characteristic function of $\sum_i^N X_i$, $N$ is a Poisson distribution

Question

I have a series of $X_i$ random variables, identically and independent distributed. $S_n=\sum_i^N X_i$, with $N$ which has a Poisson distribution and is independent from $X_i$.

I have to compute the characteristic function.

I tried to apply the law of iterated expectation:

$E[e^{it \sum_i^N X_i}]=E[E[e^{it\sum_i^N X_i}|N]]=E[N]E[e^{it\sum_i^N X_i}] =E[N]\prod_i^NE[e^{itX_i}]$

is that correct?

Answer 1

Suppose $X_i$'s characteristic function is $f(t) = E(e^{itX_i})$, then

\begin{align} E(e^{it\sum_{i=1}^N X_i}) &= E(E(e^{it\sum_{i=1}^N X_i}|N)) \\ & = E(\prod_{i=1}^NE(e^{itX_i})) \\ & = E(f^N(t)) \\ & = e^{-\lambda}\sum_{k=0}^{+\infty}\dfrac{\lambda^k}{k!}f^{k}(t)\\ & = e^{-\lambda}e^{\lambda f(t)} \\ & = e^{(f(t)-1)\lambda} \end{align} where $\lambda$ is the parameter of Poisson distribution

Answer 2

Denoting $\varphi\left(t\right)=\mathbb{E}e^{itX_{1}}$ and assuming that $N\sim Poisson\left(\lambda\right)$ :

$\mathbb{E}\left[e^{it\sum_{j=1}^{N}X_{j}}\right]=\sum_{n=0}^{\infty}\mathbb{E}\left[e^{it\sum_{j=1}^{N}X_{j}}\mid N=n\right]P\left[N=n\right]=\sum_{n=0}^{\infty}\mathbb{E}\left[e^{it\sum_{j=1}^{n}X_{j}}\right]P\left[N=n\right]=e^{-\lambda}\sum_{n=0}^{\infty}\varphi\left(t\right)^{n}\frac{\lambda^{n}}{n!}=e^{-\lambda+\lambda\varphi\left(t\right)}$