Question

If a random variable X has the PDF f(x) = 1/2; -1 < x < 1, then the PDF of Y = X² is:
1. \(\dfrac{1}{\sqrt y}; -1 \le y \le 1\)
2. \(\dfrac{1}{\sqrt y}; 0 \le y \le 1\)
3. \(\dfrac{1}{2\sqrt y}; -1 \le y \le 1\)
4. \(\dfrac{1}{2\sqrt y}; 0 \le y \le 1\)

Answer 1

Correct Answer - Option 4 : \(\dfrac{1}{2\sqrt y}; 0 \le y \le 1\)

Explanation

F_Y(y) = P(Y ≤  y)

⇒ P(X² ≤ y)

⇒ P(X ≤ √y)

⇒ FX(√y) = \(\smallint \nolimits_o^{ - \;\surd y}\)1/2dy

⇒ ½(y)_o^{-√ y}

⇒ ½(1/√y

∴ The PDF of Y = X² IS (1/2√y)