Sum of random number of random variables is negativ binomial distributed.

Question

I have problems to solve the following: if $N$ is Poisson distributed with $\lambda=-r\ln(p)$ und $Y=\sum_{k=1}^N X_k$ is negativ binomial distributed with parameters $r$ und $p$, what are $X_k$? We know, that $X_k$ are identically distributed and $X_k,N$ are independend. It seems logical to try using generating functions, but the calculations aren't easy. Is there some better way? Thanks in advance.

Answer 1

Then $X_k \sim Log(p)$ for $k = 1, ..., N$. I do not have a proof that does not involve generating functions. However, I would argue that the calculations are actually quite elegant. The probability generating function, $G_X(t) = \mathbb{E}[t^X]$, of random variable $X \sim Log(p)$ and $N \sim \text{Pois}(\lambda)$ are given by: \begin{align*} G_X(t) = \frac{\ln(1 - pt)}{\ln(1 -p)} \hspace{3mm} \text{and} \hspace{3mm} G_N(t) = \exp(\lambda(t - 1)) \end{align*} respectively. Then for Y we have, \begin{align*} G_Y(t) &= \mathbb{E}[t^{\sum_{k=1}^{N}X_k}] \\ &= \mathbb{E}[\mathbb{E}[t^{\sum_{k=1}^{N}X_k}|N]], \hspace{3mm} \text{by law of iterated expectation,} \\ &= \mathbb{E}[\mathbb{E}[\left(t^{X_1}\right)^{N}|N]], \hspace{3mm} \text{by independence,} \\ &= \mathbb{E}\left[\left(\frac{\ln(1 - pt)}{\ln(1 -p)}\right)^N\right] \\ &= \exp\left(-r\ln(1 -p)\cdot\left(\frac{\ln(1 - pt)}{\ln(1 -p)} - 1\right)\right), \hspace{3mm} \text{since, } \lambda = -r\ln(1 - p)\\ &= \exp\left(\ln\left(\frac{\ln(1 - p)}{\ln(1 -pt)}\right)^r\right) \\ &= \left(\frac{\ln(1 - p)}{\ln(1 -pt)}\right)^r. \end{align*} Which is the probability generating function of a negative binomial distribution. By uniqueness of the probability generating function we have that the $X_k$'s must be distributed according to the logarithmic distribution with parameter $p$.

Here is a nice post on the uniqueness of generating functions https://stats.stackexchange.com/questions/34956/proof-that-moment-generating-functions-uniquely-determine-probability-distributi. It focuses on the moment generating function, however, the probability generating function is just a transformation, i.e. $G_X(t) = M_X(\ln(t))$.

I know this is not a clear answer to your question, since it does not tries to circumvent the use of generating functions. Nonetheless, I think it illustrates the power and elegance of generating functions.