The random variable
$$ S^2=\sum_{i=1}^n \frac{(X_i- \overline X)^2}{(n-1)} $$
is called the sample variance.
A) Show that $ (n-1)S^2=\left[\sum_{i-1}^n ((X_i-\mu)^2 \right]-n(\overline X-\mu)^2 $
(Hint: Start with $(n-1)S^2=\sum_{i-1}^n ((X_i-\mu)+(\mu-\overline X))^2 $)
B) Use result from A) to show that $ E[S^2]=\sigma^2 $
answer for A)
Work:
I used the hint to obtain
$$ \sum_{i=1}^n [(X_i-\mu)^2+2[(X_i-\mu)(\mu-\overline X)]+(\mu-\overline X)^2] $$
$$ =\sum_{i=1}^n (X_i-\mu)^2+\sum_{i=1}^n2[(X_i-\mu)(\mu-\overline X)]+\sum_{i=1}^n(\mu-\overline X)^2 $$
$$ =\left[\sum_{i=1}^n (X_i-\mu)^2\right]+\left[2(\mu-\overline X)\sum_{i=1}^n(X_i-\mu)\right]+n(\mu-\overline X)^2 $$
$$ = \left[\sum_{i=1}^n (X_i-\mu)^2\right]+2(\mu-\overline X)(n\overline X-n
\mu)+n(\mu-\overline X)^2 $$
$$ =\left[\sum_{i=1}^n (X_i-\mu)^2\right]-2(\overline X-\mu)n(\overline X-\mu)+[(-n)(\overline X-\mu)(-1)(\overline X-\mu)] $$
$$ =\left[\sum_{i=1}^n (X_i-\mu)^2\right]-2n(\overline X - \mu)^2+n(\overline X-\mu)^2 $$
$$ (n-1)S^2=\left[\sum_{i=1}^n (X_i-\mu)^2\right]-n(\overline X - \mu)^2,\; \text{as desired.} $$
Part B) Use result from A) to show that $ E[S^2]=\sigma^2 $
I don't even know where to begin.
