Suppose that we have a model $Y_i = \beta_0 + \beta_1X_{i1} + \beta_2X_{i2} + \dots + \beta_kX_{ik} + \epsilon_i$.

Regression has a number of assumptions, such as that the errors $\epsilon_i$ should be normally distributed with mean zero and constant variance. I have been taught to check these assumptions using a normal QQ plot to test for normality of the residuals $e_i = Y_i - \hat{Y}_i$ and a residuals vs. fitted plot to check that the residuals vary around zero with constant variance.

However, these tests are all on the residuals, not the errors.

From what I understand, the errors are defined as the deviation of each observation from their 'true' mean value. So, we can write $\epsilon_i = Y_i - \mathbb{E}[Y_i]$. These errors cannot be observed by us. *

My question is this: how good of a job do the residuals do in mimicking the errors?

If the assumptions appear to be satisfied on the residuals, does this mean they are satisfied on the errors too? Are there other (better) ways to test the assumptions, like fitting the model to a testing dataset and getting the residuals from there?

* Furthermore, does this not require that the model is specified correctly? That is, that the response really does have a relationship with the predictors $X_1, X_2,$ etc. in the way specified by the model.

If we are missing some predictors (say, $X_{k+1}\ \text{to}\ X_p$), then the expectation $\mathbb{E}[Y_i] = \beta_0 + \beta_1X_{i1} + \beta_2X_{i2} + \dots + \beta_kX_{ik}$ would not even be the true mean, and further analysis on an incorrect model seems pointless.

How do we check whether the model is a correct one?

  • $\begingroup$ An intuitive look at it is by seeing residuals as a projection of the errors onto the nullspace of the design matrix. From this projection you can learn about the errors (which can't be determined directly). For instance, verify a hypothesis about the errors, or estimate the variance $\sigma^2$. A question in which this geometric view is explained/occurs is Why are the residuals in $\mathbb{R}^{n−p}$? $\endgroup$ Apr 1, 2021 at 6:40

2 Answers 2


The residuals are our estimates of the error terms

The short answer to this question is relatively simple: the assumptions in a regression model are assumptions about the behaviour of the error terms, and the residuals are our estimates of the error terms. Ipso facto, examination of the behaviour of the observed residuals tells us whether or not the assumptions about the error terms are plausible.

To understand this general line of reasoning in more detail, it helps to examine in detail the behaviour of the residuals in a standard regression model. Under a standard multiple linear regression with independent homoskedastic normal error terms, the distribution of the residual vector is known, which allows you to test the underlying distributional assumptions in the regression model. The basic idea is that you figure out the distribution of the residual vector under the regression assumptions, and then check if the residual values plausibly match this theoretical distribution. Deviations from the theoretical residual distribution show that the underlying assumed distribution of the error terms is wrong in some respect, and fortunately it is possible to diagnose any flawed assumption from different kinds of departures from the theoretical distribution.

If you use the underlying error distribution $\epsilon_i \sim \text{IID N}(0, \sigma^2)$ for a standard regression model and you use OLS estimation for the coefficients, then the distribution of the residuals can be shown to be the multivariate normal distribution:

$$\boldsymbol{r} = (\boldsymbol{I} - \boldsymbol{h}) \boldsymbol{\epsilon} \sim \text{N}(\boldsymbol{0}, \sigma^2 (\boldsymbol{I} - \boldsymbol{h})),$$

where $\boldsymbol{h} = \boldsymbol{x} (\boldsymbol{x}^{\text{T}} \boldsymbol{x})^{-1} \boldsymbol{x}^{\text{T}}$ is the hat matrix for the regression. The residual vector mimics the error vector, but the variance matrix has the additional multiplicative term $\boldsymbol{I} - \boldsymbol{h}$. To test the regression assumptions we use the studentised residuals, which have marginal T-distribution:

$$s_i \equiv \frac{r_i}{\hat{\sigma}_{\text{Ext}} \cdot (1-l_i)} \sim \text{T}(\text{df}_{\text{Res}}-1).$$

(This formula is for the externally studentised residuals, where the variance estimator excludes the variable under consideration. The values $l_i = h_{i,i}$ are the leverage values, which are the diagonal values in the hat matrix. The studentised residuals are not independent, but if $n$ is large, they are close to independent. This means that the marginal distribution is a simple known distribution but the joint distribution is complicated.) Now, if the limit $\lim_{n \rightarrow \infty} (\boldsymbol{x}^{\text{T}} \boldsymbol{x}) / n = \Delta$ exists, then it can be shown that the coefficient estimators are consistent estimators of the true regression coefficients, and the residuals are consistent estimators of the true error terms.

