# Help with the probability generating function for a conditional distribution.

I am trying to follow the following proof from Gut's book An Intermediate Course in Probability.

Let $$X$$ and $$N$$ be random variables.

$$N \sim Po(\lambda)$$ and $$X|N=n \sim\operatorname{Bin}(n,p)$$ the probability generating function is defined as $$E[t^{X}]$$ Find the probability generating function for X.

The steps taken in the book are the following. $$g_{X}(t) = E[t^{X}] = E[E[t^{X}|N=n]] = E[(q+pt)^{n}]$$ I get this step. Definition of probability generating function + law of iterated expectation. The next step is where I get stuck.

$$g_{X}(t) = E[(q+pt)^{n}] = g_{N}(q+pt) = e^{\lambda((q+pt)-1)}$$ in particular what confuses me is the step $$E[(q+pt)^{n}] = g_{N}(q+pt)$$

The only thing I can think of that reminds me of this is the convolution/multiplication duality of the transform. That this would be the PGF of an N-fold convolution of i.i.d random variables.

• The step $E[t^{X}] = E[E[t^{X}|N=n]]$ is (quite faulty and) not in the book.
– Did
Dec 17, 2016 at 11:59

I would prefer your earlier step to be written like

$$g_{X}^{\,}(t) = E_{X}^{\,}[t^{X}]= E_N^{\,}[E_{X}^{\,}[t^{X}\mid N=n]] = E_N^{\,}[(q+pt)^{N}]$$

and, since the generating function for $N$ is $g_{N}^{\,}(s) =E_N^{\,}[s^{N}]$, you can let $s=q+pt$ and thus $$g_{N}^{\,}(q+pt) =E_N^{\,}[(q+pt)^{N}]$$

implying $$g_{X}^{\,}(t) = g_{N}^{\,}(q+pt)$$

• The part $E_{X}^{\,}[t^{X}\mid N=n]$ should read $E[t^{X}\mid N]$ (and yes, you might want to drop all the subscripts $N$ or $X$ to the $E$ signs, that one may find only confusing).
– Did
Dec 17, 2016 at 11:18
• Ahhh, so obvious. Thanks. Dec 17, 2016 at 11:26
• @Did - What I intend to mean by $\displaystyle E_N^{\,}[E_{X}^{\,}[t^{X}\mid N=n]]$ is $\displaystyle \sum_{n=0}^\infty \Pr(N=n) \left(\sum_{x=0}^n \Pr(X=x \mid N=n) \,t^X \right)$ Dec 17, 2016 at 11:34
• I know, but first the appearance of $n$ in the identity $E(t^X)=E(E(t^X\mid N=n))$ is frankly wrong since $E(t^X)$ involves no specific $n$ (the correct identity being $E(t^X)=E(E(t^X\mid N))$, as already said), and second, $E(E(t^X\mid N))$ is by definition the double sum in your comment, each inner sum being $E(t^X\mid N=n)$ (that is, if $t^X$ in each inner sum is replaced by $t^x$).
– Did
Dec 17, 2016 at 11:46