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Solution to Analytical Exercise

by Qiang Gao, updated at May 17, 2017

Chapter 1 Finite-Sample Properties of OLS

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Analytical Exercise 1.2 (The annihilator associated with the vector of ones)

Let $$ \mathbf{1} $$ be the $$n$$-dimensional column vector of ones, and let $$ \mathbf{M}_1 \equiv \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' $$. That is, $$ \mathbf{M}_1 $$ is the annihilator associated with $$ \mathbf{1} $$. Prove the following:

(a) $$ \mathbf{M}_1 $$ is symmetric and idempotent.

(b) $$ \mathbf{M}_1 \mathbf{1} = \mathbf{0} $$.

(c) $$ \mathbf{M}_1 \mathbf{y} = \mathbf{y} - \bar{y} \cdot \mathbf{1} $$ where

$$ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i. $$

$$ \mathbf{M}_1 \mathbf{y} $$ is the vector of deviations from the mean.

Solution

(a) The symmetry of $$ \mathbf{M}_1 $$ is verified as

$$ \begin{align} \mathbf{M}'_1 & = \mathbf{I}'_n - ( \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' )' \ & = \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' && \text{(Because $( \mathbf{A} \mathbf{B} )' = \mathbf{B}' \mathbf{A}' $, $ ( \mathbf{A}^{-1} )' = ( \mathbf{A}' )^{-1} $)} \ & = \mathbf{M}_1. \end{align} $$

The idempotency of $$ \mathbf{M}_1 $$ is verified as

$$ \begin{align} \mathbf{M}_1 \mathbf{M}_1 & = ( \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' )( \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' ) \ & = \mathbf{I}_n

  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \ & = \mathbf{I}_n
  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}'
  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \ & = \mathbf{I}_n
  • \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \ & = \mathbf{M}_1. \end{align} $$

(b)

$$ \begin{align} \mathbf{M}_1 \mathbf{1} & = ( \mathbf{I}_n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' ) \mathbf{1} \ & = \mathbf{1} - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \mathbf{1} \ & = \mathbf{1} - \mathbf{1} \ & = \mathbf{0}. \end{align} $$

(c)

$$ \begin{align} \mathbf{M}_{1} \mathbf{y} & = ( \mathbf{I}n - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' ) \mathbf{y} \ & = \mathbf{y} - \mathbf{1} ( \mathbf{1}' \mathbf{1} )^{-1} \mathbf{1}' \mathbf{y} \ & = \mathbf{y} - \mathbf{1} \cdot \frac{1}{n} \cdot \sum{i=1}^{n} y_i \ & = \mathbf{y} - \bar{y} \cdot \mathbf{1}. \end{align} $$

Copyright ©2017 by Qiang Gao