Question

Consider a conditional Poisson process in which the rate \(L\) is, as in Example \(5.29\), gamma distributed with parameters \(m\) and \(p\). Find the conditional density function of \(L\) given that \(N(t)=n\).

Short answer

The conditional density function of L, given that N(t) = n, is: \[ f_{L|N(t) = n}(l) = C e^{-(l+p)t} (lt)^n p^m l^{m - 1} \] Where C is some normalization constant.

Step-by-step solution

Recall the Density Functions of Poisson and Gamma Distributions

The probability mass function of a Poisson distribution with rate L is given by: \[ P(N(t) = n) = e^{-Lt} \frac{(Lt)^n}{n!} \] The probability density function of a gamma distribution with parameters m and p is given by: \[ f_L(l) = \frac{p^m l^{m-1} e^{-pl}}{\Gamma(m)} \] We are looking for the conditional density function: \[ f_{L|N(t) = n}(l) = P(L = l \mid N(t) = n) \]

Use the Conditional Probability Formula

Using the conditional probability formula, we have: \[ f_{L|N(t) = n}(l) = \frac{P(N(t) = n \mid L = l) f_L(l)}{P(N(t) = n)} \] Now substitute the Poisson and Gamma distributions into the formula: \[ f_{L|N(t) = n}(l) = \frac{e^{-lt} \frac{(lt)^n}{n!} \cdot \frac{p^m l^{m-1} e^{-pl}}{\Gamma(m)}}{P(N(t) = n)} \]

Simplify the Expression

Simplify the expression by solving the product of the terms: \[ f_{L|N(t) = n}(l) = \frac{e^{-(l+p)t} (lt)^n p^m l^{m - 1}}{n! \Gamma(m) P(N(t) = n)} \]

Calculate the Conditional Density Function

The conditional density function of L, given that N(t) = n, is: \[ f_{L|N(t) = n}(l) = \frac{e^{-(l+p)t} (lt)^n p^m l^{m - 1}}{n! \Gamma(m) P(N(t) = n)} \] Since the denominator does not depend on L and we want to find a density function, we can ignore the denominator in our answer. Therefore, the conditional density function of L, given that N(t) = n, is: \[ f_{L|N(t) = n}(l) = C e^{-(l+p)t} (lt)^n p^m l^{m - 1} \] Where C is some normalization constant.