Analyzing Properties of Normal Random Variables

The question asks about the distributions of sums and squares of independent standard normal random variables ($X_i \sim Normal(0, 1)$) and related statistics.

Option 1 Analysis: Sum of Squares

Statement: $\sum_{i=1}^{n} X_i^2$ follows Chi-square distribution with $n$ degrees of freedom.

• If $X_i \sim Normal(0, 1)$ are independent standard normal variables, then the sum of their squares, $\sum_{i=1}^{n} X_i^2$, is known to follow a Chi-square distribution with $n$ degrees of freedom.
• This is a standard result in statistics. The distribution is denoted as $\chi^2(n)$.
• Therefore, this statement is correct.

Option 2 Analysis: Sum of Squared Deviations

Statement: $\sum_{i=1}^{n} (X_i - \bar{X})^2$ follows Chi-square distribution with $(n - 1)$ degrees of freedom.

• Here, $\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$ is the sample mean.
• While $\sum_{i=1}^{n} X_i^2 \sim \chi^2(n)$, the sum of squared deviations from the sample mean, $\sum_{i=1}^{n} (X_i - \bar{X})^2$, accounts for the estimation of the mean.
• For independent $X_i \sim Normal(0, 1)$, the quantity $\frac{1}{\sigma^2}\sum_{i=1}^{n} (X_i - \bar{X})^2$ follows a $\chi^2(n-1)$ distribution, where $\sigma^2$ is the variance.
• Since $\sigma^2 = 1$ in this case, $\sum_{i=1}^{n} (X_i - \bar{X})^2 \sim \chi^2(n-1)$.
• Therefore, this statement is correct.

Option 3 Analysis: Sum of Two Squares

Statement: $X_1^2 + X_n^2$ follows exponential distribution with mean 2.

• $X_1$ and $X_n$ are independent standard normal variables ($Normal(0, 1)$).
• Therefore, $X_1^2 \sim \chi^2(1)$ and $X_n^2 \sim \chi^2(1)$.
• The sum of two independent $\chi^2(1)$ variables follows a $\chi^2(1+1) = \chi^2(2)$ distribution.
• A Chi-square distribution with 2 degrees of freedom, $\chi^2(2)$, is equivalent to an Exponential distribution with rate parameter $\lambda = 1/2$.
• The mean of an Exponential distribution with rate $\lambda$ is $1/\lambda$. So, the mean is $1/(1/2) = 2$.
• Therefore, this statement is correct.

Option 4 Analysis: Squared Scaled Mean

Statement: $(\sqrt{n}\bar{X})^2$ follows Chi-square distribution with 2 degrees of freedom.

• Since $X_i \sim Normal(0, 1)$ are independent, their mean $\bar{X} \sim Normal(0, 1/n)$.
• The random variable $\sqrt{n}\bar{X} = \sqrt{n} \left( \frac{1}{n} \sum_{i=1}^{n} X_i \right) = \frac{1}{\sqrt{n}} \sum_{i=1}^{n} X_i$.
• The distribution of $\sqrt{n}\bar{X}$ is $Normal(0, n \cdot (1/n)) = Normal(0, 1)$.
• Therefore, $(\sqrt{n}\bar{X})^2$ is the square of a standard normal variable, which follows a Chi-square distribution with 1 degree of freedom, i.e., $\chi^2(1)$.
• The statement claims it follows $\chi^2(2)$, which is incorrect.
• Therefore, this statement is incorrect.

Conclusion

Based on the analysis, statements 1, 2, and 3 are correct.