Question

Suppose that the random variable \( X \) has an \( \text{Exp}\left(\frac{1}{5}\right) \) distribution, and for any \( x > 0 \), the conditional distribution of the random variable \( Y \), given \( X = x \), is \( N(x, 2) \). Then \( \text{Var}(X + Y) \) is equal to:

• A. 52
• B. 50
• C. 2
• D. 102

Answer 1

The Correct Option is D

1. Variance of \( X \): - Since \( X \sim \text{Exp}\left(\frac{1}{5}\right) \), the variance of \( X \) is: \[ \mathrm{Var}(X) = 5^2 = 25. \] 2. Conditional Variance of \( Y \): - Given \( X = x \), \( Y \sim N(x, 2) \), so the conditional variance is: \[ \mathrm{Var}(Y \mid X = x) = 2. \] 3. Unconditional Variance of \( Y \): - Using the law of total variance: \[ \mathrm{Var}(Y) = \mathbb{E}[\mathrm{Var}(Y \mid X)] + \mathrm{Var}(\mathbb{E}[Y \mid X]). \] - Substituting: \[ \mathbb{E}[Y \mid X] = X \quad \Rightarrow \quad \mathrm{Var}(\mathbb{E}[Y \mid X]) = \mathrm{Var}(X) = 25. \] \[ \mathrm{Var}(Y) = \mathbb{E}[2] + 25 = 2 + 25 = 27. \] 4. Variance of \( X + Y \): - Since \( X \) and \( Y \) are dependent: \[ \mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2\mathrm{Cov}(X, Y). \] - From the definition of conditional variance: \[ \mathrm{Cov}(X, Y) = \mathrm{Var}(X). \] - Substituting: \[ \mathrm{Var}(X + Y) = 25 + 27 + 2(25) = 102. \]