Proof that regression residual error is an unbiased estimate of error variance

Question

Consider the least squares problem $Y=X\beta +\epsilon$ while $\epsilon$ is zero mean Gaussian with $E(\epsilon) = 0$ and variance $\sigma^2$. I need to prove that

$\frac{V(\hat{\beta})}{N-(n+m)}$ is an unbiased estimate of $\sigma^2$ with $V(\beta) = ||Y-X\beta||$ .

I wasn't able to find the answer online. I just got confused by a thousand different ways to write things down.

EDIT:
$Y = \begin{pmatrix} y(0)\\ \vdots \\ y(N-1)\end{pmatrix} \quad$ $X = \begin{pmatrix} x^T(0)\\ \vdots \\ x^T(N-1)\end{pmatrix}\quad $ $\beta = \begin{pmatrix} a_1\\ \vdots \\ a_n\\ b_1 \\\vdots \\ b_m \end{pmatrix}$

Answer 1

Following your notations, we have $$V(\hat{\beta}) = \|\hat{\epsilon}\|^2 = \text{RSS}$$ i.e., the Residual Sum of Squares.

It is a fact that (cf. here) $$\frac{\text{RSS}}{\sigma²} \sim \chi_{(N-p)}^2$$ with $N$ the total sample size and $p$ the number of parameters in $\beta$ (here, $p = n + m$).

The result follows from the fact that the expectation of a chi-square random variable equals its number of degrees of freedom, i.e., $$ \text{E}\left(\frac{\text{RSS}}{\sigma²}\right) = N - p $$ which can be rewritten as $$ \text{E}\left(\frac{\text{RSS}}{N-p}\right) = \sigma² $$ since $N-p$ and $\sigma²$ are both non-random.