Question

Let X∼Normal (0, 1) and 𝑌 = |𝑋|. If the probability density function of 𝑌 is 𝑓𝑌 (y) then for 𝑦>0,\( \sqrt{\frac{\pi}{2}}\) 𝑓𝑌 (𝑦) is

• A. \(e^{-\frac{y^2}{2}}\)
• B. \(e^{\frac{y^2}{2}}\)
• C. \(e^{-y^2}\)
• D. \(e^{-\frac{y}{2}}\)

Answer 1

The Correct Option is A

To solve this problem, we need to find the probability density function (pdf) of \( Y = |X| \), where \( X \) follows a standard normal distribution, i.e., \( X \sim \text{Normal}(0, 1) \).

The probability density function of a standard normal distribution is given by:

\(f_X(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\)

Since \( Y = |X| \), the transformation involves taking the absolute value of \( X \). Therefore, for \( y > 0 \), \( Y \) can be \( +y \) or \( -y \). Hence, we need to consider both cases.

The pdf of \( Y \) can be derived as follows: 

The probability \( P(Y \leq y) = P(|X| \leq y) = P(-y \leq X \leq y) \).

Thus, the cumulative distribution function (CDF) of \( Y \) is:

1. \(F_Y(y) = P(-y \leq X \leq y) = \int_{-y}^{y} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, dx\)

Differentiate the CDF to get the pdf:

1. \(f_Y(y) = \frac{d}{dy} F_Y(y) = f_X(y) + f_X(-y)\)

For \( y > 0 \),

\(f_Y(y) = \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} + \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} = \frac{2}{\sqrt{2\pi}} e^{-\frac{y^2}{2}}\)

Given \( \sqrt{\frac{\pi}{2}} f_Y (y) \), substitute the expression for \( f_Y(y) \):

\(\sqrt{\frac{\pi}{2}} \times \frac{2}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} = e^{-\frac{y^2}{2}}\)

Thus, the correct option is:

\(e^{-\frac{y^2}{2}}\)

This final expression shows why the correct answer is \(e^{-\frac{y^2}{2}}\) and not the other options given.