2
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.