Suppose $Y \sim Negative \,Binomial \, (\mu, \theta) $
$X|Y \sim Binomial \, (Y, p) $ (with $\mu, \theta, p$ constant).
What is the distribution of $X$ ? I think $X$ must be Negative Binomial ($\mu_X, \theta_X$) with $\mu_X = \mu.p$ and $\theta_X = \theta$ but can't prove it fully.
My proof is as follow:
$P(Y = y) = \frac{\Gamma(\theta+y)}{\Gamma(\theta).y!} . \frac{\mu^y . \theta^\theta}{(\mu + \theta)^{\mu + \theta}}$ (1)
$P(X = x | Y = y) = \binom{y}{x} . p^x . (1-p)^{y-x}$
$P(X = x, Y = y) = P(X = x | Y=y). P(Y=y)$
Sum over $y = x \to \infty$ we have the marginal distribution of $X$:
$P(X) = \frac{\theta^\theta . p^x . \mu ^ x}{\Gamma(\theta) \, x! \, (\mu + \theta)^{\theta}} \sum \limits_{y = x}^{\infty} \frac{\Gamma(\theta + y) \, \mu^{y-x} \, (1-p)^{y-x}}{(\mu + \theta)^y \, (y-x)!}$
From now, I'm stuck and don't know how to manipulate the $\sum$ part to arrive at the answer.
If you have any idea or suggestion of proof, it would be very nice to me. Thank you very much!
Best Answer
As usual, get the joint pmf and sum w.r.t. $y$
$$\mathbb{P}(x,y)=\binom{\mu+y-1}{y}\binom{y}{x}\theta^\mu(1-\theta)^yp^x(1-p)^{y-x}$$
some esay passages that you know lead to
$$\frac{\theta^\mu p^x(1-\theta)^x}{x!(\mu-1)!}\sum_{y-x=0}^{+\infty}\frac{(\mu+y-1)!}{(y-x)!}[(1-\theta)(1-p)]^{y-x}$$
Here where I have been stuck for some minutes...but then I look at the known series founding this one
$$ \bbox[5px,border:2px solid black] { \sum_{i=0}^{\infty}x^i\binom{n+i}{i}=\frac{1}{(1-x)^{n+1}} \qquad (1) } $$
Thus I multiplied/divided my series for $(\mu+x-1)!$ finding
$$\frac{\theta^\mu p^x(1-\theta)^x(\mu+x-1)!}{x!(\mu-1)!}\sum_{k=0}^{+\infty}\binom{(\mu+x-1)+k}{k}[(1-\theta)(1-p)]^{k}$$
That is using the known series (1) and after some easy simplifications
EDIT: intermediate passage
$$\frac{\theta^\mu[p(1-\theta)]^x}{x!(\mu-1)!}\frac{(\mu+x-1)!}{[1-(1-\theta)(1-p)]^{\mu+x}}$$
$$\mathbb{P}[X=x]=\binom{\mu+x-1}{x}\left[ \frac{\theta}{1-(1-\theta)(1-p)} \right]^\mu\left[ \frac{p(1-\theta)}{1-(1-\theta)(1-p)} \right]^x$$
or
$$\mathbb{P}[X=x]=\binom{\mu+x-1}{x}\left[ \frac{\theta}{\theta+p(1-\theta)} \right]^\mu\left[ \frac{p(1-\theta)}{\theta+p(1-\theta)} \right]^x$$
which is evidently a negative binomial $NB(\mu;\pi)$ where $\pi=\frac{\theta}{\theta+p(1-\theta)} $
I by-passed some passages but I think you will not have problem with this easy algebra