When they assume the $\varepsilon_i$ have a normal distribution, with mean $0$ and variance $\sigma^2,$ then the estimated errors $\hat\varepsilon$ have a normal distribution with mean $0$ and variance $\operatorname{Var}(\hat\varepsilon).$

But what is the connection? Why, when errors are normally distributed, the estimated residuals also are normally distributed?

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    $\begingroup$ Because this is a statistics site and we have been around a fairly long time, you can find a great deal written about this with a site search. The key phrase "hat matrix" is especially helpful. $\endgroup$
    – whuber
    Commented Jul 13, 2022 at 13:36

1 Answer 1


The connection is made through a specific procedure: namely, ordinary least squares. When you suppose a response $Y$ is a random variable related to other variables $X$ in the form

$$Y = X\beta + \varepsilon$$

where $\varepsilon$ is assumed to have a joint Normal distribution (plus some other assumptions that don't matter here), then the least squares solution is

$$\hat\beta = (X^\prime X)^{-}X^\prime Y = (X^\prime X)^{-}X^\prime (X\beta + \epsilon) = \beta + J\varepsilon$$

for a matrix $J = (X^\prime X)^{-}X^\prime $ that is computed only from $X.$ The residuals, or estimated errors, are the differences

$$\hat\varepsilon = Y - \hat Y = (X\beta + \varepsilon) - X\hat\beta = \varepsilon - XJ\varepsilon = (\mathbb I - H)\varepsilon$$

where $H = XJ = X(X^\prime X)^{-}X^\prime$ is the "hat matrix." This exhibits the residuals as linear combinations of $\varepsilon$ (with coefficients given by $\mathbb I - H$). Linear combinations of jointly Normal variables are Normal, QED.


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