11
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ 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
    Jul 13, 2022 at 13:36

1 Answer 1

17
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.