Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 101 / 5

From the masses of the Poisson distribution, its probability generating function isAt $z=0$, the term $z^0$ is interpreted as one, so this also gives $\mathbb{P}(Z=0)=e^{-\lambda }$.

One constructive definition of a rate-$\lambda$ Poisson process uses independent and identically distributed exponential random variables $E_j$ of rate $\lambda$. Set $T_0=0$, $T_k={\sum }_{j=1}^k{E}_j$ andThe strong law of large numbers gives $T_k/k\to 1/\lambda$ almost surely, so the maximum is finite at every finite $t$. The process starts at zero, has right-continuous nondecreasing integer paths, and jumps by one at the arrival times.

For $k\ge 1$, exactly $k$ arrivals by $t$ means $T_k\le t<T_{k+1}$. Integrating the first $k$ exponential gaps and the survival probability of the next gap givesThe last factor is the volume of the $k$-dimensional simplex, obtained for example by iterated integration. For $k=0$, directly $\mathbb{P}(N_t=0)=\mathbb{P}(E_1>t)=e^{-\lambda t}$. ThusAt a fixed time $s$, conditional on the counting history, the residual time until the next arrival has rate-$\lambda$ exponential distribution, independently of that history, by the memoryless property. After that arrival all subsequent gaps are fresh independent exponentials. Thus the process of future counts $N_{s+u}-N_s$ has the same law as $N_u$ and is independent of the past. This proves stationary independent increments, the equivalent usual defining properties of the Poisson process, and gives

Conditional on $Z=k$, the summands are independent with the common Laplace transform of a nonnegative random variable $\varphi (q)$. The empty sum for $k=0$ is zero. Therefore the compound Poisson distribution hasand applying the probability generating function from part (a) givesIf $m=\mathbb{E}{X}_1<\infty$, differentiating at zero from the right gives ${\varphi }^{\prime }(0+)=-m$ and $L_Y^{\prime }(0+)=-\lambda m$. HenceThe differentiation is justified by $(1-e^{-q{X}_1})/q\le X_1$ and the dominated convergence theorem. No finite moment was assumed in the question, so the formula must also be interpreted when $m=\infty$. For any nonnegative random variable $V$, the monotone convergence theorem givesApply this to $X_1$ and $Y$, and use $1-e^{-\lambda h}\sim \lambda h$ as $h\downarrow 0$, with $h=1-\varphi (q)$; it yields the same formula with value $+\infty$ when $m=\infty$. If $m=0$, nonnegativity makes all marks zero almost surely and the result is immediate.

For the second method, condition on the count rather than differentiating a transform. For finite $m$, the tower property of conditional expectation gives $\mathbb{E}[Y\mid Z]=Zm$, and $\mathbb{E}Z=\lambda$ gives $\mathbb{E}Y=\lambda m$. More generally, Tonelli's theorem and independence giveFor $m=\infty$, already the term $j=1$ is infinite since $\mathbb{P}(Z\ge 1)>0$. Thus the two methods agree, including in the extended-valued case.

As in the intended independent-mark construction, take the entire sequence $(X_j)$ independent of the Poisson process, not merely independent of one fixed count. Then $Y$ is a Compound Poisson process. For deterministic $s<t$, its increment $Y_t-Y_s$ is independent of ${\mathcal{F}}_s=\sigma (Y_u:0\le u\le s)$. To justify this carefully, first condition on $N_s=k$. The past of $Y$ uses only the counting history up to $s$ and marks among $X_1,\dots ,X_k$. The future count has independent rate-$\lambda$ Poisson process increments, and future marks $X_{k+1},X_{k+2},\dots$ are fresh with the common law. Their joint conditional law is the same for every $k$ and every past history, so the increment is independent even of this larger marked history, hence of ${\mathcal{F}}_s$. Marks equal to zero do not cause a problem: ${\mathcal{F}}_s$ may then reveal less than the full counting history.

By part (b), with Poisson parameter $\lambda (t-s)$,The proposed process is adapted to ${\mathcal{F}}_t$, nonnegative, andIn particular, no finite expectation of a mark is required. Also $Z_s>0$, andis independent of ${\mathcal{F}}_s$ and has mean one. Therefore the exponential Laplace martingale of a compound Poisson process satisfiesThis proves every condition for a continuous-time martingale in the requested filtration. When $q=0$, $\varphi (0)=1$ and the same conclusion reduces to the constant martingale $Z_t=1$.