I have solved this issue. I was expecting some functionality on subscripted random variables that is not available.
You shouldn’t be using N are a variable name as it is a reserved function. It’s best practice to start the names of variables with lowercase letters.
Also what do you propose for when n = 0 which happens with a positive probability for a Poisson random variable?
If you consider a truncated Poisson (i.e., getting rid of 0), then Mathematica can do all of the heavy lifting.
First we find the distribution of a truncated Poisson:
d1 = TruncatedDistribution[{1, \[Infinity]}, PoissonDistribution[\[Lambda]]]
For the sum of n independent normal random variable with mean \mu and variance \sigma^2, the resulting mean and variance of that sum are n \mu and n \sigma^2. We use the ParameterMixtureDistribution function to find the distribution of the sum of a random number of normal random variables:
d2 = ParameterMixtureDistribution[NormalDistribution[n \[Mu], Sqrt[n] \[Sigma]], n \[Distributed] d1]
The mean and variance are found with the following:
mean = Mean[d2] // FunctionExpand // FullSimplify
\frac{\left(e^{\lambda }-1\right) \lambda \mu }{-\lambda +e^{\lambda }-1}
variance = Variance[d2] // FunctionExpand // FullSimplify
\frac{\lambda \left(\left(e^{\lambda }-1\right) \left(-\lambda +e^{\lambda }-1\right) \sigma ^2-e^{\lambda } \mu ^2 \left(\lambda ^2-2 \cosh (\lambda )+2\right)\right)}{\left(\lambda -e^{\lambda }+1\right)^2}
If \lambda is large (such the the probability of zero is very low), then one obtains the following for \lambda=15:
mean /. \[Lambda] -> 15 // N
(* 15.0001 \[Mu] *)
variance /. \[Lambda] -> 15 // N // Expand
(* 14.9991 \[Mu]^2 + 15.0001 \[Sigma]^2 *)
Right, thanks.
If the Poisson is 0, the sum is equal to 0.
Found the desired function! CompoundPoissonDistribution