I've noticed that when standardizing or centering the regressor(s), the p value for the intercept becomes smaller. Does this always happen? If so, why?

I know that $$ var(\hat{\beta}) = \sigma^2(X^TX)^{-1} $$

and that the t-stat is

$$ t_i = \frac{\hat{\beta}_i}{SE(\hat{\beta}_i)} $$

I can't tell from this general equation, nor the ones for simple linear regression found in https://en.wikipedia.org/wiki/Simple_linear_regression why the $t_i$ will increase and thus result in a reduction in the p-value.


1 Answer 1


When you center the predictors, the only coefficient that changes is the intercept. See this answer.

Particularly of notice, is the fact that

\begin{matrix} \hat{\beta_L} = \left[\matrix{n^{-1}\mathbf {1}_n^T \\ (X^{*T}CX^*)^{-1}X^{*T}C}\right]y & \hat{\beta_X} = \left[\matrix{ n^{-1}\mathbf {1}_n^T+n^{-1} (\mathbf {1}_n^TX^*(X^{*T}CX^*)^{-1}X^{*T})C \\ (X^{*T}CX^*)^{-1}X^{*T}C }\right]y \end{matrix}

It's not particularly obvious that the p-value will diminish, it can actually increase. But, for non-trivial transformation of the independent variables, it's almost guaranteed to change.

  • $\begingroup$ I'll check out your derivation in the linked post in a little bit, but I just want to follow up on "it can actually increase." When can it increase? $\endgroup$
    – 24n8
    May 4, 2021 at 14:36
  • 1
    $\begingroup$ Suppose that the estimated intercept for the centered model is exactly zero (there exist values of the means and variances of x and y for which this will be true). Then the centered p-value is 1, regardless of the standard deviation ... $\endgroup$
    – Ben Bolker
    May 4, 2021 at 15:20

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