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.

  • $\begingroup$ 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 $\endgroup$ – Don John Prep Mar 21 '16 at 20:04
  • $\begingroup$ I figured since N is independent from all other Xs, maybe this can be useful? $\endgroup$ – Don John Prep Mar 21 '16 at 20:05

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