Show that the variance is biased

Question

I am trying to understand the proof that the uncorrected sample variance is biased (given here)

$$ \begin{eqnarray} E[S^2] &=& E \left [ \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \bar X)^2 \right ] \\ &=& E \left [ \frac{1}{n} \sum \limits_{i=1}^{n} \left ( (X_i - \mu)- (\bar X - \mu) \right)^2 \right ] \label{eq:s2q1p2p1}\\ &=& E \left [ \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \mu)^2 - 2(\bar X -\mu) \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \mu) + (\bar X - \mu)^2 \right ] \\ &=& E \left [ \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \mu)^2 - (\bar X - \mu)^2 \right ] \\ &=& \sigma^2 - E \left [(\bar X - \mu)^2 \right ] < \sigma^2 \end{eqnarray} $$

Conceptually I do understand everything, but I don't understand how to get from

$$ E \left [ \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \mu)^2 - 2(\bar X -\mu) \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \mu) + (\bar X - \mu)^2 \right ] $$

to

$$ E \left [ \frac{1}{n} \sum \limits_{i=1}^{n} (X_i - \mu)^2 - (\bar X - \mu)^2 \right ]. $$

Answer 1

Since $$ \sum_{i=1}^n(X_i-\mu)=n(\bar{X}-\mu) $$ the second term becomes $$ -2(\bar{X}-\mu)\frac1n\sum_{i=1}^n(X_i-\mu)=-2(\bar{X}-\mu)^2. $$