Past exam of the options II course of the University of Oxford / 2010 / M / M5 / e

For $i\ge 1$, the stationarity equation iswhich rearranges to the stated recurrence. At zero,Define the adjacent probability current $J_i=a{\pi }_i-b{\pi }_{i+1}$. The recurrence says $J_i=J_{i-1}$, and the boundary equation says $J_0=0$. Hence ${\pi }_{i+1}=(a/b){\pi }_i$ for all $i$. Since $\rho =a/b<1$, normalization givesThe adjacent balance equations also verify stationarity directly. If $a=0$, this formula means the point mass at zero; otherwise the birth-death chain is irreducible and this is its unique stationary probability distribution.

Let $T_0^{+}=\inf \{n\ge 1:X_n=0\}$. For $a>0$, the chain is irreducible and its stationary probability distribution makes it positive recurrent. The mean-return-time identity givesThis counts a self-loop at zero as a return after one time unit, as “next zero” requires. Equivalently, with probability $1-a$ that immediate return occurs; with probability $a$ the chain moves to one and then has mean hitting time $1/(b-a)$ back to zero, giving $1+a/(b-a)$. If $a=0$, the return is deterministically at time one and the displayed formula still equals one.