Step 1: Concept Overview:
We are asked to find the expected value of the absolute value of a standard normal random variable, \(X \sim N(0,1)\). Let \(Y = |X|\), which follows a folded normal distribution. We aim to calculate \(E(Y) = E(|X|)\) using the definition of expected value.
Step 2: Core Formula:
The expected value of a function \(g(X)\) of a continuous random variable is:
\[ E[g(X)] = \int_{-\infty}^{\infty} g(x) f_X(x) dx \]
In our case, \(g(x) = |x|\) and \(f_X(x)\) is the standard normal probability density function (PDF).
Step 3: Detailed Solution:
Applying the expected value formula:
\[ E(Y) = E(|X|) = \int_{-\infty}^{\infty} |x| \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \]
Split the integral at 0 due to the absolute value:
\[ E(|X|) = \int_{-\infty}^{0} (-x) \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx + \int_{0}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \]
Since \(|x|e^{-x^2/2}\) is an even function, we can simplify:
\[ E(|X|) = 2 \int_{0}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \]
\[ E(|X|) = \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty} x e^{-x^2/2} dx = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} x e^{-x^2/2} dx \]
Use u-substitution: Let \(u = x^2/2\), so \(du = x \, dx\). The limits of integration remain \(0\) to \(\infty\).
\[ \int_{0}^{\infty} e^{-u} du = [-e^{-u}]_0^\infty = (-e^{-\infty}) - (-e^{-0}) = 0 - (-1) = 1 \]
Substitute back into the expression for \(E(|X|)\):
\[ E(|X|) = \sqrt{\frac{2}{\pi}} \times 1 = \sqrt{\frac{2}{\pi}} \]
Step 4: Final Result:
The expected value \(E(Y)\) is \( \sqrt{\frac{2}{\pi}} \).