Question

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?

• 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.
• D. \((\sqrt{n}\,\bar{X})^2\) follows Chi-square distribution with 2 degrees of freedom.

Hint:

Remember that a sum of k independent squared standard normals is Chi-square with k degrees of freedom, and Chi-square with 2 degrees of freedom is the same as Exponential with mean 2.

Answer 1

The Correct Option is A, B, C

Step 1: List the two facts this question leans on.
Fact 1: if $Z \sim N(0,1)$ then $Z^2 \sim \chi^2_1$ (chi-square with 1 df).
Fact 2 (additivity): if $U \sim \chi^2_{k_1}$ and $V \sim \chi^2_{k_2}$ are independent, then $U + V \sim \chi^2_{k_1+k_2}$.
Both facts follow directly from how the chi-square distribution is built as a sum of squared independent standard normals.

Step 2: Option (A) by direct application of Fact 1 and Fact 2.
Each $X_i \sim N(0,1)$, so each $X_i^2 \sim \chi^2_1$ by Fact 1. Since the $X_i$ are independent, repeated use of Fact 2 gives $\sum_{i=1}^n X_i^2 \sim \chi^2_n$. This matches option (A), so it is true.

Step 3: Option (B) via the standard sample variance result.
For iid normal data, $\bar{X}$ and the deviations $(X_i - \bar{X})$ are independent, and $\sum (X_i-\bar{X})^2/\sigma^2$ has one fewer degree of freedom than $\sum X_i^2 / \sigma^2$ because estimating $\bar{X}$ removes one degree of freedom. With $\sigma^2 = 1$, this gives $\sum_{i=1}^n(X_i - \bar{X})^2 \sim \chi^2_{n-1}$ directly, so option (B) is true.

Step 4: Option (C) using the Gamma-Exponential link.
By independence, $X_1^2 \sim \chi^2_1$ and $X_n^2 \sim \chi^2_1$ independently, so $X_1^2 + X_n^2 \sim \chi^2_2$ by Fact 2. A $\chi^2_2$ random variable has probability density $f(x) = \frac{1}{2}e^{-x/2}$ for $x \geq 0$, which is exactly the density of an Exponential distribution with rate $\frac{1}{2}$, i.e. mean $\frac{1}{1/2} = 2$. So option (C) is true.

Step 5: Option (D) by computing the distribution of $\bar{X}$ directly.
$\bar{X}$ is a linear combination of independent normals, so it is itself normal, with mean $E[\bar{X}] = 0$ and variance $Var(\bar{X}) = \frac{1}{n^2}\sum Var(X_i) = \frac{1}{n^2}\cdot n = \frac{1}{n}$.
So $\sqrt{n}\,\bar{X}$ has mean 0 and variance $n \cdot \frac{1}{n} = 1$, meaning $\sqrt{n}\,\bar{X} \sim N(0,1)$.
Squaring a single $N(0,1)$ variable gives $\chi^2_1$, not $\chi^2_2$, so option (D) is false.

Final Answer:
Statements (A), (B) and (C) are correct; (D) is not. $$\boxed{\text{(A), (B), (C)}}$$