Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / M / M4

The moment generating function is $\boxed{M_X(\theta )=\mathbb{E}[e^{\theta X}]}$ for real $\theta$ where this expectation is finite. Existence on an open interval about zero determines the distribution and permits differentiation there to obtain moments, when the usual local integrability conditions hold.

A Poisson process of rate $\lambda >0$ is a counting process with $N_0=0$, independent stationary increments, and $N_t-N_s\sim \mathrm{Poisson}(\lambda (t-s))$ for $0\le s<t$. It has nondecreasing right-continuous integer-valued paths with unit jumps almost surely. The independent increment requirement distinguishes the process from merely having Poisson marginals.

The interarrival times are independent and identically distributed with law $\mathrm{Exponential}(\lambda )$. In particular any finite subcollection of the first $r$ has joint density ${\lambda }^r\exp (-\lambda {\sum }_{j=1}^r{x}_j)$ on $x_j>0$. Direct integration gives

Under the intended independent-marking interpretation, the coach colors are independent also of the arrival process. Let $K$ be the index of the first white coach. Then $P(K=k)=p(1-p{)}^{k-1}$, $k\ge 1$, independently of the exponential interarrival times. For $p>0$, $T={\sum }_{j=1}^K{X}_j$, so conditioning on $K$ givesThus $\boxed{T\sim \mathrm{Exponential}(\lambda p)}$, the waiting time in the thinned Poisson process. If $p=0$, no white coach arrives and $T=+\infty$. Independence of colors from arrival times is needed for this distribution; mutual independence of colors alone would not rule out marks chosen using the interarrival durations.

The total is the independent random sum $Z_t={\sum }_{j=1}^{N_t}{Y}_j$, with the empty sum zero. Conditional on $N_t=n$, its moment generating function is $M_Y(\theta {)}^n$. The Poisson probability generating function therefore givesAssuming a moment generating function in a neighborhood of zero, write $\mu =\mathbb{E}[Y]$ and $s^2=\mathrm{Var}(Y)$. Conditional expectation and variance yieldThe second contribution to the variance is due to randomness in the number of arriving coaches, not just randomness in the occupancy of each coach.

Since a Poisson count has mean and variance $\lambda t$, substitute the centering and scaling into its moment generating function:This is finite for every real $\theta$ when $t>0$.

For fixed $\theta$, Taylor expansion at zero gives $e^h-1-h=h^2/2+O(h^3)$. With $h=\theta /\sqrt{\lambda t}$, the logarithm of the moment generating function tends to ${\theta }^2/2$. Thus $M_{U_t}(\theta )\to e^{{\theta }^2/2}$ on a neighborhood of zero. The continuity theorem for moment generating functions identifies the limit asThis is the centered Poisson normal limit with its exact variance scaling.