# Combining geometric distribution with binomial

Let $N \sim \operatorname{Geo}(p)$, so $\mathbb{P}(N=k)=(1-p)^k p$, $k=0,1,2,\dots$, and $N_R | N \sim \operatorname{Bin}(\alpha,N)$.

I am trying to find a distribution of $N_R$, it is fairly easy to write down $$\mathbb{P}(N_R=k)=\sum_{n=k}^\infty \binom{n}{k} \alpha^k(1-\alpha)^k(1-p)^n p,$$ but I do not see an easy way of evaluating it.

Another approach is to look at the probability generating functions, to deduce that $$G_{N_R}(z)=G_N(\alpha z+1-\alpha)=\frac{p}{1-(1-p)(\alpha z+1-\alpha)},$$ and I can't recognise this as a PGF of anything familiar.

Is the distribution of $N_R$ well-known distribution (or a mixture of two well-known distributions), but I just don't see it?

Thank you.

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Hint: There exists $q$ depending on $p$ and $\alpha$ such that $G_{N_R}(z)=\dfrac{q}{1-(1-q)z}$, hence $N_R$ is geometric with parameter $q$.

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Is there a way to see that such q exists straight away? – Tom Artiom Fiodorov May 9 '12 at 16:13
Yes, the pgf is a rational function with constant numerator and linear denominator, like that of the geometric rv. This, along with pgf(1) = 1 fixes all the coefficients uniquely. – Sasha May 9 '12 at 16:17
Of course, Maclaurin expansion will make it a geometric RV! – Tom Artiom Fiodorov May 9 '12 at 16:20