Sum of negative binomial variables with transformed parameters

Question

Let $X_i:i=1,2,...,n$ be independent ~ $NegBin(\mu,\alpha)$random variables such that $E(X_i) = \mu$, $Var(X_i) = \mu + \frac{\mu^2}{\alpha}$. (i) Find the mean and variance of $Y=∑(X_i)$. (ii) Find the MGF of Y and argue that Y ~ $NegBin (n*\mu, n*\alpha)$

The first part is straightforward - $E(Y) =n*\mu$ and $var(Y) = n*(\mu + \frac{mu^2}{\alpha})$

For the second part, the MGF of Y will simply be the product of the MGFs of all the Xs. I know we need to find the pdf of $Y$ or the pdf of each of the X's in terms of $\mu$ and $\alpha$ - This is where I am stuck. The Mean of a Neg Bin variable is $\frac{pr}{(1-p)}$ and its Variance is $\frac{pr}{(1-p)^2}$. Also the MGF is $$\\(\frac{1-p}{1-p*exp(t)})^r\\$$. Assuming that all X's have the same p and r (since no other info is given), the MGF of Y will be $\frac{(1-p)^{nr}}{(1-p*exp(t))^{nr}}$. Now, how does one convert this into terms of $\mu$ and alpha and show that Y ~ $NegBin(n*\mu, n*\alpha)$.

Thank you.

Answer 1

MGF of $Y$ is still in Negative Binomial form, i.e. $Y$ ~ NegBin($p$,$r'=nr$):

$$M_Y(t)=\left( \frac{1-p}{1-pe^t} \right)^{nr}=\left( \frac{1-p}{1-pe^t} \right)^{r'}$$

Now we'll find $\mu_Y$ and $\alpha_Y$ from $p$ and $r'$. $\mu_Y=\frac{pr'}{1-p}=n\frac{pr}{1-p}=n\mu$. Then, we first write $\alpha$ in terms of $\sigma^2$. Based on your first sentence: $\alpha=\frac{\mu^2}{\sigma^2-\mu}$. So, $$\alpha_Y=\frac{\mu_Y^2}{\sigma_Y^2-\mu_Y}=\frac{n^2\mu^2}{\frac{pr'}{(1-p)^2}-n\mu}=\frac{n^2\mu^2}{\frac{npr}{(1-p)^2}-n\mu}=n\frac{\mu^2}{\frac{pr}{(1-p)^2}-\mu}=n\frac{\mu^2}{\sigma^2-\mu}=n\alpha$$