# A sum of a random number of Poisson random variables

in my probability class I was given this question on which I am stuck concerning a sum of random number of Poisson random variables:

Let us define the countable set of independent random variables $X_i \sim \mathrm{Pois}(\lambda _i)$ and the random variable $N$ independent from the rest of the $X_i$ which also has a Poisson distribution $N \sim \mathrm{Pois}(\lambda)$. We are asked to check if the following sum $Y = \sum_{i=1}^N X_i$ also has a Poisson distribution and if so, with what parameter? As a hint we are asked to look at the characteristic function of the variable to check.

I know the deterministic finite sum of Poisson random variables is again a Poisson random variable with the sum of parameters, but I cannot solve it for a random number of summands that is also Poisson-ly distributed, I know the characteristic function of a Poisson distributed random variable $\Phi(t) = e^{\lambda (e^{it}-1)}$. I thank all helpers who can show me a way out of this.

In the case where all $\lambda_j$ are equal, let's say to $\mu$, you can compute
$$\mathbb E\left[ \exp(itY) \mid N \right] = \exp(\mu (e^{it}-1) N)$$ so that $$\mathbb E[\exp(itY)] = \mathbb E [\exp(\mu(e^{it}-1)N] = \exp(\lambda ( \exp(\mu (e^{it}-1))-1))$$ which is not the characteristic function of a Poisson distribution. However, things are more complicated if the $\lambda_j$ are not all equal.
EDIT: In general, let $s_n = \sum_{j=1}^n \mu_j$. In particular $s_0 = 0$. Then $$\mathbb E \left[\exp(itY) \mid N \right] = \mathbb E \left[\exp(s_N (e^{it}-1))\right] = \exp(-\lambda) \sum_{n=0}^\infty \dfrac{\lambda^n}{n!} \exp(s_n (e^{it}-1))$$ Letting $z = 1 - e^{it}$, we want to know if there is a formula $$\exp(-\lambda) \sum_{n=0}^\infty \dfrac{\lambda^n}{n!} \exp(-s_n z) = \exp(-\rho z)$$ where $\rho > 0$. If this is valid for $z = 1 - e^{it}$, by analyticity it must be valid in the right half plane. Now as $z \to +\infty$, the limit of the left side is $\exp(-\lambda)$ (since $s_0 = 0$), while the limit of the right side is $0$. So this is impossible: the distribution is not Poisson.
• Thank you Robert the case that all parameters are equal is now clear to me but is there a simple method to determine the distribution if the $\lambda$s are not equal? Could it be a Poisson distribution – Don John Prep Mar 21 '16 at 20:04