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The variance of residuals in a linear regression is given by : $$Var(e_i)=(1-h_{ii})\sigma^2$$ This means that residuals have a lower variance than the error terms, and the variance of residuals is smaller for observations with higher leverage. I see that from the equation above but I can't get an intuitive explanation of these facts. The only explanation I can think of for the 2nd point is : high leverage makes the fitted value $\hat y_i$ close to $y_i$, which means that their difference (= $e_i$) cannot vary a lot. Are there better explanations.

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    $\begingroup$ The flexibility afforded by fitting a regression function reduces the variability. One way to see this clearly is to imagine extreme cases of overfitting. Another way is to contemplate the decomposition into sums of squares: part of the error variance is manifested in the variance of the estimated regression function. Another way is to work through the simplest cases, such as the model $y_i = \mu+\varepsilon_i,$ which you can relate to well-understood facts about unbiased estimates of variance. It comes down to what kind of explanation you perceive as "better." $\endgroup$
    – whuber
    Nov 3, 2023 at 22:29

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