Question

A continuous random variable \(X\) has the density function \[ f(x)= \begin{cases} e^{-x}, & x>0, 0, & \text{elsewhere}, \end{cases} \] then the expected value of \[ g(x)=e^{\frac{2x}{3}} \] is

• A. \(1\)
• B. \(2\)
• C. \(0\)
• D. \(3\)

Hint:

For an exponential random variable with \[ f(x)=e^{-x},\;x>0, \] \[ \boxed{ E[g(X)] = \int_{0}^{\infty}g(x)e^{-x}\,dx. } \]

Answer 1

The Correct Option is D

Step 1: Use the expectation formula. \[ E[g(X)] = \int_{0}^{\infty}g(x)f(x)\,dx. \] Substituting, \[ E\!\left(e^{\frac{2X}{3}}\right) = \int_{0}^{\infty} e^{\frac{2x}{3}}e^{-x}\,dx. \]

Step 2: Evaluate the integral. \[ = \int_{0}^{\infty} e^{-x/3}\,dx = \left[-3e^{-x/3}\right]_{0}^{\infty} =3. \] Hence, \[ \boxed{E\!\left(e^{\frac{2X}{3}}\right)=3.} \] Therefore, \[ \boxed{(D)} \] is the correct answer.