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!


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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