I was revisiting my notes about the classical linear regression model, $Y = X \beta$.

If we want to estimate the variance of the least squares estimator we usually suppose that the outputs $y_{i}$ are uncorrelated and have fixed variance $\sigma ^{2}$.

We then have $Var(\hat{\beta} ^{2}) = (X^{T}X)^{-1} \sigma ^{2}$ and we estimate $\sigma^{2}$ using $\hat{\sigma}^{2} = \frac{1}{N-p-1}\sum_{i=1}^{N}{y_{i}-\hat{y_{i}}}$. Where $N$ is the number of training examples and $p$ the number of predictors.

Why does the estimation of the variance of the outputs contains elements from the model ($p$ and $\hat{y}$)?

  • $\begingroup$ What do you expect it should be? $\endgroup$ Oct 3 at 5:14
  • $\begingroup$ @User1865345 I thought it would be $\frac{1}{N-1} \sum_{i=1}^{N}{(y_{i} - \bar{y})^{2}}$ where $\bar{y}$ is the sample mean. $\endgroup$
    – Toshi Mint
    Oct 3 at 5:30

1 Answer 1


$\DeclareMathOperator{\X}{\mathbf X^\mathsf T\mathbf X}\DeclareMathOperator{\ep}{\boldsymbol\varepsilon}$

Let $\mathbf M := \mathbf I-\mathbf X(\X)^{-1}\mathbf X^\mathsf T$ be the residual maker. Then

$$\mathbf e =\mathbf M\mathbf y. \tag 1$$

Now $\mathbf e^\mathsf T\mathbf e=\ep^\mathsf T\mathbf M\ep.$ Therefore its expectation

\begin{align}\mathbb E\left[\mathbf e^\mathsf T\mathbf e|\mathbf X\right]&= \mathbb E\left[\ep^\mathsf T\mathbf M\ep|\mathbf X\right]\\&= \mathbb E\left[\operatorname{tr}(\ep^\mathsf T\mathbf M\ep)|\mathbf X\right]\\ &= \mathbb E\left[\operatorname{tr}(\mathbf M\ep\ep^\mathsf T)|\mathbf X\right]\\ &= \operatorname{tr} \left(\mathbf M~\mathbb E\left[(\ep\ep^\mathsf T)|\mathbf X\right]\right)\\&= \operatorname{tr}(\mathbf M~\sigma^2\mathbf I)\\&=(n-P)\sigma^2\tag 2\label 2\end{align}

From $\eqref{2},$ one can deduce a natural (unbiased) estimator of $\sigma^2$ that is $(n-P)^{-1}(\mathbf e^\mathsf T\mathbf e). $


$[\rm I]$ Econometric Analysis, William H. Greene, Pearson Education, $2018, $ chapter $4, $ p. $62.$

  • $\begingroup$ Thanks a lot for your answer. It was very helpful. Unfortunately my vote isn't taken into account because I need at least 15 reputation but it said that my feedback has been recorded. Just a little precision if you permit, may I say that this leads us to deduce an unbiased estimator of $\sigma ^{2}$ taking into account our knowledge of the data, and not the usual unbiased sample variance (because it doesn't incorporate the data)? Because we're conditioning on $\textbf{X}$. $\endgroup$
    – Toshi Mint
    Oct 3 at 15:29
  • $\begingroup$ That's correct assessment. $\endgroup$ Oct 3 at 15:30
  • 1
    $\begingroup$ Thank you again. You have clarified a huge misunderstanding of mine. $\endgroup$
    – Toshi Mint
    Oct 3 at 15:30

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