Question

Let \( X \) and \( Y \) be jointly distributed continuous random variables, where \( Y \) is positive valued with \( E(Y^2) = 6 \). If the conditional distribution of \( X \) given \( Y = y \) is \( U(1 - y, 1 + y) \), then \( \text{Var}(X) \) equals .................

Hint:

The law of total variance states that the total variance of a random variable can be decomposed into the variance of its conditional expectation and the expected variance of the conditional distribution.

Answer 1

Correct Answer: 2

Step 1: Recall the formula for conditional variance.
The variance of \( X \) given \( Y = y \) is: \[ \text{Var}(X \mid Y = y) = \frac{(1 + y) - (1 - y)}{2} = y. \] Thus, \( \text{Var}(X \mid Y = y) = y \).
Step 2: Apply the law of total variance.
The total variance of \( X \) is: \[ \text{Var}(X) = E(\text{Var}(X \mid Y)) + \text{Var}(E(X \mid Y)). \] We already know that \( \text{Var}(X \mid Y = y) = y \), so we need to compute \( E(Y) \) and \( \text{Var}(Y) \).
Step 3: Compute the expected value and variance of \( Y \).
From the given information \( E(Y^2) = 6 \), and since \( Y \) is positive, we assume \( E(Y) = 0 \) (as the exact distribution is not given but implied from the problem). Thus: \[ E(\text{Var}(X \mid Y)) = E(Y) = 0, \] and \[ \text{Var}(E(X \mid Y)) = \text{Var}(Y) = 6. \]
Step 4: Compute the total variance.
Thus, the variance of \( X \) is: \[ \text{Var}(X) = 0 + 6 = 6. \]
Final Answer: \[ \boxed{6}. \]