Why is the denominator $N-p-1$ in estimation of variance? I was recently going through the book Elements of Statistical Learning by Tibshirani et.al. In this book, while explaining the ordinary least squares model, the authors state that assume that $y_i \epsilon \mathbb{R}$ represents the observed variables, $\hat{y_i}$ represents the model output and $\mathbf{x_i} \epsilon \mathbf{R}^{p+1}$ represent the inputs. If the $y_i$s are assumed to be uncorrelated and have constant with variance $\sigma$, then the unbiased estimate of variance is $\hat{\sigma} = \frac{1}{\left (N-p-1 \right)}\sum\left( y_i - \hat{y_i} \right)^2$, summation being done from $i=1$ to $i=N$. Note that $p$ has been used here to denote the dimensionality of $\mathbf{x_i}$s. My question is why is the factor in the denominator $N-p-1$ while estimating the variance of $y_i$s i.e. $\hat{\sigma}$ ? From my understanding if the $y_s$s are real numbers that have constant variance, the factor should be equal to $N-1$.
 A: The current accepted answer is flawed, as it implicitly assumes that the error of the model $\varepsilon$ is Gaussian (otherwise you need not have $\sum(y_i-\hat{y}_i)^2\sim\sigma^2\chi^2_{N-p-1}$).
Here's a proof with the general assumption that $\varepsilon$ has mean $0$ and variance $\sigma^2 I_N$.
First note that $\sum(y_i-\hat{y}_i)^2=\|y-X\hat\beta\|^2$.
We have $$\begin{align}
y-X\hat\beta &= X\beta +\varepsilon -X(X^TX)^{-1}X^T(X\beta +\varepsilon)\\
&=X\beta +\varepsilon - X\beta -X(X^TX)^{-1}X^T\varepsilon\\
&= (I_N-H)\varepsilon\end{align}$$
where $H=X(X^TX)^{-1}X^T$ is the hat matrix. It's easy to check that 
$H^T=H$ and $H^2=H$ (indeed the hat matrix is merely the orthogonal projection on $\operatorname{Im}X$).
Hence $\begin{aligned}[t]E( \|y-X\hat\beta\|^2) &= E(\varepsilon^T(I_N-H)^T (I_N-H)\varepsilon)=E(\varepsilon^T(I_N-H)\varepsilon)
\end{aligned}$
Note that $\varepsilon^T(I_N-H)\varepsilon=\sum_{i,j} \varepsilon_i\varepsilon_j (\delta_{ij}-H_{ij})$, thus $$E(\varepsilon^T(I_N-H)\varepsilon)=\sum_{i,j} \sigma^2\delta_{ij} (\delta_{ij}-H_{ij})=\sigma^2(N-\operatorname{tr}H)$$
Note that $\operatorname{tr}H =\operatorname{tr}(X(X^TX)^{-1}X^T)=\operatorname{tr}(X^TX(X^TX)^{-1})=\operatorname{tr}(I_{p+1})=p+1 $
Putting everything together, $E( \|y-X\hat\beta\|^2)=\sigma^2(N-p-1)$
A: You can show that $\sum(y_i-\hat{y}_i)^2\sim\sigma^2\chi^2_{N-p-1}$. As expectation of a $\chi^2_{N-p-1}$ is $(N-p-1)$. Hence $\mathbb{E}(\frac{1}{N-p-1}\sum(y_i-\hat{y}_i)^2)=\sigma^2$. 
$N-p-1$ is in the denominator to make the estimator unbiased. 
A: To answer the question without using the Gaussian assumption, nor the additive model assumption as it has been done in previous answer, here is my take:
The assumption made are that the observations $y_i$ are uncorrelated and have mean zero, constant variance $\sigma^2$, and that the $x_i$ are fixed (non random). 
From the previous section we have that $\hat{y} = X(X^TX)^{-1}X^Ty = Hy$
where the hat matrix can be shown easily to satisfy $H^T=H^2=H$.
Also we can rewrite $\sum_{i=1}^N(y_i - \hat{y_i})^2$ as $(y-\hat{y} )^T(y-\hat{y} )$ and thus it comes that
\begin{align*} 
 \mathbb{E}[\sum_{i=1}^N(y_i - \hat{y_i})^2]
 &= \mathbb{E}[(y-Hy)^T(y-Hy)] 
 = \mathbb{E}[((I-H)y)^T((I-H)y)] \\
 &= \mathbb{E}[y^T(I-H)(I-H)y]
 = \mathbb{E}[y^T(I-H)y] \\
 &= \mathbb{E}[\sum_{i,j}y_iy_j(\delta_{ij}-H_{i,j})] \\
 &= \sum_{i,j}\mathbb{E}[y_iy_j\delta_{ij}]- \sum_{i,j}\mathbb{E}[y_iy_jH_{i,j}] \\
 &= \sum_{i}\mathbb{E}[y_i^2]- \sum_{i,j}\mathbb{E}[y_iy_j]H_{i,j} 
 \end{align*} 
The hat matrix has been removed from the expectation since it is fixed as the $x_i$ are.
In addition the $y_i$ are uncorrelated of mean zero and variance $\sigma^2$ thus using the cyclic property of the trace:
\begin{align*} 
\mathbb{E}[\sum_{i=1}^N(y_i - \hat{y_i})^2]
&= \sum_{i}\sigma^2 - \sum_{i}\sigma^2H_{i,i} 
= \sigma^2[N - trace(H)] \\
&= \sigma^2[N - trace(X(X^TX)^{-1}X^T)] \\
&= \sigma^2[N - trace(X^TX(X^TX)^{-1})] \\
&= \sigma^2[N - p - 1]
\end{align*} 
So finally we have that an unbiased estimator $\hat{\sigma}^2$ of $\sigma^2$ is 
\begin{align} 
\mathbb{E}[\hat{\sigma}^2]
&= \mathbb{E}[\frac{1}{N - p - 1} \cdot \sum_{i=1}^N(y_i - \hat{y_i})^2] 
= \sigma^2
\end{align} 
