Past exam of the probability course of the University of Oxford / 2021 / 3 / b / iv

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The binomial distribution giveswhere the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence ${\sum }_{n\ge 0}{\mathbb{P}}_0(S_n=0)=\infty$. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let $L_1,L_2,\dots$ be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives $\mathbb{E}{L}_1=\infty$. For each positive integer $K$, the strong law of large numbers applied to $\min (L_j,K)$ givesTake a common event of probability one for these countably many $K$. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.