Question

For each fixed \(\lambda>0\) let \(X\) have a Poisson distribution with parameter \(\lambda\). Suppose \(\lambda\) itself is a random variable following a gamma distribution (i.e., with density $$ f(\lambda)= \begin{cases}\frac{1}{\Gamma(n)} \lambda^{n-1} e^{-\lambda}, & \lambda \geq 0 \\ 0, & \lambda<0\end{cases} $$where \(n\) is a fixed positive constant). Show that now $$ \operatorname{Pr}\\{X=k\\}=\frac{\Gamma(k+n)}{\Gamma(n) \Gamma(k+1)}\left(\frac{1}{2}\right)^{k+n}, \quad k=0,1, \ldots $$ When \(n\) is an integer this is the negative binomial distribution with \(p=\frac{1}{2}\).

Short answer

The value of \(\Pr(X = k)\) must be \(\frac{\Gamma(k + n)}{\Gamma(n) \Gamma(k + 1)} (1/2)^{k + n}\), which is the negative binomial distribution when \(n\) is an integer.

Step-by-step solution

Analyze the Probability Density Function

Notice that the parameter \(\lambda\) of the Poisson distribution is itself a random variable with a gamma distribution. The probability density function \(f(\lambda)\) of the gamma distribution is given, which involves the gamma function \(\Gamma(n)\), the parameter \(\lambda\), as well as the constant \(n\).

Compute the Probability Distribution for X

Since X has a Poisson distribution with parameter \(\lambda\), the probability mass function of X is given by \(\Pr(X = k) = e^{-\lambda} \lambda^k / k!\) for \(k = 0, 1, 2, ...\). However, now \(\lambda\) itself is a random variable, so the probability of \(X = k\) can be given by the integral over all possible values of \(\lambda\) of the product of the Poisson density with the gamma density.

Evaluate the Integral

The integral can be evaluated by recognizing it as a gamma function. The integral evaluation results in \(\Pr(X = k) = \frac{\Gamma(k + n)}{\Gamma(n) \Gamma(k + 1)} (1/2)^{k + n}\), which is the desired result.

Recognize the Negative Binomial Distribution

Finally, recognize that when \(n\) is an integer, the result is the negative binomial distribution with parameter \(p = 1/2\).