Question

Suppose that \(X\) and \(Y\) are independent random variables with probability densities \[ f_X(x)= \begin{cases} \dfrac{8}{x^3}, & x>2,[2mm] 0, & \text{elsewhere}, \end{cases} \qquad f_Y(y)= \begin{cases} 2y, & 0 0, & \text{elsewhere}. \end{cases} \] Then the expected value of \(Z=XY\) is

• A. \(\dfrac{4}{3}\)
• B. \(\dfrac{8}{3}\)
• C. \(\dfrac{1}{3}\)
• D. \(1\)

Hint:

If \(X\) and \(Y\) are independent, \[ \boxed{ E(XY)=E(X)\,E(Y). } \]

Answer 1

The Correct Option is B

Step 1: Use independence. Since \(X\) and \(Y\) are independent, \[ E(XY)=E(X)\,E(Y). \]

Step 2: Compute \(E(X)\) and \(E(Y)\). For \(X\), \[ E(X) = \int_{2}^{\infty} x\cdot\frac{8}{x^3}\,dx = 8\int_{2}^{\infty}\frac{1}{x^2}\,dx = 8\left[-\frac1x\right]_{2}^{\infty} =4. \] For \(Y\), \[ E(Y) = \int_{0}^{1}y(2y)\,dy = 2\int_{0}^{1}y^2\,dy = \frac23. \] Hence, \[ E(Z) = E(XY) = 4\cdot\frac23 = \frac83. \] Therefore, \[ \boxed{\frac83} \] is the correct answer. Hence, \[ \boxed{(B)} \] is the correct answer.