GATE 2026 DA Q.53: solution and answer

GATE 2026 · DAQ.53 · 2 marks · MSQ

Let $X_{1}, X_{2}, \dots, X_{n}$ be $n$ independent random variables. Each of the random variables follows $Normal (μ = 0, σ^{2} = 1)$ distribution. Define $\overset{ˉ}{X} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ .

Which of the following statements is/are correct?

(A)
$\sum_{i = 1}^{n} X_{i}^{2}$ follows Chi-square distribution with $n$ degrees of freedom.
(B)
$\sum_{i = 1}^{n} (X_{i} - \overset{ˉ}{X})^{2}$ follows Chi-square distribution with $(n - 1)$ degrees of freedom.
(C)
$X_{1}^{2} + X_{n}^{2}$ follows exponential distribution with mean 2.
(D)
$(n \overset{ˉ}{X})^{2}$ follows Chi-square distribution with 2 degrees of freedom.

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Answer and solution

ANSWER

Statements (A), (B) and (C) are correct. (D) is wrong because $(n \overset{ˉ}{X})^{2}$ has 1 degree of freedom, not 2.

Official answer: A, B, CSource: Official answer key, GATE 2026 DA (IIT Guwahati)

Each $X_{i} \sim N (0, 1)$ independently, so $X_{i}^{2} \sim χ_{1}^{2}$ . A sum of independent chi-square variables is chi-square with the degrees of freedom added.
(A) $\sum_{i = 1}^{n} X_{i}^{2}$ is a sum of $n$ independent $χ_{1}^{2}$ variables, so it follows $χ_{n}^{2}$ . This is the definition of the chi-square distribution with $n$ degrees of freedom. (A) is correct.
(B) Because $σ^{2} = 1$ , the sample-variance theorem gives $\sum_{i = 1}^{n} (X_{i} - \overset{ˉ}{X})^{2} = (n - 1) S^{2} \sim χ_{n - 1}^{2}$ . One degree of freedom is lost because the deviations $X_{i} - \overset{ˉ}{X}$ always sum to zero. (B) is correct.
(C) For $n \geq 2$ , $X_{1}$ and $X_{n}$ are different, independent variables, so $X_{1}^{2} + X_{n}^{2} \sim χ_{2}^{2}$ . Its density is $f (x) = \frac{1}{2} e^{- x /2}$ for $x > 0$ . This is the exponential distribution with rate $\frac{1}{2}$ , whose mean is $2$ . (C) is correct.
(D) $\overset{ˉ}{X} \sim N (0, 1/ n)$ , so $n \overset{ˉ}{X} \sim N (0, 1)$ . Hence $(n \overset{ˉ}{X})^{2} \sim χ_{1}^{2}$ , with 1 degree of freedom, not 2. (D) is incorrect.
Correct options: A, B and C.

KEY IDEA

A sum of squares of independent standard normals is chi-square with one degree of freedom per independent term. Replacing the true mean with $\overset{ˉ}{X}$ costs one degree of freedom. $χ_{2}^{2}$ is exponential with mean 2.

WATCH OUT FOR

Judging (D) from the factor $n$ or from the sum, and giving it 2 or $n$ degrees of freedom. $n \overset{ˉ}{X}$ is a single standard normal, so its square is $χ_{1}^{2}$ . Another slip is to treat (B) as $χ_{n}^{2}$ and forget the lost degree of freedom.

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GATE 2026 DA Q.53: solution and answer

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