 Solution

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.