Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / M / M5 / e

The stationarity equation at an interior state collects arrivals from its two neighbors and itself:This rearranges to the required recurrence. At zero, stationarity gives ${\pi }_0=(1-a){\pi }_0+b{\pi }_1$, hence $\boxed{b{\pi }_1=a{\pi }_0}$.

The recurrence says the flux $a{\pi }_{i-1}-b{\pi }_i$ equals $a{\pi }_i-b{\pi }_{i+1}$. The boundary equation makes that flux zero, so ${\pi }_i=(a/b{)}^i{\pi }_0$. Since $0\le a/b<1$, normalization givesWhen $a=0$ this means ${\pi }_0=1$ and ${\pi }_i=0$ for $i>0$. The probability assumptions also require $a,b\ge 0$, $a+b\le 1$, as implicit in the transition rules.

For $a>0$ the chain is irreducible, and the normalized stationary distribution in (i) makes it positive recurrent. The mean-return-time formula givesHere $T_0^{+}=\inf \{n\ge 1:X_n=0\}$ includes a return at time one when the chain stays at zero. This convention is essential. Equivalently the mean hitting time of zero from one is $1/(b-a)$, so a first-step decomposition gives $1+a/(b-a)$, the same result. For $a=0$ the population remains zero and the next zero is at time one, also agreeing with the formula.