# Sum of negative binomial variables with transformed parameters

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.

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, $$\alpha_Y=\frac{\mu_Y^2}{\sigma_Y^2-\mu_Y}=\frac{n^2\mu^2}{\frac{pr'}{(1-p)^2}-n\mu}=\frac{n^2\mu^2}{\frac{npr}{(1-p)^2}-n\mu}=n\frac{\mu^2}{\frac{pr}{(1-p)^2}-\mu}=n\frac{\mu^2}{\sigma^2-\mu}=n\alpha$$