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

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.