Question

If \(f(X) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}; -\infty<x<\infty\) and \(Y = |X|\), then E(Y) is

• A. \( \frac{1}{\sqrt{\pi}} \)
• B. \( \sqrt{\frac{2}{\pi}} \)
• C. \( \sqrt{2} \)
• D. \( \frac{2}{\sqrt{\pi}} \)

Hint:

The expected value of the absolute value of a standard normal variable is the mean of the standard half-normal distribution. Remembering this value, \( \sqrt{2/\pi} \), can save time on calculations for this common problem.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem asks for the expected value of the absolute value of a standard normal random variable, \(X \sim N(0,1)\). The random variable \(Y = |X|\) follows a folded normal distribution. We need to compute \(E(Y) = E(|X|)\) using the definition of expected value for a continuous random variable.

Step 2: Key Formula or Approach:
The expected value of a function of a continuous random variable, \(g(X)\), is given by: \[ E[g(X)] = \int_{-\infty}^{\infty} g(x) f_X(x) dx \] Here, \(g(x) = |x|\) and \(f_X(x)\) is the standard normal PDF.

Step 3: Detailed Explanation:
Using the formula for expected value: \[ E(Y) = E(|X|) = \int_{-\infty}^{\infty} |x| \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \] We can split the integral because of the absolute value function: \[ E(|X|) = \int_{-\infty}^{0} (-x) \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx + \int_{0}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \] The integrand \(|x|e^{-x^2/2}\) is an even function. Therefore, the integral from \(-\infty\) to \(\infty\) is twice the integral from \(0\) to \(\infty\). \[ E(|X|) = 2 \int_{0}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \] \[ E(|X|) = \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty} x e^{-x^2/2} dx = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} x e^{-x^2/2} dx \] To solve the integral, we use u-substitution. Let \(u = x^2/2\). Then \(du = x \, dx\). The limits of integration remain the same: when \(x=0, u=0\); when \(x \to \infty, u \to \infty\). \[ \int_{0}^{\infty} e^{-u} du = [-e^{-u}]_0^\infty = (-e^{-\infty}) - (-e^{-0}) = 0 - (-1) = 1 \] Substituting this result back into our expression for E(|X|): \[ E(|X|) = \sqrt{\frac{2}{\pi}} \times 1 = \sqrt{\frac{2}{\pi}} \]
Step 4: Final Answer:
The expected value E(Y) is \( \sqrt{\frac{2}{\pi}} \).