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I'm reading a paper about linear regression and in some point they define: $$ w_{ij}=\frac{h^{2}_{ji}}{ph_{ii}(1-h_{jj})^{2}}$$, where $h_{ij}$ are the elements of the hat matrix.

The problem is that they propose a bound for $w_{ij}$ and is not clear why the bound and the approximation are valid. $$w_{ij} \leq \frac{h_{jj}}{p(1-h_{jj})^{2}} \simeq \frac{1}{p}h_{jj}(1+2h_{jj})$$

We know that $$h_{ii}= h^{2}_{ii} + \sum_{i \neq j} h^{2}_{ij}$$, so $h_{ii} \ge h_{ij}^{2} $, but the bound proposed imply that, $ h_{ii}h_{jj}\ge h_{ij}^{2}$. ¿ that's valid?.

Please. i hope your help

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Yes, $h_{ii}h_{jj}\geq h_{ij}^2$ is true. Note that $h_{ii} = \mathbf{e}_i^T H \mathbf{e}_{i}$, $h_{jj} = \mathbf{e}_j^T H \mathbf{e}_{j}$, and $h_{ij} = \mathbf{e}_i^T H \mathbf{e}_{j}$, where $\mathbf{e}_{k}$ refers to the $k$-th standard basis vector of $\mathbb{R}^{n}$ and $H$ is your hat matrix. Since $H$ is symmetric positive semi-definite, the result follows by using the Cauchy-Schwarz inequality (see Is this equivalent to Cauchy-Schwarz Inequality?).

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  • $\begingroup$ ok, got it and thanks a lot. mmm what can you say about the approximation they take to the bound?. $\endgroup$ – JohanR Feb 14 at 13:39
  • $\begingroup$ For that, they are using the fact that $\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 +\cdots$ for $|x| < 1$ (you can show this by differentiating the series $\frac{1}{1-x} = 1 + x+x^2 + x^3+\cdots$). Hence if $0< x < 1$, the higher order terms will become negligible and $\frac{1}{(1-x)^2} \approx 1+2x$. And remember that $h_{jj}$, as a diagonal entry of a hat matrix, is always between $0$ and $1$ (in practice strictly less than $1$). $\endgroup$ – Minus One-Twelfth Feb 14 at 13:46
  • $\begingroup$ thank you very much for your help. $\endgroup$ – JohanR Feb 14 at 14:38
  • $\begingroup$ i want to apologize, but i'm dealing with other difficulties with the paper i'm trying to understand. is possible you help me with that?. again i feel so embarrased but is a little frustrated not advance. $\endgroup$ – JohanR Feb 15 at 13:03
  • $\begingroup$ You're welcome! And you can post another question if you have more trouble, and someone will probably help. $\endgroup$ – Minus One-Twelfth Feb 15 at 13:05

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