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! 😀
Best Answer
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: $$ \begin{align} \mathbb{E}(e^{iuY_1})&=e^{\lambda \int (e^{iux}-1)f(x)dx} \end{align} $$