Proof of MSE is unbiased estimator in Regression I am trying to prove that in multivariate linear regression $MSE = (n-2)\sigma^2 $
Here is my approach: 
Under the usual notation, 
$$  Y = X\beta + \epsilon \\
$$ 
$$ \hat Y = X\hat\beta \\
$$ 
$$  \hat\beta = (X'X)^{-1}X'Y \\ \\
\implies \hat\beta' = Y'X(X'X)^{-1}
$$ 
Now, 
\begin{align}
\Sigma (Y_i - \hat Y_i)^2  & = (Y_i - \hat Y_i)'(Y_i - \hat Y_i) \\
 & = (X(\beta - \hat \beta) + \epsilon)' (X(\beta - \hat \beta) + \epsilon)\\
 & = \underbrace {(\beta - \hat \beta)'X'X(\beta - \hat \beta)}_{term1} + \underbrace {\epsilon'X (\beta - \hat \beta)}_{term2}\\ 
 & + \underbrace {(\beta - \hat \beta)'X'\epsilon}_{term3} + \epsilon'\epsilon \\
\end{align}
Simplifying the individual terms
Term 1: \begin{align}
(\beta - \hat \beta)'X'X(\beta - \hat \beta) &= (\beta - (X'X)^{-1}X'Y)'X'X(\beta - (X'X)^{-1}X'Y)\\
& = (\beta' - Y'X(X'X)^{-1})X'X(\beta - (X'X)^{-1}X'Y) \\
& = \beta'X'X\beta - Y'X\beta - \beta'(X'X)(X'X)^{-1}X'Y + Y'X(X'X)^{-1}X'Y \\
& = \beta'X'X\beta - (\beta'X' + \epsilon')X\beta - \beta'(X'X)(X'X)^{-1}X'Y + \\ & (\beta'X' + \epsilon')X(X'X)^{-1}X'Y  \quad \text{(substituting the value of }Y') \\
& = - \epsilon'X\beta + \epsilon'X(X'X)^{-1}X'Y \quad \text {some terms get cancelled} \\
& = - \epsilon'X\beta + \epsilon'X(X'X)^{-1}X'( X\beta + \epsilon) \quad \text {substituting the value of } Y \\
& = \epsilon'X(X'X)^{-1}X'\epsilon
\end{align}
Term 2 : 
\begin{align}
\epsilon'X (\beta - \hat \beta) &= \epsilon'X(\beta - (X'X)^{-1}X'Y)\\
& = \epsilon'X(\beta - (X'X)^{-1}X'X\beta)\quad \text {substituting the value of } Y \\\\
& = 0
\end{align}
As Term 3 is transpose of Term 2, Term 3 = 0
\begin{align}
\Sigma (Y_i - \hat Y_i)^2  & = \epsilon'X(X'X)^{-1}X'\epsilon + \epsilon'\epsilon \\
  E(\Sigma (Y_i - \hat Y_i)^2)  & = E(\epsilon'X(X'X)^{-1}X'\epsilon + \epsilon'\epsilon) \\
\end{align}
I'm stuck here, unable to make any further simplifications. Can someone please help. 
What further baffles me is the RHS term is greater than $n\sigma^2$ as $E(\epsilon'\epsilon) = n*\sigma$
 A: Martijn Weterings's commnet is very useful. Your derivation of term 2 is wrong.
$\epsilon'X (\beta - \hat \beta) \\= \epsilon'X(\beta - (X'X)^{-1}X'Y) \\=\epsilon'X\left\{\beta - (X'X)^{-1}X'(X\beta+\epsilon)\right\}\\=\epsilon'X \left\{\beta-(X'X)^{-1}X'X\beta -(X'X)^{-1}X'\epsilon\right\}\\=-\epsilon'X(X'X)^{-1}X'\epsilon$
Now 
$\Sigma (Y_i - \hat Y_i)^2\\=\epsilon'X(X'X)^{-1}X'\epsilon-\epsilon'X(X'X)^{-1}X'\epsilon-\epsilon'X(X'X)^{-1}X'\epsilon+\epsilon'\epsilon\\=\epsilon'\epsilon-\epsilon'X(X'X)^{-1}X'\epsilon\\=\epsilon'\epsilon-\epsilon'P\epsilon$
$P$ is the projection matrix which is symmetric and idempotent 
Now calculate the expectation.
$E[\Sigma (Y_i - \hat Y_i)^2]\\=E(\epsilon'\epsilon-\epsilon'P\epsilon)\\=E(\epsilon'\epsilon)-E(\epsilon'P\epsilon)\\=n\sigma^2-\sigma^2trace(P) \\\text{(Suppose tarace(P)}=k)$
$=(n-k)\sigma^2$
$\therefore \frac{\Sigma (Y_i - \hat Y_i)^2}{n-k}$ is the unbiased estimator of $\sigma^2$, $k$ is the number of parameters you want to estimate,such as you want to estimate $\beta_0$ for intercept and $\beta_1$ for one predictor, the $k$ will be equal to 2.
A: A less computationally intensive method would be
$$
\begin{aligned}
e&=y-\hat{y}\\
\Sigma(y[k]-\hat{y}[k])^2&=e^Te\\
y-\hat{y}&=\phi\theta-\phi\hat{\theta}+\epsilon\\
&=\phi\theta-\phi(\phi^T\phi)^{-1}\phi^Ty+\epsilon\\
&=\phi\theta-\phi(\phi^T\phi)^{-1}\phi^T(\phi\theta+\epsilon)+\epsilon\\
&=-\phi(\phi^T\phi)^{-1}\phi^T\epsilon+\epsilon\\
e&=(I-P)\epsilon\\
e^Te&=\epsilon^T(I-P^T-P+P^TP),\;\{P=P^T=P^TP\}\\
&=\epsilon^T\epsilon-\epsilon^TP\epsilon
\end{aligned}
$$
This is an easier method to get to the necessary step. You can then proceed further as explained by the previous answer.
