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So if the model is $y_i = \beta_0 + \beta_1 x_{i1} + \beta_2 x_{i2} + \beta_{12} x_{i1} x_{i2} + \epsilon_i$, how can it be proved that the estimate and standard error for $\beta_{12}$ the same if we center $x_{.1}$ and $x_{.2}$ ?

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  • $\begingroup$ This is a problem I am going through in a class I am auditing and I am unable to prove it. The result is supposed to hold ONLY with centering, not standardising as well. Ie in lm(Sepal.Length ~ scale(Petal.Length, scale = F) * scale(Petal.Width,scale = F), iris) will have the same interaction term as lm(Sepal.Length ~ scale(Petal.Length) * scale(Petal.Width), iris) $\endgroup$
    – Aditya Kaushik
    Commented Jun 28, 2020 at 23:51
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    $\begingroup$ @FransRodenburg the question is about centering alone, not centering and scaling to unit standard deviation as in your code. (Default is center=TRUE, scale=TRUE). $\endgroup$
    – EdM
    Commented Jun 28, 2020 at 23:51
  • $\begingroup$ Does this answer your question? Mean centering interaction terms. The example there is for centering only one of 2 interacting variables, but just try centering both in the first equation presented in the answer. $\endgroup$
    – EdM
    Commented Jun 28, 2020 at 23:53
  • $\begingroup$ @EdM Yeah that does answer the question, thanks! $\endgroup$
    – Aditya Kaushik
    Commented Jun 28, 2020 at 23:59

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