Essentially, this means that you test the underlying distributional assumptions for the error terms by comparing the studentised residuals to the T-distribution. Each of the underlying properties of the error distribution (linearity, homoskedasticity, uncorrelated errors, normality) can be tested by using the analogous properties of the distribuion of the studentised residuals. If the model is correctly specified, then for large $n$ the residuals should be close to the true error terms, and they have a similar distributional form.

Omission of an explanatory variable from the regression model leads to omitted variable bias in the coefficient estimators and this affects the residual distribution. Both the mean and variance of the residual vector are affected by the omitted variable. If the omitted terms in the regression are $\boldsymbol{Z} \boldsymbol{\delta}$ then the residual vector becomes $\boldsymbol{r} = (\boldsymbol{I} - \boldsymbol{h}) (\boldsymbol{Z \delta} + \boldsymbol{\epsilon})$. If the data vectors in the omitted matrix $\boldsymbol{Z}$ are IID normal vectors and independent of the error terms then $\boldsymbol{Z \delta} + \boldsymbol{\epsilon} \sim \text{N} (\mu \boldsymbol{1}, \sigma_*^2 \boldsymbol{I})$ so that the residual distribution becomes:

$$\boldsymbol{r} = (\boldsymbol{I} - \boldsymbol{h}) (\boldsymbol{Z \delta} + \boldsymbol{\epsilon}) \sim \text{N} \Big( \mu (\boldsymbol{I} - \boldsymbol{h}) \boldsymbol{1}, \sigma_*^2 (\boldsymbol{I} - \boldsymbol{h}) \Big).$$

If there is already an intercept term in the model (i.e., if the unit vector $\boldsymbol{1}$ is in the design matrix) then $(\boldsymbol{I} - \boldsymbol{h}) \boldsymbol{1} = \boldsymbol{0}$, which means that the standard distributional form of the residuals is preserved. If there is no intercept term in the model then the omitted variable may give a non-zero mean for the residuals. Alternatively, if the omitted variable is not IID normal then it can lead to other deviations from the standard residual distribution. In this latter case, the residual tests are unlikely to detect anything resulting from the presence of an omitted variable; it is not usually possible to determine whether deviations from the theoretical residual distribution occurs as a result of an omitted variable, or merely because of an ill-posed relationship with the included variables (and arguably these are the same thing in any case).

  • 1
    $\begingroup$ Thank you for the comprehensive response. Can I ask where you got $r=(I−h)ϵ$? It seems to me that $r=Y-\hat{Y}=(I-h)Y$ $\endgroup$
    – mai
    Apr 1, 2018 at 17:39
  • 1
    $\begingroup$ Since $\boldsymbol{h} \boldsymbol {x} = \boldsymbol {x}$ you have $(\boldsymbol {I} - \boldsymbol {h}) \boldsymbol {x} = \boldsymbol {0}$ so that $\boldsymbol {r} = (\boldsymbol {I} - \boldsymbol {h}) \boldsymbol {Y} = (\boldsymbol {I} - \boldsymbol {h}) (\boldsymbol {x} \boldsymbol {\beta} + \boldsymbol {\epsilon} ) = (\boldsymbol {I} - \boldsymbol {h}) \boldsymbol {\epsilon}$. $\endgroup$
    – Ben
    Apr 1, 2018 at 22:21

Usually, the terms residuals and errors mean the same thing. If your model has no predictors, E(Y) is indeed the mean of Y. With predictors (as in your model), E(Y) is the value of Y predicted from each X. So the residuals are the difference between each observed and predicted Y.

  • 4
    $\begingroup$ "Usually, the terms residuals and errors mean the same thing." I don't think this is true - as far as I understand, residuals measure the difference between the observed value and the predicted value, while errors measure the difference between the observed value and the true mean value. $\endgroup$
    – mai
    Apr 1, 2018 at 5:32
  • 2
    $\begingroup$ Strictly speaking errors and residuals are not synonyms. The former are random variables, the latter are realizations. $\endgroup$ Apr 1, 2018 at 5:32

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