What is the distribution of a random variable whose pmf is a function of other pmfs? I'm interested to find out what is the resulting distribution for a discrete random variable $Y$ whose probability mass function is defined as:
$$
Pr(Y = k) = C \cdot  \frac{Pr(X_1 = k) \cdot Pr(X_2 = k)}{Pr(X_3 = k)}  
$$
for $k = 0, 1, 2,\ldots$, where each of the (independent) $X_i$'s follows a Negative Binomial distribution with $X_i \sim NB(r_i, p_i)$ for $i = 1, 2, 3$, while $C$ is the normalizing constant.
I've managed to show that when $X_i \sim Poisson(λ_i)$ then $Y \sim Poisson(\frac{λ_1 λ_2}{λ_3})$ but have so far failed to get somewhere with the negative binomial case.
Starting from:
$$
Pr(X_i = k) = {k + r_i - 1 \choose k} {p_i}^{r_i} (1 - p_i)^k
$$
I've managed to find (I think!) that the pmf of $Y$ would include this quantity:
$$
\frac{{p_1}^{r_1} {p_2}^{r_2}}{{p_3}^{r_3}} \cdot \bigg[\frac{(1-p_1)(1-p_2)}{(1-p_3)}\bigg]^k 
$$
and other factorials / Gamma functions involving $r_1, r_2, r_3$ and $k$, but couldn't reach the end. My gut feeling (backed by some simulations) is that $Y$ will also follow a negative binomial distribution but can't prove if that's the case and, if so, what the corresponding expressions for $r_y$ and $p_y$ (in terms of the $r_i$'s and the $p_i$'s) will be.
Any help will be much appreciated. Thanks!  
 A: On the face of it it seems  your suggestion could work for the negative binomial if $r_1=r_2=r_3$ even if $p_1,p_2,p_3$ differ so long as $p_3$ is small enough, but might not work if $r_1,r_2,r_3$ differ.  

Suppose $r_1=r_2=r_3=r$. Then 
$$\Pr(Y = k) = C \cdot  \frac{\Pr(X_1 = k) \cdot \Pr(X_2 = k)}{\Pr(X_3 = k)}  \\ = C \cdot  \frac{{k + r - 1 \choose k} {p_1}^{r} (1 - p_1)^k \cdot {k + r - 1 \choose k} {p_2}^{r} (1 - p_2)^k}{{k + r - 1 \choose k} {p_3}^{r} (1 - p_3)^k} \\ = {k + r - 1 \choose k} C \cdot  \frac{{p_1}^{r}  \cdot  {p_2}^{r} }{ {p_3}^{r} }\left(\frac{(1 - p_1) \cdot   (1 - p_2)}{ (1 - p_3)}\right)^k $$ 
and now let $p=1-\frac{(1 - p_1) \cdot   (1 - p_2)}{ (1 - p_3)}$ and $C=\frac{ {p}^{r}  \cdot{p_3}^{r} }{{p_1}^{r}  \cdot  {p_2}^{r} }$ so $$\Pr(Y = k) = {k + r - 1 \choose k} p^r \left(1-p\right)^k$$ which is a negative binomial, at least when $0 < p < 1$ which requires $1-p_3 > (1-p_1)\cdot (1-p_2)$; you need that condition anyway as otherwise  $\frac{\Pr(X_1 = k) \cdot \Pr(X_2 = k)}{\Pr(X_3 = k)}$ would be an increasing function of $k$ and so have an infinite sum making $C$ zero.

Now for a counterexample where $r_1,r_2,r_3$ differ.  Suppose $r_1=r_2=1$ but $r_3=2$, and $p_1=p_2=p_3=\frac12$.  Then   
$$\Pr(Y = k) =  C \cdot  \frac{{k \choose k} \frac12 \frac1{2^{k}}\cdot {k \choose k} \frac12 \frac1{2^{k}}}{{k+1 \choose k} \frac1{2^2} \frac1{2^{k}}} = C \cdot  \frac{ 1}{(k+1){2^{k}} }$$
which will require $C=\frac{1}{2\log_e(2)}$.  This $Y$ will have mean $\frac{1}{\log_e(2)}-1\approx 0.4427$ and variance  $\frac{2}{\log_e(2)}- \frac{1}{\log_e(2)^2}\approx 0.8040$ and $\Pr(Y = 0) = C \approx 0.7213$.
If this $Y$ were negative binomial then this mean and variance would suggest $p = \frac{\log_e(2)\left(1-\log_e(2)\right)}{2\log_e(2)-1}\approx 0.5506$ and $r=\frac{(1-\log_e(2))^2}{\log_e(2)^2 +\log_e(2)-1} \approx 0.5424$.  That $r$ is not an integer, but even if we allowed that, those $p$ and $r$ would suggest $\Pr(Y = 0)=p^r \approx 0.7235$ which is wrong, so we can conclude that this $Y$ does not have a negative binomial distribution
