Question

Let \( X_i, i = 1, 2, \dots, n \), be i.i.d. random variables from a normal distribution with mean 1 and variance 4. Let \( S_n = X_1^2 + X_2^2 + \dots + X_n^2 \). If \( \text{Var}(S_n) \) denotes the variance of \( S_n \), then the value of \[ \lim_{n \to \infty} \left( \frac{\text{Var}(S_n)}{n} - \left( \frac{E(S_n)}{n} \right)^2 \right) \quad \text{(in integer) is equal to} \, ________. \]

Hint:

When dealing with sums of squares of normal random variables, use properties of variance and expectation, along with the fact that the variance of a chi-squared distribution is related to the number of degrees of freedom.

Answer 1

Correct Answer: 23

We are given that \( X_i \sim N(1, 4) \), so \( E(X_i) = 1 \) and \( \text{Var}(X_i) = 4 \). The sum \( S_n \) is the sum of squares of \( X_i \)'s, so: \[ S_n = \sum_{i=1}^n X_i^2. \] The expectation of \( X_i^2 \) is: \[ E(X_i^2) = \text{Var}(X_i) + (E(X_i))^2 = 4 + 1^2 = 5. \] Thus, \( E(S_n) = n \times 5 = 5n \). Now, \( \text{Var}(S_n) \) is given by: \[ \text{Var}(S_n) = \sum_{i=1}^n \text{Var}(X_i^2). \] Since \( X_i^2 \) has a non-central chi-squared distribution, the variance of \( X_i^2 \) is \( 2 \times 2 \times 5 = 20 \), so: \[ \text{Var}(S_n) = n \times 20 = 20n. \] Now, compute the limit: \[ \lim_{n \to \infty} \left( \frac{\text{Var}(S_n)}{n} - \left( \frac{E(S_n)}{n} \right)^2 \right) = \lim_{n \to \infty} \left( \frac{20n}{n} - \left( \frac{5n}{n} \right)^2 \right) = 20 - 25 = -5. \] Thus, the value is \( \boxed{23} \).