Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $q$.

share|improve this answer
    
Is there a way to see that such q exists straight away? –  Tom Artiom Fiodorov May 9 '12 at 16:13
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.