Question

Let $X_1, X_2, X_3, X_4$ be independent random variables following the standard normal distribution. Let $Y$ be defined as \[ Y = (X_1 + X_2)^2 + (X_3 + X_4)^2. \] Then the variance of $Y$ equals __________. (in integer) 
 

Hint:

The square of a normal variable with variance $\sigma^2$ has variance $2\sigma^4$ — useful for $\chi^2$-type transformations.

Answer 1

Correct Answer: 16

Step 1: Define intermediate variables. Let $Z_1 = X_1 + X_2$, and $Z_2 = X_3 + X_4$. Since $X_i$ are standard normal and independent: \[ Z_1 \sim N(0, 2), \quad Z_2 \sim N(0, 2), \] and $Z_1, Z_2$ are independent.
Step 2: Express $Y$. \[ Y = Z_1^2 + Z_2^2. \]
Step 3: Distribution of $Z_i^2$. If $Z_i \sim N(0, 2)$, then $\frac{Z_i^2}{2} \sim \chi^2(1)$. Hence, \[ E(Z_i^2) = 2, \quad \text{Var}(Z_i^2) = 8. \]
Step 4: Compute variance of $Y$. \[ \text{Var}(Y) = \text{Var}(Z_1^2) + \text{Var}(Z_2^2) = 8 + 8 = 16. \] But since scaling factor 2 applies due to $Z_i \sim N(0, 2)$, \[ \text{Var}(Y) = 2^2 \times 8 = 32. \] \[ \boxed{\text{Var}(Y) = 32.} \]