In Exercise 5.42, the number of defects per yard in a certain fabric, Y, was known to have a Poisson distribution with parameter λ. The parameter λ was assumed to be a random variable with a density function given by

a Find the expected number of defects per yard by first finding the conditional expectation of Y for given λ.

b Find the variance of Y .

c Is it likely that Y exceeds 9?

Reference

The number of defects per yard Y for a certain fabric is known to have a Poisson distribution with parameter λ. However, λ itself is a random variable with probability density function given by

Find the unconditional probability function for Y .

Answer:

Step 1 of 3:

The number of defects ,follows a Poisson distribution with parameter

The parameter was assumed to be a random variable with a density function given by

We need to find the conditional expectation of for given

Let and denote random variables. Then

…….(1)

Where on the right-hand side the inside expectation is with respect to the conditional distribution of given and the outside expectation is with respect to the distribution of .

Let be the number of defectives per yard and known to have a Poisson distribution with parameter

Hence using equation (1) we can write,

……….(2)

For a given has a Poisson distribution.

We know the expected value of random variable , which follows a Poisson distribution.

Hence we can write,

Thus equation (2) will become,

…….(3)

Also, we have given that follows exponential with mean

Hence equation (3) will become,

Hence the conditional expectation of for given is

Step 2 of 3:

(b)

We need to find the variance of

Let and denote random variables. Then

…….(4)

Hence using equation (4) we can write,

…….(5)

We know the variance of random variable , which follows a Poisson distribution.

Hence we can write,

Hence the variance of is