Suppose we have a linear model $Y=X \beta + \epsilon$ where $\epsilon \sim N(0, \sigma^2 I)$ and let $H$ be the projection matrix into the column space of $X$. We define the residual $e$ as the projection of $Y$ into span$(X)$ (i.e. $e=(I-H)\epsilon \text{ }$).

A statement says that : by construction, $X^T e=0$ and it implies that no correlation should appear between explanatory variables and residuals.

I can convince myself that the statement is true by saying that if $X_i$ is a column of $X$, then using the fact that $E(e)=0$, we have $Cov(X_i,e)=\Bbb E(X_i. e)- \Bbb E(X_i )\Bbb E(e)=0 \implies Corr(X_i,e)=0$.

However, I don't really intuitively see why there should or not be any correlation between explanatory variables and residuals. Can someone explain the intuition behind the statement ?

  • 2
    $\begingroup$ The residuals are what you haven't explained with your explanatory variables. That's how they are defined. If there were a better explanation using those variables, meaning with a coefficient that would work better, then the machinery you used didn't find it. In reverse, what would count as intuitive here? Do you seek a different explanation in terms of algebra or geometry, for example? Often when people say that something is intuitive, they just mean familiar. "The user interface is intuitive" $=$ you'll get used to it, or it's like that in other programs you should have used. $\endgroup$
    – Nick Cox
    Oct 30, 2022 at 12:57
  • $\begingroup$ @NickCox Your explanation is a nice. I am open for any aspects of explanations : algebra, geometry, etc. What I meant by intuition is for example that a 0-torsion curve is in a plane (maybe not the best example illustrating what I want to say) but I am looking for an explanation that does not need to be very formal, but that explains why we should expect things to behave like this. What happens if we observe a correlation ? What happens if the correlation is 0 but there is still a non-linear fashion when plotting things ? $\endgroup$
    – Kilkik
    Oct 30, 2022 at 15:50

1 Answer 1


Nick Cox answered OP's query in the comment:

The residuals are what you haven't explained with your explanatory variables. That's how they are defined.

Take a quick glance at what the least square does geometrically:

enter image description here

In the figure, $\Omega := \mathcal C(\mathbf X). $ One needs to seek $\boldsymbol\theta \in \mathcal C(\mathbf X) $ such that $\rm AB$ is minimum. From the figure, it can be noticed $\hat{\boldsymbol\theta}$ does the job. Method of least squares yields the desired $\hat{\boldsymbol\theta}$ which is the projection of $\mathbf Y$ on $\Omega.$ Again noticing the figure dictates us the following:

$$\left(\mathbf Y-\hat{\boldsymbol\theta}\right) \perp \Omega.\tag 1\label 1$$

What is $\mathbf Y-\hat{\boldsymbol\theta}? $ It is the residual vector $\mathbf e. $ So, what $\eqref 1$ is implying is that $\mathbf e$ is orthogonal to $\mathcal C(\mathbf X) $ that is, $\mathbf e$ cannot be expressed as the linear combination of the columns of $\mathbf X. $

Source of the figure:

$\rm [I]$ Linear Regression Analysis, George A. F. Seber, Alan J. Lee, John Wiley & Sons, $2003, $ Fig. $3.1, $ p. $37.$


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