Past exam of the options II course of the University of Oxford / 2009 / M / M4

The moment-generating function is $M_X(\theta )=\mathbb{E}(e^{\theta X})$ for those real $\theta$ where this expectation is finite. When it exists on a neighbourhood of zero it determines the distribution, and its derivatives at zero give the corresponding moments.

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 $0\le s<t$, with the usual right-continuous unit-jump sample paths.

The interarrival times are independent and identically distributed with exponential distribution of rate $\lambda$: for any finite $k$ their joint density is ${\lambda }^k\exp (-\lambda {\sum }_{j=1}^k{x}_j)$ on $x_j>0$. In particular

Use independent colour marks, also independent of arrival times, as in the standard marked-process model. Let $K$ be the index of the first white arrival. Then $\mathbb{P}(K=k)=p(1-p{)}^{k-1}$ and $K$ is independent of the interarrival times. Thus $T={\sum }_{j=1}^K{X}_j$. Conditional on $K$, independence makes its moment-generating function a power of $M_X$, and summing the geometric series givesThis is the exponential distribution of rate $p\lambda$, so $T\sim \mathrm{Exp}(p\lambda )$. The calculation assumes $p>0$; if $p=0$, there is no white arrival and $T=+\infty$ almost surely.

Write $Z_t={\sum }_{j=1}^{N_t}{Y}_j$, with an empty sum equal to zero. Conditioning on the independent Poisson random variable $N_t$, the moment-generating function isThis is the compound Poisson distribution. Differentiating its log moment-generating function at zero, when the first two moments are finite, givesThe variance includes variability in the number of coaches, and so uses the second moment, not merely $\mathrm{Var}(Y)$.

Since $\mathbb{E}{N}_t=\mathrm{Var}(N_t)=\lambda t$, use the Poisson moment-generating function to get, for $t>0$,This is finite for every real $\theta$.

For each fixed $\theta$, the Taylor expansion of the exponential yieldsThe limiting moment-generating function $e^{{\theta }^2/2}$ is that of the standard normal distribution and is finite around zero. The moment-generating function continuity theorem therefore gives