Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 28J (Applied Probability)

Superpose the two independent Poisson processes. The combined process has rate $9$, and each arrival is independently a Number 1 bus with probability $1/9$ and a Number 8 bus with probability $8/9$. Therefore the number of Number 8 buses before the first Number 1 has the geometric distribution on $\{0,1,2,\dots \}$, andThis is number of one Poisson type before another with rates $1$ and $8$.

For $0<j_1<\cdots <j_n<t$, the probability of one jump in each infinitesimal interval around $j_1,\dots ,j_n$ and no other jump by time $t$ gives the joint densityfor $(J_1,ldots,J_n)$ together with the event $N_t=n$. Sincethe conditional density ison that ordered simplex.

Now $n$ independent uniform points on $[0,t]$ have joint density $t^{-n}$ on the cube. Each ordered vector has $n!$ permutations, so their order statistics have density $n!/t^n$ on the same simplex. The densities agree, proving Poisson process conditional arrival times.

Condition on $N_t=n$. By Poisson process conditional arrival times, the jump times are the ordered version of independent $U_1,\dots ,U_n$ uniformly distributed on $[0,t]$. The sum is symmetric in the marked pairs, and the marks are independent, soWrite the bracket as $a$. Averaging over the Poisson distribution of $N_t$ givesThis is the exponential formula for a marked Poisson sum.

Use the convention $J_0=0$, so $A(t)=t$ when $N_t=0$. Conditional on $N_t=n\ge 1$, the last jump $J_n$ is the largest of $n$ independent uniform points on $[0,t]$. The stated order statistic formula givesand hence, for every $n\ge 0$,Putting $\mu =\lambda t$ and averaging over $N_t$,This agrees with age of a Poisson process.