Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 26J (Applied Probability)

Let $S_1,S_2,\dots$ be independent exponential random variables of rate $\lambda$, let $T_n=S_1+\cdots +S_n$ and $T_0=0$, and define $N_t=\max \{n:T_n\le t\}$. The strong law of large numbers gives $T_n/n\to 1/\lambda$, so there is no finite-time explosion. This defines the Poisson process of rate $\lambda$.

For $n\ge 1$, repeated convolution gives $T_n$ the gamma distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$. Independence of $S_{n+1}$ givesFor $n=0$, the probability is $\mathbb{P}(S_1>t)=e^{-\lambda t}$. Hence $\boxed{N_t\sim \mathrm{Poisson}(\lambda t)}$ for all $t\ge 0$, including the degenerate law at $t=0$.

Convolve three rate-$\lambda$ exponential distributions. The first two have convolution density ${\lambda }^{2}x{e}^{-\lambda x}{1}_{x>0}$; convolving once more givesThus $X$ is an Erlang distribution of shape three, equal in law to the sum of three independent exponentials. Addition of independent means and variances yieldsFor $t\ge \lambda$ the moment-generating function is infinite.

Realize each holding time as a consecutive block of three exponential holding times from one Poisson process $N_t$. The renewal count is then $X_t=\lfloor N_t/3\rfloor$ pathwise under this coupling. If $R_t$ is the remainder of $N_t$ modulo three, $N_t=3X_t+R_t$ and $\mathbb{E}{R}_t=p_1(t)+2p_2(t)$. Taking expectations givesThis proves the formula without assuming that the renewal count itself is Poisson.

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 $\{1,2,3\}$.