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The equilibrium-delay theorem for a renewal process with positive holding time $S$, finite mean $\mu$, and distribution function $F$ says that the first future renewal must have the equilibrium residual-life distribution,independently of subsequent ordinary holding times. This is the residual lifetime seen at a stationary observation time, rather than an ordinary inter-renewal lifetime.

Here $\mu =3/\lambda$ and $1-F(t)=e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. Thus the required initial-delay density isEquivalently, choose with equal probability a sum of one, two, or three independent rate-$\lambda$ exponential random variables. The three Erlang distributions have densities that average to the displayed density. A concrete stationary construction chooses the phase modulo three of a stationary Poisson clock uniformly; the number of exponential stages until its next third-stage renewal is then uniform on $\{1,2,3\}$.

For a renewal holding time $T$ with mean $m$, the equilibrium residual-life distribution has survivor $m^{-1}{\int }_t^{\infty }\mathbb{P}(T>s)ds$. For C, and also the common long-run B/C renewal structure, $m=2/r$ and the Erlang survivor giveFor A the exponential law is already its own equilibrium delay.

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.