Let us say we have a linear model $$ Y=\beta_0+\beta_1X+\varepsilon, \quad \varepsilon|X\sim i.i.N(0,\sigma^2). $$ Suppose estimation yields $(\hat\beta_0,\hat\beta_1,\hat\sigma^2)=(5,2,9)$.

Question 1: Shall one write $$ Y=5+2X+\hat\varepsilon $$ or $$ Y=5+2X+\varepsilon? $$ I believe the former makes more sense, since the coefficient estimates $(5,2)$ need not perfectly coincide with the true coefficients $(\beta_0,\beta_1)$ and so $\hat\varepsilon$ will not perfectly coincide with $\varepsilon$. On the other hand, I do not think I commonly see $\hat\varepsilon$ in such situations...

Question 2: What can be said about the estimated counterpart of $\varepsilon|X\sim i.i.N(0,\sigma^2)$? I think that $\hat\varepsilon$s are dependent, so one $i.$ from $i.i.N$ drops. Not sure whether conditioning on $X$ makes sense and if it does, what role it plays. The distribution has an estimated imposed mean of $0$ and variance of $9$, but other than that, can we say anything more? What is the most we can say that would be correct?


2 Answers 2


Q1: It is true that $y_i = \hat\beta_0 + \hat\beta_1 x_i + \hat\epsilon_i$ for each observation $i$, largely because this is the definition the estimated residual $\hat\epsilon_i$. Note the estimated residual is not a parameter which is estimated--instead it is constructed. On the other hand, the latter expression assumes you've exactly found the error $\epsilon$, which is of course not true.

Q2: In the case that $y \sim \mathcal{N}(X\beta, \sigma^2 I)$, which is a default linear model, the estimated residual $\hat{\epsilon} \sim \mathcal{N} \left(0, \sigma^2 \left( I - H \right) \right)$, where $H=X(X^TX)^{-1}X^T$ is the hat matrix. This fully characterizes the distribution of the estimated residuals.

  • $\begingroup$ That is very helpful. Some follow ups: 1. Does it make sense to use $\hat\varepsilon|X$ instead of $\hat\varepsilon$? 2. Does the $i.i.$ drop only because the residuals are dependent or are there more reasons for that? 3. What if the normality assumption is violated? Since we rarely can be sure it holds, I was hesitant to use $N$ for $\hat\varepsilon$. While I did not mind that for $\varepsilon$ since there it serves as an explicit assumption and nothing more, I am concerned about using it for $\hat\varepsilon$ it would serve as a statement of fact (a finding). $\endgroup$ Commented May 28, 2021 at 16:44
  • $\begingroup$ @RichardHardy 1. Yes, that's fine to use. Most people treat the covariates as fixed, so it's mostly a matter of preference. 2. my response to Q2 is describing a vector, not each entry. 3. In that case, the distribution of the residuals is just a linear transformation of the distribution of y, whatever it is. So its $(I-H) \times (\text{distribution of y})$. $\endgroup$
    – user257566
    Commented May 28, 2021 at 19:20
  • $\begingroup$ @RichardHardy does that make sense? $\endgroup$
    – user257566
    Commented Jun 6, 2021 at 22:37
  • $\begingroup$ Yes, I think so. How does your vector notation deal with i.i.d. though? It is an assumption I would not like to skip, I would like it to be explicit. $\endgroup$ Commented Jun 7, 2021 at 6:27
  • $\begingroup$ @RichardHardy It automatically deals with that. To help explain vector notation, let's briefly talk about the normal distribution, which may be more concrete. You could have $y \sim \mathcal{N}(X \beta, \sigma^2 I)$, in which case the entries of $y$ are independent and have constant variance. You could also have $y \sim \mathcal{N}(X \beta, \Sigma)$, where $\Sigma$ is an arbitrary positive semi definite matrix; in this case, the covariance between the entries of $y$ is whatever you specify. $\endgroup$
    – user257566
    Commented Jun 7, 2021 at 14:49


A simple way to circumvent this is to write instead:

$$\hat{y} = 5 + 2x$$

This avoids any confusion over $\epsilon \neq \text{residuals}$. The next part is a matter of opinion, but I also think it looks better, because $\hat{\epsilon}$ implies the error is estimated, while it is $\boldsymbol{\beta}$ that is estimated and the residual is just a remainder of that estimating procedure.


The assumptions of ordinary linear regression include that the error is independently, identically distributed and independent of $X$.$^{[1]}$ Conditioning on $X$ is therefore not necessary and you can simply write $\epsilon$ in place of $\epsilon|X$.

In a textbook, the $i.i.$ part might be elucidating, but for a paper, using normal linear regression implies the errors are independent. If they weren't, any standard errors and $p$-values generated by the model wouldn't be correct anyway.

Lastly, a small point, but the error mean is assumed to be zero and the residuals will by definition sum to $0$ when using ordinary least squares, so I wouldn't say that the mean was estimated to be $0$.

$[1]$: See e.g. Fox, John. Applied regression analysis and generalized linear models. Los Angeles: Sage, 2008

  • $\begingroup$ Excellent minor point. I would like to press a bit harder on Q2: is $\hat\sigma^2=9$ the most we can say as the counterpart of the entire $\varepsilon|X\sim i.i.N(0,\sigma^2)$ part? $\endgroup$ Commented May 28, 2021 at 16:34
  • $\begingroup$ Also, regarding the first paragraph under Q2, my notation makes the assumptions you have listed explicit. I wonder if concealing one of the assumptions by skipping $|X$ is justified. If it is, why not conceal any other assumption by omitting some other part in the notation (e.g. omitting $i.i.$)? I think it would be most consistent to keep the entire thing. $\endgroup$ Commented May 28, 2021 at 16:42
  • $\begingroup$ Ping........... $\endgroup$ Commented Jun 7, 2021 at 6:27

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