Question

Suppose that W, the amount of moisture in the air on a given day, is a gamma random variable with parameters (t, β). That is, its density is f(w) = βe−βw(βw)t−1/(t), w > 0. Suppose also that given that W = w, the number of accidents during that day—call it N—has a Poisson distribution with mean w. Show that the conditional distribution of W given that N = n is the gamma distribution with parameters (t + n, β + 1)

Short answer

In order to obtain the required conditional distribution, use the definition of conditional PDF.

Step-by-step solution

Content Introduction

A random variable is a variable with an unknown value or a function that gives values to each of the results of an experiment. It's possible for a random variable to be discrete or continuous.

Content Explanation

We are required to find the distribution of W given that N = n. We have that,

$$\begin{gathered} f_{W,N}(w,n)=\frac{f_{W,N}(w,n)}{P(N=n)} \\ =\frac{P(N=n|W=w\left)f_W\right(w)}{P(N=n)} \end{gathered}$$

Since, we have that $N/W=w~Pois\left(w\right)$we have that

$P(N=n,W=w)=\frac{w^n}{n!}{e}^{-w}$

Thus,

$\frac{P(N=n,W=w)f_W\left(w\right)}{P(N=n)}=\frac{{\displaystyle \frac{w^n}{n!}}{e}^{-w}.{\displaystyle \frac{\beta e^{-\beta w}{\left(\beta w\right)}^{t-1}}{\tau \left(t\right)}}}{P(N=n)}$