I want to show that the sample variance $$S^2 := \frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar{X_n})^2$$ is unbiased, i.e. $\mathbb{E}[S^2] = \sigma^2$.
I know that the fastest way of showing this is by adding and subtracting the population mean $\mu$ within $(X_i - \bar{X_n})$ but this has lead me to some confusion.
On the one hand, if I consider the quantity $$\sum (X_i - \mu)^2$$ then I have $$\sum(X_i - \mu)^2 = \sum(X_i - \bar{X} + \bar{X} - \mu)^2 = \sum \left\{(X_i - \bar{X})^2 + 2(X_i - \bar{X})(\bar{X}-\mu) + (\bar{X} - \mu)^2 \right\}$$ and with the second term I can pull out the factor $(\bar{X} - \mu)$ and since $\mathbb{E}[X_i] = \mu = \mathbb{E}[\bar{X}]$, the second term vanishes and so I have that $$\sum (X_i - \bar{X})^2 = \sum (X_i - \mu)^2 - n(\bar{X} - \mu)^2$$
On the other hand, I have $$\sum (X_i - \bar{X})^2 = \sum(X_i - \mu + \mu - \bar{X})^2 = \sum \left\{(X_i - \mu)^2 + 2(X_i - \mu)(\mu - \bar{X}) + (\mu - \bar{X})^2 \right\}$$ and with the second term I can pull out the factor $(\mu - \bar{X})$ and since $\mathbb{E}[X_i] = \mu$ I have that $$\sum(X_i - \bar{X}^2) = \sum (X_i - \mu)^2 + n(\mu - \bar{X})^2$$ so I've ended up with a differing in sign. I'm confused by this and would appreciate any help.
