GATE 2016 | MATHS | Q-33

Question

Suppose \( X \) and \( Y \) are two random variables such that \( aX + bY \) is a normal random variable for all \( a, b \) in \( \mathbf{R} \). Consider the following statements P, Q, R, and S:
(P) : \( X \) is a standard normal random variable.
(Q) : The conditional distribution of \( X \) given \( Y \) is normal.
(R) : The conditional distribution of \( X + Y\) is normal.
(S) : \( X - Y \) has mean 0.

Which of the above statements ALWAYS hold TRUE?
(A) both P and Q

(B) both Q and R

(C) both Q and S

(D) both P and S

Answer 1

(B), both Q and R always hold true.

 

Statement P:

\(X\) doesn't necessarily have to be a standard normal random variable (mean of 0 and variance of 1).
It could have a different mean or variance and still satisfy the condition that \(aX + bY\) is normal for all \(a, b \in \mathbb{R}\).
Therefore, P is not always true.

Statement Q:

If \(aX + bY\) is normal for all \(a, b\), then in particular, \(Y = 0 \cdot X + 1 \cdot Y\) is normal.
This implies that the conditional distribution of \(X\) given \(Y\) is also normal.
Therefore, Q is always true.

Statement R:

Similar to Q, if \(aX + bY\) is normal for all \(a, b\), then \(X + Y = 1 \cdot X + 1 \cdot Y\) is normal.
This implies that the conditional distribution of \(X + Y\) is normal.
Therefore, R is always true.

Statement S:

There's no guarantee that \(X - Y\) has mean 0.
It could have a non-zero mean and still satisfy the condition that \(aX + bY\) is normal for all \(a, b\).
Therefore, S is not always true.