In Exercise 5.42, the number of defects per yard in a

Step 1 of 3

The number of defects ,Y, follows a Poisson distribution with parameter \(\lambda\)

The parameter \(\lambda\) was assumed to be a random variable with a density function given by

\(f(\lambda)=e^{-\lambda}, \quad \lambda \geq 0,0, \text { elsewhere }\)

We need to find the conditional expectation of Y for given \(\lambda\)

Let \(Y_{1}\) and \(Y_{2}\) denote random variables. Then

\(E\left(Y_{1}\right)=E\left[E\left(Y_{1} \mid Y_{2}\right)\right] \dots (1)\)

Where on the right-hand side the inside expectation is with respect to the conditional distribution of \(Y_{1}\) given \(Y_{2}\) and the outside expectation is with respect to the distribution of \(Y_{2}\).

Let Y be the number of defectives per yard and known to have a Poisson distribution with parameter \(\lambda\).

Hence using equation (1) we can write,

\(E(Y)=E[E(Y \mid \lambda)] \dots (2)\)

For a given \(\lambda, Y\) has a Poisson distribution.

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

\(E(Y)=\mu=\lambda\)

Hence we can write,

\(E(Y \mid \lambda)=\mu=\lambda\)

Thus equation (2) will become,

\(E(Y)=E[\lambda] \dots (3)\)

Also, we have given that \(\lambda\) follows exponential with mean 1.

\(f(\lambda)=e^{-\lambda}, \quad \lambda \geq 0,0, \text { elsewhere }\)

Hence equation (3) will become,

\(\begin{array}{c}
E(Y)=E[\lambda]=\int_{0}^{\infty} \lambda f(\lambda) d \lambda \\
E(Y)=E[\lambda]=\int_{0}^{\infty} \lambda e^{-\lambda} d \lambda \\
E(Y)=E[\lambda]=1
\end{array}\)

Hence the conditional expectation of Y for given \(\lambda\) is 1.