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I have fit a logistic regression model with original y and standardized x variables. Slope coefficients can be easily converted back to their original scale by $\beta^*_j/\sigma_{x_j}$ where $\beta^*_j$ is the output from the fitted model. However, I am not sure how to get the original intercept estimate from the output. Any help is appreciated.

Thanks

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Start with a simple logistic regression:

$\,\,\text{logit}(\mu) \,= \beta_0 + \beta_1 x\quad\quad$ (original)

$\quad\quad\quad\quad= \beta_0^* + \beta_1^* (x-\bar{x})/s_x\quad\quad$ (standardized x)

$\quad\quad\quad\quad= (\beta_0^* -\beta_1^*\bar{x}/s_x)+ (\beta_1^*/s_x) x$

So $\beta_1=\beta_1^*/s_x$ and $\beta_0=\beta_0^* -\beta_1^*\bar{x}/s_x$

More generally, the same manipulation yields

$\beta_0=\beta_0^* -\sum_{i=1}^p\beta_i^*\bar{x_i}/s_{x_i}$ and $\beta_i=\beta_i^*/s_{x_i}$

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  • $\begingroup$ Thanks Glen. Looks alright. However I did not mention in my question that after I standardized the X variables, I fitted the model without intercept, i.e., $\beta^*_0=0$. This is generally done in multiple linear regression. I am not sure whether this trick hold for logistic regression as well or not. If holds, then intercept estimate will be obtained easily by using your formula. But if not, then I have to do my work. $\endgroup$
    – user3755127
    Jun 24, 2014 at 5:00
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    $\begingroup$ Yes, it is the same $\endgroup$
    – Glen_b
    Aug 29, 2014 at 23:13

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