Answer to Question #203755 in Statistics and Probability for Wilfred Muendo

Question

Show that the adjusted sample variance is a consistent estimator of the true variance

Answer 1

Solution:

Let $X_{1}, X_{2} \ldots \ldots \ldots X_{n}$ be the independent observations from a population with mean $\mu$ and variance $\sigma^{2}$

$E\left(X_{i}\right)=\mu, \operatorname{Var}\left(X_{i}\right)=\sigma^{2} \\E\left(X^{2}\right)=\sigma^{2}+\mu^{2} \\\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2} \\E\left(\bar{X}^{2}\right)=\frac{\sigma^{2}}{n}+\mu^{2}$

Now, we show that

${E}\left(s^{2}\right)={E}\left(\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{n-1}\right)=\sigma^{2}$

$\sum$ is going from 1 to n

${E}\left(\sum\left(X_{i}-\bar{X}\right)^{2}\right)={E}\left(\sum X_{i}^{2}-2 \bar{X} \sum X_{i}+n \bar{X}^{2}\right)=\sum {E}\left(X_{i}^{2}\right)-{E}\left(n \bar{X}^{2}\right) \\\sum {E}\left(X_{i}^{2}\right)-{E}\left(n \bar{X}^{2}\right)=\sum {E}\left(X_{i}^{2}\right)-n {E}\left(\bar{X}^{2}\right)=n \sigma^{2}+n \mu^{2}-\sigma^{2}-n \mu^{2}$

It will be simplified to $(\mathrm{n}-1) \sigma^{2}$

Therefore,

${E}\left(\sum\left(X_{i}-\bar{X}\right)^{2}\right)=(n-1) \sigma^{2} \\{E}\left(s^{2}\right)={E}\left(\frac{\Sigma\left(X_{i}-\bar{X}\right)^{2}}{n-1}\right)=\frac{1}{n-1} {E}\left(\sum\left(X_{i}-\bar{X}\right)^{2}\right)$

${E}(s^2)=\dfrac{(n-1) \sigma^{2}}{n-1}=\sigma^{2}$

Hence, it is proved that that the sample variance is an unbiased estimator of the population variance. Replacing X with the Y

$E[\dfrac1{n-1}\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)]={\sigma}^2$

Hence, proved.