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The only closed communicating class is $\{b,d\}$; the class $\{a,c\}$ is open and transient. Its one-step probability of staying inside is at most $2/3$, so if $\tau$ is the entrance time into $\{b,d\}$, then $\mathbb{P}(\tau >n)\le (2/3{)}^n$ for an initial state in $\{a,c\}$. Hence entrance occurs almost surely in finite time.

On $\{b,d\}$ it is an irreducible aperiodic Markov chain, because both states have positive self-loop probabilities. The finite irreducible aperiodic countable-state Markov chain convergence theorem says that its transition probabilities converge to its unique stationary distribution. The balance equation ${\pi }_b(3/4)={\pi }_d(2/3)$ with ${\pi }_b+{\pi }_d=1$ gives ${\pi }_b=8/17$ and ${\pi }_d=9/17$.

This limit also holds from $a,c$: condition on the finite entrance time and entrance state. For $\tau \le M$, convergence in the closed class applies to a finite sum; the remaining probability is bounded by $\mathbb{P}(\tau >M)$, which tends to zero as $M\to \infty$. Thus in the order $a,b,c,d$,

Every reachable state $i$ reaches zero deterministically in $i$ steps. For $j\in E$, choose a supported lifetime $k>j$; zero can jump to $k-1$ and then count down to $j$. Thus the chain is an irreducible Markov chain on $E$. Its positive-probability first return lengths are exactly the support of $Y_1$; all other return lengths are sums of such lengths. Their greatest common divisor is therefore one by assumption, so the chain is an aperiodic Markov chain.

The equilibrium residual-life distribution isIt is a probability distribution because the tail-sum identity gives ${\sum }_{j\ge 0}\mathbb{P}(Y_1>j)=\mathbb{E}{Y}_1=\mu$. With zero masses outside $E$,so it is a stationary distribution. This also displays why finite mean is needed for normalization.

Use the countable-state Markov chain convergence theorem: an irreducible, aperiodic, positive recurrent discrete-time chain on a countable state space has a unique stationary distribution and $P^n(i,j)\to {\pi }_j$ for every pair of states. All hypotheses were verified, including positive recurrence in part (b). Since $X_0=0$ and $\mathbb{P}(Y_1>0)=1$,The finite-state convergence theorem alone would not suffice when the lifetimes are unbounded. The gcd hypothesis is also necessary for this ordinary limit: deterministic lifetime 2, for example, produces alternating renewal probabilities rather than convergence.