Central Limit Theorem (CLT) and Uniformly Minimum Variance Unbiased Estimator (UMVUE)

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:

$$Y_n=\frac{1}{n}\sum _{i=1}^n{X}_{2i-1},Z_n=\frac{1}{n}\sum _{i=1}^n{X}_{2i}$$

The asymptotic distribution (as $n\to \mathrm{\infty }$) 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:

$$Y_n=\frac{1}{n}\sum _{i=1}^n{X}_{2i-1},Z_n=\frac{1}{n}\sum _{i=1}^n{X}_{2i}$$

The expectations are:

$$E[Y_n]=E[Z_n]=E[X]$$

Variances:

$$\text{Var}(Y_n)=\frac{{\sigma }^2}{n},\text{Var}(Z_n)=\frac{{\sigma }^2}{n}$$

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:

$$\sqrt{n}(Y_n–Z_n)\stackrel{d}{\to }N(0,2{\sigma }^2)$$

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:

$$X_i\sim N(0,{\sigma }^2),0<{\sigma }^2<\mathrm{\infty }$$

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) $\frac{1}{n}\sum _{i=1}^n(X_i–\overline{X}{)}^2$
• (d) $\frac{1}{n-1}\sum _{i=1}^n(X_i–\overline{X}{)}^2$

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) $\frac{1}{n-1}\sum _{i=1}^n{X}_i^2$.

PDF

Video

Reach PostNetwork Academy

• Website: www.postnetwork.co
• YouTube: www.youtube.com/@postnetworkacademy
• Facebook: www.facebook.com/postnetworkacademy
• LinkedIn: www.linkedin.com/company/postnetworkacademy

Thank You!