Let $X_n$ be a sequence of i.i.d. random variables with $E(X_i)=\mu,Var(X_i)=\sigma^2$ and define $\mu_n:=\frac{1}{n} \sum_{i=1}^n X_i,\sigma_n^2:=\sum_{i=1}^n(X_i-\mu_n)^2.$
I have already shown that $Var(\mu_n)=\frac{\sigma^2}{n}$. Now to show that $\sigma_n^2$ is an unbiased estimator for $\sigma^2$ I need to show $E(\sigma_n^2)=\sigma^2$. What I have so far is:
$E(\sum_{i=1}^n(X_i-\mu_n)^2)=\sum_{i=1}^nE((X_i-\mu_n)^2)=\sum_{i=1}^nVar(X_i-\mu_n)-(E(X_i-\mu_n))^2=\sum_{i=1}^nVar(X_i)+Var(\mu_n)-\sum_{i=1}^nE((X_i-\mu_n)^2)=\sum_{i=1}^n \sigma^2+\sigma^2/n-\sum_{i=1}^nE((X_i-\mu_n)^2)=n\sigma^2+\sigma^2-E(\sum_{i=1}^n(X_i-\mu_n)^2)$
Here I have used the fact that the variance of the sum of two independent variables is the sum of the single variances, the product of expectations of independent variables is the expectation of the product and the sum of the expectations is the expectation of the sum.
Adding the identical sums on both sides togehter I get $2E(\sum_{i=1}^n(X_i-\mu_n)^2)=n\sigma^2+\sigma^2$. Dividing by $2$ unfortunately doesn't give me $\sigma^2$ due to the $n$ on the right side which makes me think that some of my conclusions were wrong.
Any help would be appreciated!
Define: $$ \sigma_n^2 \;=\; \frac{1}{n-1}\sum_{i=1}^n{\left(X_i - \mu_n\right)}^2 $$ Then: \begin{align} E(\sigma_n^2) &\;=\; \frac{1}{n-1}\sum_{i=1}^n E\left[{\left(X_i - \mu_n\right)}^2\right]\\ &\;=\; \frac{1}{n-1}\sum_{i=1}^n E\left(X_i^2 - 2X_i\mu_n + \mu_n^2\right)\\ &\;=\; \frac{1}{n-1}\sum_{i=1}^n\Bigl[E(X_i^2) - 2 E(X_i\mu_n) + E(\mu_n^2)\Bigr]\qquad (1) \end{align} The three terms inside the sum are: \begin{align} \bullet\; E(X_i^2) &\;=\;\sigma^2 + \mu^2 \end{align} \begin{align} \bullet\; E(X_i \mu_n) &\;=\;\frac{1}{n}E\left(X_i \sum_{j = 1}^n X_j\right)\\ &\;=\; \frac{1}{n}E\left(X_i^2 + \sum_{j\neq i} X_iX_j\right)\\ &\;=\; \frac{1}{n}\left[\sigma^2 + \mu^2 + (n-1)\mu^2\right]\\ &\;=\; \frac{\sigma^2}{n} + \mu^2 \end{align} \begin{align} \bullet\; E(\mu_n^2) &\;=\;\frac{1}{n^2}E\left(\sum_{j=1}^n\sum_{k=1}^n X_j X_k\right)\\ &\;=\; \frac{1}{n^2}\left[\sum_{j=1}^nE(X_j^2) + \sum_{j=1}^n\sum_{k\neq j}E(X_j)E(X_k)\right]\\ &\;=\; \frac{1}{n^2}\left[n(\sigma^2 +\mu^2) + n(n-1)\mu^2\right]\\ &\;=\; \frac{\sigma^2}{n} +\mu^2 \end{align}
If you substitute these three expressions into (1) and simplify, you do indeed get $E(\sigma_n^2) = \sigma^2$.