# Derivation of a conditional probability mass function which involves Geometric and Gamma random variables

conditional probabilitygamma distributionprobability

In the $$6$$th chapter of the textbook A first course in probability by Ross, there is the following question:

Let $$N$$ be a geometric random variable with parameter $$p.$$ Suppose that the conditional distribution of $$X$$ given that $$N = n$$ is the gamma distribution with parameters $$n$$ and $$\lambda.$$ Find the conditional probability mass function of $$N$$ given that $$X = x.$$

At first, I'll setup notation and a solution. Then, I'll start towards another solution, which I cannot complete at the moment.

We are given $$f_{X|N=n}(x) = \frac{\lambda e^{- \lambda x} (\lambda x)^{n-1}}{\Gamma(n)} = \frac{\lambda e^{- \lambda x} (\lambda x)^{n-1}}{(n-1)!},$$ since $$n$$ is a positive integer; and we want to find $$P(N=n | X=x).$$ So, we write
$$P(N=n | X=x) = \frac{f_{X|N=n}(x) P(N=n)}{f_{X}(x)},$$
which is equal to
$$c \frac{(\lambda(1-p)x)^{n-1}}{(n-1)!},$$
where $$c$$ is a term that does not depend upon $$n.$$ Now, we know that
$$1 = c \sum_{n=1}^{\infty} \frac{(\lambda(1-p)x)^{n-1}}{(n-1)!} = c e^{\lambda(1-p)x},$$ so $$c$$ must be equal to $$e^{- \lambda(1-p)x}.$$ Thus,
$$P(N=n | X=x) = e^{- \lambda(1-p)x} \frac{(\lambda(1-p)x)^{n-1}}{(n-1)!}$$ is equivalent to a Poisson probability mass function, where the parameter is $$\lambda(1-p)x$$ and $$n=1$$ corresponds to $$1 – 1 = 0,$$ and so on, in terms of the numerical values.

Now, the alternative solution starts the same, but aims to avoid the constant term $$c.$$ That is,
$$P(N=n | X=x) = \frac{f_{X|N=n}(x) P(N=n)}{f_{X}(x)} = \frac{\frac{\lambda e^{- \lambda x} (\lambda x)^{n-1}}{(n-1)!} (1-p)^{n-1}p}{\sum_{n=1}^{\infty} \frac{\lambda e^{- \lambda x} (\lambda x)^{n-1}}{(n-1)!}},$$
which simplifies to
$$\bigg(\frac{p}{\sum_{n=1}^{\infty} \frac{(\lambda x)^{n-1}}{(n-1)!}} \bigg) \bigg(\frac{(\lambda(1-p)x)^{n-1}}{(n-1)!}\bigg).$$

So, the first term in parenthesis is equivalent to $$c.$$ Moreover, it is equal to
$$\frac{p}{\sum_{m=0}^{\infty} \frac{(\lambda x)^{m}}{(m)!}} = \frac{p}{e^{\lambda x}} = p e^{- \lambda x}.$$ But I do not see how to proceed from here, or, where I should trace back to in the derivation and proceed from there.

The infinite series in the denominator is missing a factor of $$\left(1-p\right)^{n-1}p$$. Basically you need to add up the value from the numerator over all possible $$n$$. After including this factor, you can cancel out the $$p$$ in the numerator, and the series will sum to $$e^{\lambda\left(1-p\right)x}$$, which is what you need.