Central Limit Theorem (CLT) and Uniformly Minimum Variance Unbiased Estimator (UMVUE)
By: Bindeshwar Singh Kushwaha
Institute: PostNetwork Academy
Question 1
Suppose \( X_1, X_2, \dots \) is an i.i.d. sequence of random variables with common variance \( \sigma^2 > 0 \). Define:
\[
Y_n = \frac{1}{n} \sum_{i=1}^{n} X_{2i-1}, \quad Z_n = \frac{1}{n} \sum_{i=1}^{n} X_{2i}
\]
The asymptotic distribution (as \( n \to \infty \)) of \( \sqrt{n} (Y_n – Z_n) \) is:
- (a) \( N(0,1) \)
- (b) \( N(0, \sigma^2) \)
- (c) \( N(0, 2\sigma^2) \)
- (d) Degenerate at 0
Solution
We define:
\[
Y_n = \frac{1}{n} \sum_{i=1}^{n} X_{2i-1}, \quad Z_n = \frac{1}{n} \sum_{i=1}^{n} X_{2i}
\]
The expectations are:
\[
E[Y_n] = E[Z_n] = E[X]
\]
Variances:
\[
\text{Var}(Y_n) = \frac{\sigma^2}{n}, \quad \text{Var}(Z_n) = \frac{\sigma^2}{n}
\]
Since \( Y_n \) and \( Z_n \) are independent, their difference has variance:
\[
\text{Var}(Y_n – Z_n) = \text{Var}(Y_n) + \text{Var}(Z_n) = \frac{2\sigma^2}{n}
\]
Multiplying by \( \sqrt{n} \), we get:
\[
\text{Var}(\sqrt{n} (Y_n – Z_n)) = 2\sigma^2
\]
By the Central Limit Theorem:
\[
\sqrt{n} (Y_n – Z_n) \xrightarrow{d} N(0, 2\sigma^2)
\]
Thus, the correct answer is (c) \( N(0, 2\sigma^2) \).
Question 2
Let \( X_1, X_2, \dots, X_n \) be i.i.d. observations from:
\[
X_i \sim N(0, \sigma^2), \quad 0 < \sigma^2 < \infty
\]
Find the Uniformly Minimum Variance Unbiased Estimator (UMVUE) for \( \sigma^2 \).
- (a) \( \frac{1}{n} \sum_{i=1}^{n} X_i^2 \)
- (b) \( \frac{1}{n-1} \sum_{i=1}^{n} X_i^2 \)
- (c) \( \frac{1}{n} \sum_{i=1}^{n} (X_i – \bar{X})^2 \)
- (d) \( \frac{1}{n-1} \sum_{i=1}^{n} (X_i – \bar{X})^2 \)
Solution
The sample variance is:
\[
S^2 = \frac{1}{n} \sum_{i=1}^{n} X_i^2
\]
Expectation:
\[
E[S^2] = \frac{n-1}{n} \sigma^2
\]
\( S^2 \) is a biased estimator for \( \sigma^2 \).
To correct this bias, we use:
\[
\hat{\sigma}^2 = \frac{n}{n-1} S^2 = \frac{1}{n-1} \sum_{i=1}^{n} X_i^2
\]
Thus, the correct answer is (b) \( \frac{1}{n-1} \sum_{i=1}^{n} X_i^2 \).
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