GATE 2025 DA Q.36: solution and answer

GATE 2025 · DAQ.36 · 2 marks · MCQ

Let $Y = Z^{2}$ , $Z = \frac{X - μ}{σ}$ , where $X$ is a normal random variable with mean $μ$ and variance $σ^{2}$ . The variance of $Y$ is

(A)
1
(B)
2
(C)
3
(D)
4

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The variance of $Y = Z^{2}$ is 2, option (B).

Official answer: BSource: Official answer key, GATE 2025 DA (IIT Roorkee)

Standardising gives $Z = \frac{X - μ}{σ} \sim N (0, 1)$ , so $Y = Z^{2}$ is a chi-square variable with 1 degree of freedom. We need $Var (Y)$ .
Use $Var (Y) = E [Y^{2}] - (E [Y])^{2} = E [Z^{4}] - (E [Z^{2}])^{2}$ .
$E [Z^{2}] = Var (Z) + (E [Z])^{2} = 1 + 0 = 1$ .
For the fourth moment, integrate by parts with $ϕ^{'} (z) = - z ϕ (z)$ : $E [Z^{4}] = \int_{- \infty}^{\infty} z^{3} \cdot z ϕ (z) d z = [- z^{3} ϕ (z)]_{- \infty}^{\infty} + 3 \int_{- \infty}^{\infty} z^{2} ϕ (z) d z = 0 + 3 E [Z^{2}] = 3.$ This agrees with the even-moment formula $E [Z^{2 k}] = (2 k - 1)!!$ , which gives $3!! = 3$ for $k = 2$ .
Therefore $Var (Y) = 3 - 1^{2} = 2$ . This matches the chi-square variance $2 k$ with $k = 1$ . The answer is option (B).

KEY IDEA

A standardised normal $Z$ has $E [Z^{2}] = 1$ and $E [Z^{4}] = 3$ , so $Var (Z^{2}) = 3 - 1 = 2$ . Equivalently, $Z^{2}$ is chi-square with 1 degree of freedom, whose variance is $2 k = 2$ .

WATCH OUT FOR

Taking $E [Z^{4}] = 1$ or treating $Z^{2}$ like a variable with variance $Var (Z)^{2} = 1$ (option A). The fourth moment of a standard normal is 3, not 1, so $Var (Z^{2}) = 3 - 1 = 2$ .

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 8 Random variables questions, with notes

GATE 2025 DA Q.36: solution and answer

Answer and solution

More Random variables questions