Answer: In Exercise 5.16, Y1 and Y2 denoted the

Step 1 of 2:

In Exercise  \(Y_{1}\) and \(Y_{2}\) denoted the proportions of time that employees I and II actually spent working on their assigned tasks during a workday.

The joint probability density function of  \(Y_{1}\) and \(Y_{2}\) is given by,

\(f\left(y_{1}, y_{2}\right)=y_{1}+y_{2}, \quad 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \text {, and } 0 \text { elsewhere }\)

The Employee I has a higher productivity rating than employee II and a measure of the total productivity of the pair of employees is \(30 Y_{1}+25 Y_{2}\)

Find the expected value of this measure of productivity.

We need to find \(E\left(30 Y_{1}+25 Y_{2}\right)\).

Let \(Y_{1}\) and \(Y_{2}\) jointly continuous random variables with the joint (or bivariate) probability function \(f\left(y_{1}, y_{2}\right)\). Then the marginal density functions of \(Y_{1}\) and \(Y_{2}\) respectively, are given by

\(f_{1}\left(y_{1}\right)=\int_{-\infty}^{\infty} f\left(y_{1}, y_{2}\right) d y_{2}\) and \(f_{2}\left(y_{2}\right)=\int_{-\infty}^{\infty} f\left(y_{1}, y_{2}\right) d y_{1}\)

Hence the marginal density functions for \(Y_{1}\) is,

\(f_{1}\left(y_{1}\right)=\int_{0}^{1}\left(y_{1}+y_{2}\right) d y_{2}=y_{1}+\frac{1}{2} \quad 0 \leq y_{1} \leq 1\)

The marginal density functions for \(Y_{2}\) is,

\(f_{2}\left(y_{2}\right)=\int_{0}^{1}\left(y_{1}+y_{2}\right) d y_{1}=y_{2}+\frac{1}{2} \quad 0 \leq y_{2} \leq 1\)