Question

Let \( X_1, X_2, X_3 \) be independent standard normal random variables, and let \[ Y_1 = X_1 - X_2, \quad Y_2 = X_1 + X_2 - 2X_3, \quad Y_3 = X_1 + X_2 + X_3. \] Then which of the following options is/are correct?

• A. \( Y_1, Y_2, Y_3 \) are independently distributed
• B. \( Y_1^2 + Y_2^2 + Y_3^2 \sim \chi_3^2 \)
• C. \( \frac{2Y_3}{\sqrt{3Y_1^2 + Y_2^2}} \sim t_2 \)
• D. \( \frac{3Y_1^2 + 2Y_3^2}{2Y_2^2} \sim F_{1,1} \)

Hint:

For sums of squares and ratios of quadratic forms in normal variables, the resulting distributions often follow chi-squared or t-distributions. Carefully analyze the dependencies and relationships between the variables involved.

Answer 1

The Correct Option is A, C

Step 1: Checking the Independence of \( Y_1, Y_2, Y_3 \)
The random variables \( Y_1, Y_2, Y_3 \) are linear combinations of \( X_1, X_2, X_3 \). Since \( X_1, X_2, X_3 \) are independent, it follows that \( Y_1, Y_2, Y_3 \) are also independent.
Thus, Option (A) is correct.
Step 2: Distribution of \( Y_1^2 + Y_2^2 + Y_3^2 \)
Since \( Y_1, Y_2, Y_3 \) are linear combinations of independent normal random variables, the sum of their squares follows a chi-squared distribution with 3 degrees of freedom.
Thus, Option (B) is incorrect because the sum of squares does follow \( \chi_3^2 \), but this was not required.
Step 3: Distribution of \( \frac{2Y_3}{\sqrt{3Y_1^2 + Y_2^2}} \)
This expression involves the ratio of quadratic forms in normal variables, which results in a t-distribution with 2 degrees of freedom.
Thus, Option (C) is correct.
Step 4: Distribution of \( \frac{3Y_1^2 + 2Y_3^2}{2Y_2^2} \)
This expression does not result in an \( F \)-distribution with 1 and 1 degrees of freedom, so Option (D) is incorrect.
Final Answer:
The correct answers are \( \boxed{A \text{ and } C} \).