1.3.7.md

Solution to Review Question

by Qiang Gao, updated at May 8, 2017

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Chapter 1 Finite-Sample Properties of OLS

Section 3 Finite-Sample Properties of OLS

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Review Question 1.3.7

Prove (1.2.21),

$$ 0 \le p_i \le 1 \text{ and } \sum_{i=1}^{n} p_i = K, \tag{1.2.21} $$

where

$$ p_i \equiv \mathbf{x}_i' ( \mathbf{X}' \mathbf{X} )^{-1} \mathbf{x}_i. \tag{1.2.20} $$

Solution

Because

$$ \mathbf{P} \equiv \mathbf{X} ( \mathbf{X}' \mathbf{X} )^{-1} \mathbf{ X }', $$

$$ p_i $$ is the $$i$$-th row and $$i$$-th column of $$ \mathbf{P} $$. Because $$ \mathbf{P} $$ is positive semidefinite, $$ p_i \ge 0 $$. Similarly, because

$$ \mathbf{M} \equiv \mathbf{I} - \mathbf{X} ( \mathbf{X}' \mathbf{X} )^{-1} \mathbf{ X }', $$

$$ 1 - p_i $$ is the $$i$$-th row and $$i$$-th column of $$ \mathbf{M} $$. Because $$ \mathbf{M} $$ is positive semidefinite, $$ 1 - p_i \ge 0 $$, $$ p_i \le 1 $$.

Finally,

$$ \begin{align} \sum_{i=1}^n p_i & = \mathrm{trace} ( \mathbf{P} ) \ & = \mathrm{trace} ( \mathbf{X} ( \mathbf{X}' \mathbf{X} )^{-1} \mathbf{ X }' ) \ & = \mathrm{trace} ( ( \mathbf{X}' \mathbf{X} )^{-1} \mathbf{ X }' \mathbf{X} ) \ & = \mathrm{trace} ( \mathbf{I}_K ) \ & = K. \end{align} $$

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Copyright ©2017 by Qiang Gao