb{E}(f(S_0))=10^2 \cdot5^2 $$
This implies that $$ \mathbb{P}(S_{\tau}=100)=\frac{1}{4} $$ Which gives you your solution.