Originally Posted by

**utopiaNow** Thankyou Laurent for your detailed explanation. Really helped my understanding. I'm interested in the proof you mentioned. ie. Proving that $\displaystyle N_d $ is a poisson r.v. with parameter $\displaystyle \lambda q $. The only generating function I've learned is the moment generating function. And I don't have a good enough conceptual understanding what exactly the m.g.f is other than the definition I've been given to really use it in a proof. So is there a link where they prove this that I could look at? Or perhaps if you have time you could outline the basic principles of such a proof?

Thanks again for the explanation.

I'd rather give you the sketch of a more direct proof ('cause it's easier to explain...):

We need to find $\displaystyle P(N_d=n)$, and we know both $\displaystyle P(N=n)$ (Poisson distribution with parameter $\displaystyle \lambda$) and $\displaystyle P(N_d=n|N=m)$ (Binomial distribution with parameter $\displaystyle (m,q)$). To that aim, let's decompose the event $\displaystyle \{N_d=n\}$ depending on the value of $\displaystyle N$. Of course, $\displaystyle N$ must be larger than $\displaystyle n$. We have:

$\displaystyle P(N_d=n)=\sum_{m=n}^\infty P(N_d=n\mbox{ and }N=m)=\sum_{m=n}^\infty P(N_d=n|N=m)P(N=m)$.

Let replace by the known quantities:

$\displaystyle P(N_d=n)=\sum_{m=n}^\infty {m\choose n}q^n (1-q)^{m-n} e^{-\lambda}\frac{\lambda^m}{m!}$.

Now, write $\displaystyle {m\choose n}=\frac{m!}{n!(m-n)!}$ so that the $\displaystyle m!$ simplifies out, and the term in the sum depend not on $\displaystyle m$ but rather on $\displaystyle m-n$ (write $\displaystyle \lambda^m=\lambda^n\lambda^{m-n}$ also). Since the sum begins at $\displaystyle n$, change the index into $\displaystyle m'=m-n$ (and put the terms independent of $\displaystyle m$ outside). You should recognize the series expansion of the exponential function, and finally get $\displaystyle P(N_d=n)=e^{-\lambda q}\frac{(\lambda q)^n}{n!}$. This is what you wanted.