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We have a linear model \begin{align} Y=X\beta + \varepsilon \end{align} where $X$ is $n \times p$ ($n > p$) matrix of full rank. All assumptions of linear model hold.

I have to prove inequality which is connected with the elements $h_{ij}$ of the hat matrix $H=X(X'X)^{-1}X'$. The inequality is: \begin{align} 0 \le h_{ij}^{2} \le 0.25, \end{align} where $i \neq j$. Do you have any ideas how to deal with it?

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  • $\begingroup$ Do you mean that for an arbitrary projection matrix this inequality holds, or you need to show it for a specific projection matrix, in which case you must know something specific about the $X$ matrix? $\endgroup$ Nov 10, 2013 at 16:14
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    $\begingroup$ Use the fact that $H=H^2$ -write out what that means. $\endgroup$ Nov 10, 2013 at 16:52
  • $\begingroup$ This question has been asked also in math SE, here : math.stackexchange.com/questions/561287/…. I have voted to close. $\endgroup$ Nov 10, 2013 at 18:38
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    $\begingroup$ Please do not cross-post; this is against SE policy. Figure out where you want this question to be & delete the other version. $\endgroup$ Nov 10, 2013 at 19:10
  • $\begingroup$ Ok, sorry for this. I have already deleted it from stackexchange. $\endgroup$
    – John Snow
    Nov 10, 2013 at 19:23

1 Answer 1

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Let $H$ be an idempotent symmetric matrix. Then $H^2=H$ and for each $i=1,...,n$ we have

$$\sum_{j=1}^nh_{ij}^2=h_{ii},$$

where we simply write the elements of the main diagonal $H^2$. We can rewrite this equality as

$$\sum_{j\neq i}h_{ij}^2=h_{ii}-h_{ii}^2 \>.$$

The quantity $h_{ii}-h_{ii}^2$ can be at most $1/4$ and we immediately get our desired result.

P.S. The inequality cannot be improved without additional assumptions, as illustrated by the following idempotent matrix

$$\begin{bmatrix}1/2 & 1/2\\1/2 & 1/2\end{bmatrix} \>.$$

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  • $\begingroup$ And $h^2_{ij} \ge 0$, for $i \ne j$, follows trivially from $h_{ii}= ||\mathbf{H}\mathbf{e}_i||^2$. $\endgroup$ Nov 11, 2013 at 13:21
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    $\begingroup$ @Skull: Or more trivially...because it is the square of a real number. :-) $\endgroup$
    – cardinal
    Nov 11, 2013 at 13:23
  • $\begingroup$ ;) Sure. So remember my comment when someone happens to ask why we have $h_{ii} \ge 0$ ... $\endgroup$ Nov 11, 2013 at 13:28
  • $\begingroup$ @Skullduggery note you write $h_{ij}$ for $i\neq j$ (sic!) follows from $h_{ii}=...$. Nevertheless the fact that $h_{ii}\ge 0$ is a useful one. $\endgroup$
    – mpiktas
    Nov 11, 2013 at 13:30
  • $\begingroup$ @mpiktas: Thanks for being clear. So please enter my statement directly after the second equation of your answer and bear in mind that $h_{ii} \le 1$ follows trivially from $||\mathbf{e}_1||=1$. $\endgroup$ Nov 11, 2013 at 13:33

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