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Is the high leverage sample observation considered a subset of the population, forcing its squared residual to be less than or equal to the population variance? Or am I misinterpreting the question? I'm trying to determine why the answer is false to the below question.

Midterm Review Q

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  • $\begingroup$ Do you know what leverage measures, in an intuitive sense? $\endgroup$ – Glen_b Feb 9 '14 at 21:30
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    $\begingroup$ Yes, horizontal distance of an observation from the mean of the data. So is the answer that a high leverage point could have an squared residual of 0 which could be less than population variance ? $\endgroup$ – Info5ek Feb 9 '14 at 21:43
  • $\begingroup$ "Distance from the mean" is not intuitively "leverage". Why is that called "leverage"? $\endgroup$ – Glen_b Feb 9 '14 at 22:33
  • $\begingroup$ That's how it was explained to me. Intuition is subjective. $\endgroup$ – Info5ek Feb 9 '14 at 23:25
  • $\begingroup$ Again, why leverage? What is this quantity designed to measure? Note the second paragraph on the Wikipedia page on leverage $\endgroup$ – Glen_b Feb 10 '14 at 0:05
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and

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Which precisely matches what I guessed initially, that any squared residual must be less than or equal to the population variance. These formulas also suggest that higher leverage will result in a lower squared residual.

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    $\begingroup$ This answer begins with a useful and relevant observation but then appears to conflate a "squared residual" with variance. The two differ; a squared residual can be a lot larger than its variance even for influential data. $\endgroup$ – whuber Feb 9 '14 at 23:02
  • $\begingroup$ aren't squared residuals an estimator for the population variance? $\endgroup$ – Info5ek Feb 9 '14 at 23:14
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    $\begingroup$ The estimator and the estimand are different things. In this case we must clearly distinguish $e_i^2$ (the squared residual itself) from $\mathbb{E}(e_i^2)$ (its expectation--a population property) from $\sigma^2$ (another population property). $\endgroup$ – whuber Feb 9 '14 at 23:32
  • $\begingroup$ In that case what's the proper way to describe the answer using the above formulas? $\endgroup$ – Info5ek Feb 9 '14 at 23:36
  • $\begingroup$ My understanding is that $E (e_i)=0$ because the above analysis is (implicitly) conditioning on the configuration of the "X matrix". So $E (e_i^2)=var (e_i) $ holds in this case. $\endgroup$ – probabilityislogic Apr 24 '16 at 16:07

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