# Multivariate (?) Probability Generating Function

We have variables $X_1,X_2,X_3..$ which are i.i.d Bernoulli RVs with parameter $p$. Also, $N \sim Po(\lambda)$.

With this given, how can I find the PGF of $Z = \sum_{i=1}^{N}X_i$?

Using rules of PGF, I worked the math out and I have

$G_z(s)=E(s^z)=E(s^{x_1+x_2+..+x_N})$

Then,

$E(s^{x_1+x_2+..+x_N}) = \prod_{k=1}^{N}E(s^{X_k})$

From this, we now see

$G_{X_1}(s) * G_{X_2}(s) * ... * G_{X_N}(s) = [G_X(s)]^N$

Solving this, I obtain $[G_X(s)]^N$, where $G_X(s)$ is a PDF of bernoulli.

I guess my question is, I'm not quite sure how to incorporate the fact that N is a poisson random variable. Obviously the exponent $N$ is dependent on poission... then "should" the exponent be $\lambda$ since our expected value for a poission RV is its parameter $\lambda$?

Also with this, then I think that $Z$ is a binomial random variable, but I'm not quite sure if $N$ makes any difference.

• Your essential mistake is the equality $E(s^{X_1+\cdots+X_N})=\prod_{k=1}^NE(s^{X_k})$. It is wrong because $N$ is a random variable, and not a constant. On LHS you find an expectation (which is just some real number, or function is $s$ if you like) and on RHS you find a random variable. That does not match. It is comparable with (wrongly) saying that $\mathbb EXY=X\mathbb EY$ where $X$ and $Y$ both denote random variables. Mar 20, 2018 at 20:52

$$\mathbb Es^Z=\sum_{n=0}^{\infty}\mathbb E(s^Z\mid N=n)P(N=n)=\sum_{n=0}^{\infty}\mathbb E(s^{X_1+\cdots+X_n})P(N=n)=$$$$\sum_{n=0}^{\infty}G_X(s)^nP(N=n)=\mathbb E\left[G_X(s)^N\right]$$where $X\sim\mathsf{Bernoulli}(p)$ and $N\sim\mathsf{Po}(\lambda)$.
The second equality is based on independence of $N$ wrt the $X_i$. and the third on independence of the $X_i$.
Can you find $G_X(s)$ and $\mathbb E\left[G_X(s)^N\right]$ yourself?
• I worked out $G_X(s)$ and I have $G_X(s) = 1-p+2p$. I don't really follow why I need to find $E[G_X(s)^N]$ and not $[G_X(s)]^N$ as I have suggested above Mar 20, 2018 at 16:30
• You are asked to find the PGF of $Z$ which is a function. Not a random function. Mar 20, 2018 at 16:49
• The correct expression for $G_X(s)$ is $\mathbb Es^X=s^1P(X=1)+s^0P(X=0)=sp+1-p$. I hope that the $2$ in your comment is a typo. There are $4$ equalities in my answer. Do you understand (and agree with) all of them? If not then let me know which one you do not understand. Mar 20, 2018 at 17:31
• Yes, my 2 was a typo - I meant $1-p+sp$ as you have suggested. I follow your explanations and equalities, but the results do not seem to match with the way I worked it out - I'll edit my answer to fully show my thought process. Mar 20, 2018 at 18:53