Expectation of sample variance

Question

Let $s^2$ be sample variance, $\sigma^2$ be population variance

$E[\frac{(n-1)s^2}{\sigma^2}] = E[\chi^2_{n-1}] = (n-1) \implies \frac{(n-1)E[s^2]}{\sigma^2} = n-1 \implies E[s^2]=\sigma^2$

But if i do in followng way, i am getting wrong answer

$E[s^2]=E[\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar x)^2] \\ = \frac{1}{n-1} \sum_i (E[x_i^2] + E(\bar x^2) - 2E[\bar x]E[x_i])$

I know that

$E[\bar x] = \mu$, population mean

$E[x_i^2] = var(x_i)+E[x_i]^2 = \sigma^2+\mu^2 \\ E[\bar x^2] = var(\bar x)+E[\bar x]^2 = \frac{\sigma^2}{n}+\mu^2$

Plugging these in above boldface equation

$E[s^2] = \frac{1}{n-1} (n(\sigma^2+\mu^2)+n(\frac{\sigma^2}{n}+\mu^2)-2\mu^2) \\ \frac{n+1}{n-1} \sigma^2$

I know this is wrong. But i donot know where i made the mistake.

Answer 1

the mistake is that $$ \mathbb{E}[X_i \bar{X}] \neq\mathbb{E}[X_i] \mathbb{E}[\bar{X}] $$ since this would imply that $X_i$ and $\bar{X}$ are independent, which they clearly are not. To solve this, expand $X_i \bar{X} = X_i \frac{1}{n} \sum_{j=1}^nX_j$ to show that

\begin{align*} \mathbb{E}[X_i \bar{X}] &= \frac{1}{n} ((n-1)\mathbb{E}[X_1]\mathbb{E}[X_2] + \mathbb{E}[X_1^2])\\ &= \frac{1}{n} ((n-1)\mu^2 + (\sigma^2 + \mu^2))\\ &= \mu^2 + \frac{\sigma^2}{n} \end{align*}

Answer 2

$$\mathbb{E}[S^2]=\frac{1}{n-1}\mathbb{E}[\Sigma_i(X_i-\overline{X}_n)^2]=$$

$$=\frac{1}{n-1}\mathbb{E}[\Sigma_i(X_i-\mu)^2-n(\overline{X}_n-\mu)^2]=$$

$$=\frac{1}{n-1}\left\{\Sigma_i\underbrace{\mathbb{E}[(X_1-\mu)^2]}_{\text{var pop}=\sigma^2}-n\underbrace{\mathbb{E}[(\overline{X}_n-\mu)^2]}_{\mathbb{V}[\overline{X}_n]=\sigma^2/n}\right\}=$$

$$=\frac{1}{n-1}\left[n\sigma^2-n\frac{\sigma^2}{n}\right]=\sigma^2$$