# Characteristic Function of a Compound Poisson Process

The definition of a compound Poisson process and its characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T$$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$\mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx}$$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

\begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\prod_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\quad\text{(by independence)}\\ &=\sum_n P(N=n)\prod_{j=1}\int e^{iux}f(x)dx\quad\text{(by uniform)} \end{align}

I am not sure how to proceed. Any tips? Thanks for helping! :D

• As a hint, it looks like you can combine the Poisson and Uniform parts of your overall process into one Poisson process. That will greatly simplify things, as you will then just have a sum of i.i.d. random variates with the number of elements of the sum distributed Poisson... – jbowman Aug 23 '16 at 19:52
• @jbowman thank you for the tip, I will try to combine them! – Guilherme Salomé Aug 23 '16 at 19:58
• If you don't see how to do it, I'll help out more later, here's a followup hint. Look carefully at the definition of $Y_t$ and how $U_i$ is used. The question is really very well done, the essential fact that makes it simple is not immediately obvious. – jbowman Aug 23 '16 at 20:04
• @jbowman ok! I'm trying to solve it by myself, so the tip helps a lot. I already found something new using that. I will update it when I understand it better. Thank you very much! :D – Guilherme Salomé Aug 23 '16 at 20:08
• @jbowman I have been trying to solve it for the past few hours, but I think I got very stuck, and I feel like I am wasting time. I found this on combining poisson and uniform. It seemed related to your hint, but I couldn't prove the theorem, even for the case $n=1$. I then proceeded to trying to find the pdf of $Y_1$, but it involves a bunch of convolutions, so I don't think that's the way you intended me to follow. – Guilherme Salomé Aug 24 '16 at 0:17

I was missing the knowledge of the exponential series: $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$ I also made a mistake when I separated the uniform in the expectation. Fixing these problems: \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\prod_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\quad\text{(by independence)}\\ &=\sum_n P(N=n)\left(\mathbb{E}(e^{iu\mathbb{1}_{\{U_1\leq 1\}}X_1})\right)^n\quad\text{(by i.i.d.)}\\ &=\sum_{n=0}^\infty\frac{(\lambda T)^n e^{-(\lambda T)}}{n!}\left(\mathbb{E}(e^{iu\mathbb{1}_{\{U_1\leq 1\}}X_1})\right)^n\quad\text{(by Poisson)}\\ &=e^{-(\lambda T)}\cdot e^{(\lambda T)\mathbb{E}(e^{iu\mathbb{1}_{\{U_1\leq 1\}}X_1})}\quad\text{(by the exponential series)} \end{align} We can calculate the expectation by conditioning on the uniform: $$\mathbb{E}(e^{iu\mathbb{1}_{\{U_1\leq 1\}}X_1})=\frac{T-1}{T}+\frac{1}{T}\int e^{iux}f(x)dx$$ Substituting and doing some algebra we get the answer: \begin{align} \mathbb{E}(e^{iuY_1})&=e^{\lambda \int (e^{iux}-1)f(x)dx} \end{align}

Here's another approach that uses a common trick with characteristic functions to avoid having to work out the sums / integrals.

I'll set $\lambda = 1$ without loss of generality, it simplifies notation and can be put back in in obvious ways in what follows. This means all the "$\lambda$"s below are not related to the $\lambda$ in the problem statement, until the very end where I include it again.

First, note that the definition of $Y_t$ involves the sum of $X_i$ corresponding to $U_i \leq t$. This can be thought of as summing "observed" $X_i$, where an $X_i$ is "observed" with a probability $p = t/T$ that is the same across all $i$.. The number of "observed" $X_i$, label it $n$, is therefore distributed Poisson$(pT)$, which is the same as Poisson$(t)$.

The proof of this is straightforward. The number of observed $X_i$, label it $n$, conditional upon $N$ is clearly distributed Binomial$(N, t/T)$. Now, let's look at the characteristic function (ch.f.) of the Binomial distribution:

$\phi_{n|N}(i\theta) = (1-p+p\text{e}^{i\theta})^N$

We will want to integrate out $N$ w.r.t. the Poisson distribution to get the ch.f. of $n$. The simple way to do this is to note that:

$\phi_{n|N}(i\theta) = \exp(N*\log(1-p+p\text{e}^{i\theta}))$

Writing out the integration (summation) gives us:

$\phi_n(i\theta) = \sum_N \exp(N*\log(1-p+p\text{e}^{i\theta})) p(N|\lambda)$

Looking at this, we can see this will have the same form as the ch.f. of a Poisson distribution ($\exp(\lambda(\text{e}^{i\theta}-1))$), just with $\log(1-p+p\text{e}^{i\theta})$ substituted in wherever $i\theta$ appears in the ch.f. Making this substitution gives us:

$\phi_n(i\theta) = \exp(\lambda \text{e}^{\log(1-p+p\text{e}^{i\theta})} - \lambda)$

which quickly reduces to:

$\phi_n(i\theta) = \exp(\lambda(1-p+p\text{e}^{i\theta}) - \lambda)$

which can be rearranged to:

$\phi_n(i\theta) = \exp(p\lambda(\text{e}^{i\theta}-1))$

which is the ch.f. of a Poisson variate with mean $p\lambda$. Substituting $t/T$ for $p$ and $T$ for $\lambda$ gives us the result.

On to step 2. Now we have the ch.f. of the number of elements in the sum $n$. Let's define $\phi_Y(i\theta)$ as the ch.f. of $Y_t$, $\phi_\Sigma(i\theta)$ as the ch.f. of the sum of $n$ $X_i$ and $\phi_X(i\theta)$ as the ch.f. of a single $X_i$. Since the elements are i.i.d., we know that, conditional upon $n$,

$\phi_\Sigma(i\theta) = \phi_X^n(i\theta) = \exp\{n \log \phi_X(i\theta)\}$

We can apply exactly the same approach as above to integrate out $n$:

$\phi_Y(i\theta) = \sum_n \exp\{n \log \phi_X(i\theta)\} p(n | t)$

where we know that $p(n|t)$ is a Poisson distribution. This will be the ch.f. of a Poisson$(t)$ distribution with $\log \phi_X(i\theta)$ substituted for $i\theta$:

$\phi_Y(i\theta) = \exp\{t(\phi_x(i\theta)-1)\}$

Adding the $\lambda$ from the original problem statement gives the answer:

$\phi_Y(i\theta) = \exp\{t\lambda(\phi_x(i\theta)-1)\}$