Past exam of the options II course of the University of Oxford / 2009 / M / M5 / a / ii / Solution

First assume $0<q<p<1$. Stop the nearest-neighbour walk on reaching zero or a finite upper level $L>i$. Its probability $h_i^{(L)}$ of reaching zero first solves $h_i^{(L)}=p{h}_{i+1}^{(L)}+q{h}_{i-1}^{(L)}$, with boundary values one and zero. Solving this second-order recurrence givesAbsorption in the finite interval occurs almost surely: from every interior state there is a uniformly positive probability of an absorbing run within $L$ steps. As $L\to \infty$, the events of hitting zero before $L$ increase to the event of ever hitting zero, since a path reaching zero at a finite time has a finite maximum before then. ConsequentlyThis limiting argument selects the actual hitting probability, rather than just an arbitrary bounded solution of the infinite recurrence. If $q=0$, the walk moves deterministically upwards from positive states and the probability is zero there; from state zero it is one. These agree with the formula interpreted separately at $i=0$.