Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 26J (Applied Probability)

An Inhomogeneous Poisson process with deterministic, nonnegative locally integrable rate $\lambda (t)$ is a counting process $N$ with $N(0)=0$, independent increments, and increment distributionEquivalently, for continuous rates it is a simple counting process with independent increments and infinitesimal probabilities $\mathbb{P}(N(t+h)-N(t)=1)=\lambda (t)h+o(h)$ and $\mathbb{P}(N(t+h)-N(t)\ge 2)=o(h)$. For merely locally integrable rates use the integrated-rate formulation, which avoids a pointwise regularity assumption.

To derive the distribution from the infinitesimal characterization, let $p_j(t)=\mathbb{P}(N(t)=j)$. Independent increments give the forward equationsWith $L(t)={\int }_0^t\lambda (u)du$ and $p_0(0)=1$, $p_j(0)=0$ for $j>0$, these have solutionSubstitution verifies each equation, and the probabilities sum to one. Alternatively the probability generating function solves ${\partial }_{t}G=\lambda (t)(z-1)G$, $G(z,0)=1$, giving $G=e^{(z-1)L(t)}$. For a general locally integrable rate, time-changing a unit-rate Poisson process by $L(t)$ gives the same formula.

Put $L_t={\int }_0^t\Lambda (u)du$, finite almost surely. Conditional on the intensity process, $N(t)$ is Poisson with mean and variance $L_t$. The law of total expectation and law of total variance giveConsequently for a Cox process,These identities apply when the moments are finite; if the mean is finite but the integrated intensity has infinite variance, the count variance is infinite and the inequality still holds.

Let $P=N(t)\sim \mathrm{Poisson}(x)$, where $x=\lambda t$. Deleting odd arrivals leaves $M=\lfloor P/2\rfloor$. Put $Q=1_{\{P\text{ odd}\}}$, so $M=(P-Q)/2$. From the probability generating function $\mathbb{E}{z}^P=e^{x(z-1)}$,Thus $\mathrm{Var}Q=(1-e^{-4x})/4$ and $\mathrm{Cov}(P,Q)=x{e}^{-2x}$. ThereforeTo establish strict underdispersion for every $x>0$, subtract the mean:Here $f(0)=0$ and $f^{\prime }(x)=4[-1+4x{e}^{-2x}+e^{-4x}]<0$ for $x>0$, because $4x{e}^{-2x}<1-e^{-4x}$ is equivalent to $2x<\sinh (2x)$. Hencewhich excludes a doubly-stochastic Poisson representation by part (iii).