I am looking for a definition of a binomial distribution where the number of observation is itself binomial. That is: $X \sim binom(N, q)$ When $N \sim binom(n, p)$.
Is this a known distribution? And how can I find its probability function?
Thanks.
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Say $N$ is Binomial$(n,p)$, and given $N$, $X$ is Binomial$(N,q)$. You can get $N$ by flipping $n$ independent $p$-coins and counting the number of heads; so $X$ corresponds to flipping an additional $q$-coin for each of the original heads, and counting the number of double heads.
This implies that $X$ is Binomial$(n,pq)$.
We can check this by using the formula for the probability generating function for the binomial twice: $$ \mathbb{E}[a^X|N] = \sum_{k=1}^N {N \choose k} q^k (1-q)^{N-k} a^k = (1+aq-q)^N $$ and so $$ \mathbb{E}[a^X] = (1+(1+aq-q)p-p)^n = (1+apq-pq)^n ,$$ which matches the moment generating function for the Binomial$(n,pq)$.
petrelharp
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In other words: for each of $n$ trials, you're counting a success if the $q$-coin and $p$-coin both ame up heads, which is just a fancy way of flipping a $$pq$-coin. – Robert Israel Dec 17 '13 at 14:17
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This is the pgf you are computing ---> not the mgf. – wolfies Dec 17 '13 at 15:12
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oh yeah.... +1... time to delete my own answer... – Lost1 Dec 17 '13 at 19:07
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Thank you Petrelharp. I understood the reasoning, but not the formulation. Could you expand on them a bit (and/or give more links) - to help me better get how you've done it? Thanks. – Tal Galili Dec 18 '13 at 00:14
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@wolfies good point, oops. edited to fix that. – petrelharp Dec 19 '13 at 03:52
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A way to find its probability function is:
$P\left\{ X=k\right\} =\mathbb{E}\left(\binom{n}{k}p^{k}q^{n-k}\right)$ where $n\sim\text{Bin}\left(N,p_{n}\right)$.
(Here $\binom{n}{k}:=0$ if $n<k$.)
This because:
$P\left\{ X=k\mid n\right\} =\binom{n}{k}p^{k}q^{n-k}$ if $k\leq n$ (and $0$ otherwise).
I am not familiar with the distribution.
drhab
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