Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 20C / b

On the negative branch set $Y=-X$. Its outward step probability is $3/4$ and inward probability is $1/4$. The same finite gambler's ruin recurrence as in part (a), with root $(1/4)/(3/4)=1/3$, gives ${\mathrm{Pr}}_{-1}(T_0<\infty )=1/3$. Hence, following one departure from zero, the probability of returning isAt each return the Strong Markov property restarts the same excursion experiment. The probability of at least $m$ returns is $r^m$, tending to zero. Thus almost surely there is a last visit to zero.

After that visit the walk remains strictly positive or strictly negative, because nearest-neighbour steps cannot change sign without crossing zero. On the positive branch it can be coupled, up to any return, to independent steps of mean $2/3-1/3=1/3$. The strong law of large numbers gives a walk tending to $+\infty$ almost surely. This conclusion remains true on the event that the excursion never returns: conditioning on a positive-probability event cannot make a probability-zero failure of the law occur. On the negative branch the mean step is $1/4-3/4=-1/2$, giving $-\infty$ in the same way. Apply this at each possible departure time, a countable family, or use the Strong Markov property at the successive returns. ThereforeFinite visits to zero alone would not prove divergence on an arbitrary chain; the outward-drift argument is the additional step needed here.