Show that in a simple linear regression y=Xbeta+epsilon, when the mean is subtracted from any feature (i.e. some column of the matrix X), the t-statistics of all the non-intercept coefficients do not change.

ebendasqc

ebendasqc

Answered question

2022-10-29

Show that in a simple linear regression y = X β + ϵ, when the mean is subtracted from any feature (i.e. some column of the matrix X), the t-statistics of all the non-intercept coefficients do not change.

Answer & Explanation

inmholtau5

inmholtau5

Beginner2022-10-30Added 16 answers

Write X = ( x i j ) i = 1 , . . . , n ; j = 1 , . . . , p . The first column j = 1 ) contains only ( 1 , 1. , , , 1 ) R n because this is the intercept. Say, you center all the columns j = 2 , . . . , p, i.e. you build the matrix X ~ = ( x ~ i j ) with x ~ i j = x i j x ¯ j where x ¯ j is the mean 1 n i = 1 n x i j . The question is whether the i-th element of ( X ~ T X ~ ) 1 is the same as the i-th element of ( X T X ) 1 .
So X e 1 is the vector of size n with all entries equal to one, and let x ¯ R p the vector with entries 0 , x ¯ 2 , . . . , x ¯ p . Then the construction of X ~ is equivalent to X ~ = X ( X e 1 ) x ¯ T = X ( I p e 1 x ¯ T ).
Now X ~ T X ~ = ( I p x ¯ e 1 T ) X T X ( I p e 1 x ¯ T ) and one can check that ( I p x ¯ e 1 T ) is invertible, because the image of canonical basis vectors are linearly independent: ( I p x ¯ e 1 T ) e j = e j for j 2 and ( I p x ¯ e 1 T ) e 1 has the first component equal to 1 due to x ¯ having 0 in the first component.
For j 2,
e j T ( X ~ T X ~ ) 1 e j = e j T ( I p e 1 x ¯ T ) 1 ( X T X ) 1 ( I p x ¯ e 1 T ) 1 e j
and one can check that ( I p x ¯ e 1 T ) 1 e j = e j because e 1 T e j = 0. This implies that X ~ T X ~ and X T X have the same diagonal elements for j 2.

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