Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / M / M5 / a

For $i\notin A$, condition on the first step. The Markov property makes the chance of eventual entry after reaching $j$ equal to $h_j$, including $h_j=1$ when $j\in A$. Total probability givesThe countable sum is justified by its nonnegative terms. This first-step identity alone need not uniquely determine a hitting probability on an infinite state space; one must choose the minimal nonnegative solution or supply the appropriate boundary behavior.

For a finite upper barrier $L>i$, let $h_i^{(L)}$ be the probability of hitting zero before $L$. It satisfies $h_0^{(L)}=1$, $h_L^{(L)}=0$, and $h_i^{(L)}=p{h}_{i+1}^{(L)}+q{h}_{i-1}^{(L)}$. For $0<q<p$, solving the second-order recurrence with roots $1$ and $r=q/p$ givesAs $L\to \infty$, the events increase to eventual hitting of zero: any finite path reaching zero has a finite maximal height. Since $r<1$, this yields $\boxed{h_i=(q/p{)}^i}$, with $h_0=1$. In the endpoint case $q=0$, a positive initial state only moves upward and the hitting probability is zero, agreeing with the positive-state formula. The finite-barrier argument selects the correct solution rather than the also-harmonic constant one.

After the first step to $+1$, the right-hand walk moves away with probability $3/4$ and toward zero with probability $1/4$. Part (ii) therefore gives return probability $1/3$ from that side. After a first step to $-1$, reverse the sign of the coordinate: movement away from zero has probability $2/3$ and toward it probability $1/3$, giving return probability $1/2$. Weight by the probabilities of the first step from zero:There is no holding step at zero, so every positive-time return is included by this decomposition.