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Let's say my model is this: $y = -0.372 + 0.045x_1 + 0.03x_2 - 0.205x_3 + 0.114x_4$, and my standardized model is this: $y = 0.635β_1 + 0.618β_2 - 0.466β_3 + 0.232β_4$.

Why is there no intercept in the model with standardized predictors?

Thank you very much for you answers!

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    $\begingroup$ There is a confusion of notation here. For example, $\beta_1$ is (or is estimated to be) $0.635$. What it multiplies is a standardised version of $x_1$. I won't edit this because it may be part of what you don't understand and need to think through. $\endgroup$ – Nick Cox Feb 10 '15 at 15:56
  • $\begingroup$ I just copied this from another website. Do you mean that it should mean: 0.635 * x1? $\endgroup$ – 00schneider Feb 10 '15 at 16:34
  • $\begingroup$ No; you need another symbol as $x_1$ is already in use for your original unstandardized variable. (Don't trust the other website if this is typical.) $\endgroup$ – Nick Cox Feb 10 '15 at 16:36
  • $\begingroup$ so, should I use z1? $\endgroup$ – 00schneider Feb 10 '15 at 21:44
  • $\begingroup$ Your choice, but $z_1$ is certainly a possibility. $\endgroup$ – Nick Cox Feb 10 '15 at 23:19
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Think of just one predictor. If we scale to $y' = (y -$ mean of $y$) $/$ sd of $y$ and similarly for $x$ then necessarily $y' = b'x'$. With several predictors it is still true (because the regression hyperplane goes through the intersection of the means of all variables) that the intercept vanishes.

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