Which books / notes contain a (nice) proof for $E[(\text{TSS}-\text{RSS})/p]=\sigma^2$?
(Some context: multiple linear regression, total sum of squares, residual sum of squares.)
I found a nice geometric proof for $\sum_i(X_{i}-{\bar {X}})^{2}\sim \sigma ^{2}\chi _{n-1}^{2}$ (with independent, $\cal{N}(\mu,0)$-distributed $X_i$), from which one can obtain $E[\text{RSS}/(n-1)]=\sigma^2$ in the case $p=0$, where I think of $[X_1-\bar{X},\dots,X_n-\bar{X}]$ as the orthonormal projection of $[X_1,\dots,X_n]$ on the hyperplane $\sum_i x_i=0$. Then I use rotationally invariance of the normal distribution, so the sum of squares on that hyperplane is $\sigma^2\chi^2_{n-1}$-distributed.
Skipping a few details here, but I really like the picture. Couldn't come up with a similar intuition for above equation though. I'd be happy about either proof, geometric in nature or not.
When doing linear regression on the model $y = X \beta^* + \epsilon$, you are essentially projecting the the i.i.d. noise $\epsilon_i \sim N(0, \sigma^2)$ onto the subspace spanned by the columns of $X$. (In the case $p=0$, this is a one-dimensional subspace spanned by $(1,\ldots,1)$.)
By properties of the Gaussian distribution, the projection of $\epsilon$ onto this subspace has the distribution of a $(p+1)$-dimensional Gaussian distribution, so the sum of squares has distribution $\sigma^2 \chi^2_{p+1}$. By a Pythagorean-type theorem, one can show that this sum of squares is precisely $\text{TSS} - \text{RSS}$, so $(\text{TSS}-\text{RSS})/(p+1) \sim \sigma^2 \chi^2_{p+1}$ and $E[(\text{TSS} - \text{RSS})/(p+1)] = \sigma^2$.
Similarly, one can show that the projection of $\epsilon$ on the orthogonal complement of the span of the columns of $X$ has the distribution of a $(n-p-1)$-dimensional Gaussian distribution, so its sum of squares has distribution $\sigma^2 \chi^2_{n-p-1}$. This sum of squares is $\text{RSS}$, so you get $\text{RSS}\sim \sigma^2 \chi^2_{n-p-1}$ and $E[\text{RSS}/(n-p-1)] = \sigma^2$, which recovers your computation in the special case $p=0$.