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: Recall the definition of the Chi-square distribution.
If \(Z_1, Z_2, \ldots, Z_k\) are independent standard normal random variables, then by definition \(\sum_{i=1}^{k} Z_i^2\) follows a Chi-square distribution with \(k\) degrees of freedom. Here each \(X_i \sim N(0,1)\) is already standard normal, since \(\mu=0\) and \(\sigma^2=1\).

Step 2: Check option (A).
\(X_1, \ldots, X_n\) are \(n\) independent standard normal variables, so by the definition in Step 1, \(\sum_{i=1}^{n} X_i^2\) follows Chi-square with \(n\) degrees of freedom exactly. Option (A) is correct.

Step 3: Check option (B).
This is a classical result used to build the sample variance: when \(X_1, \ldots, X_n\) are iid \(N(\mu, \sigma^2)\), the sum of squared deviations from the sample mean, \(\sum_{i=1}^{n}(X_i - \bar{X})^2 / \sigma^2\), follows a Chi-square distribution with \((n-1)\) degrees of freedom, one degree of freedom is used up in estimating \(\bar{X}\) from the data. Here \(\sigma^2 = 1\), so \(\sum_{i=1}^{n}(X_i-\bar{X})^2\) itself follows Chi-square with \((n-1)\) degrees of freedom. Option (B) is correct.

Step 4: Check option (C).
Each \(X_i^2\) individually follows Chi-square with 1 degree of freedom (a single squared standard normal). The sum of two independent Chi-square variables with 1 degree of freedom each gives a Chi-square with \(1+1=2\) degrees of freedom, so \(X_1^2 + X_n^2 \sim \chi^2_2\) (note \(X_1\) and \(X_n\) are independent since all \(X_i\) are independent). A Chi-square distribution with 2 degrees of freedom is a special case of the Gamma distribution with shape 1 and scale 2, which is exactly the Exponential distribution with scale 2, whose mean is 2. So \(X_1^2 + X_n^2\) does follow an Exponential distribution with mean 2. Option (C) is correct.

Step 5: Check option (D).
\(\sqrt{n}\,\bar{X} = \sqrt{n}\cdot\frac{1}{n}\sum_{i=1}^n X_i = \frac{1}{\sqrt{n}}\sum_{i=1}^n X_i\). Since the \(X_i\) are iid \(N(0,1)\), \(\sum X_i \sim N(0,n)\), so dividing by \(\sqrt{n}\) gives \(\sqrt{n}\,\bar{X} \sim N(0,1)\), a single standard normal variable. Squaring a single standard normal gives a Chi-square with 1 degree of freedom, not 2. Option (D) is incorrect.

Final Answer:
Options (A), (B), and (C) are correct. \[ \boxed{\text{(A), (B), (C)}} \]