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GATE 2026 DA – Question 53

Probability and Statistics · Random variables, discrete and continuous distributions · 2 marks · Multiple select

Let $X_1, X_2, \ldots, X_n$ be $n$ independent random variables. Each of the random variables follows $Normal(\mu = 0, \sigma^2 = 1)$ distribution. Define $\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i$.

Which of the following statements is/are correct?

  1. $\sum_{i=1}^{n} X_i^2$ follows Chi-square distribution with $n$ degrees of freedom.
  2. $\sum_{i=1}^{n} (X_i - \bar{X})^2$ follows Chi-square distribution with $(n-1)$ degrees of freedom.
  3. $X_1^2 + X_n^2$ follows exponential distribution with mean 2.
  4. $\left(\sqrt{n}\,\bar{X}\right)^2$ follows Chi-square distribution with 2 degrees of freedom.

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Show answer and explanation

Correct answer: (A) $\sum_{i=1}^{n} X_i^2$ follows Chi-square distribution with $n$ degrees of freedom.; (B) $\sum_{i=1}^{n} (X_i - \bar{X})^2$ follows Chi-square distribution with $(n-1)$ degrees of freedom.; (C) $X_1^2 + X_n^2$ follows exponential distribution with mean 2.

Explanation

A sum of $n$ squared independent standard normals is chi-square with $n$ degrees of freedom (A). Using the sample mean takes away one degree of freedom, so $\sum (X_i - \bar{X})^2$ is chi-square with $n - 1$ (B). The sum $X_1^2 + X_n^2$ is chi-square with 2 degrees of freedom, whose density is $\frac{1}{2}e^{-x/2}$, an exponential distribution with mean 2 (C). Finally $\sqrt{n}\,\bar{X}$ is a standard normal, so its square has 1 degree of freedom, not 2 (D is false).