# how to bound the hat matrix?

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?.

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?).
• 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 $1$). – Minus One-Twelfth Feb 14 at 13:46