# How cam I use linear regression coefficients to estimate logistic regression coefficients?

I am trying to come up with predictions for a logistic regression for the purpose of a power analysis.

All I have to go off of is a linear regression result.

Lets say there are 50 pictures and I want to know how likely someone is to remember them. (This is just a made up example.) I measure the brightness of the pictures and use that to predict the percentage of people who remembered each one.

Lets say b = .5.

Now, in my new experiment, I will be making predictions as to whether each image is remembered by a given person, using logistic regression.

How do I go from the linear regression coefficient to the logistic regression coefficient?

Again this is just for the purpose of estimating effect size for a power analysis.

Edit: Could I do something like: compare the percentage of items remembered at baseline (form the intercept) to the amount remembered when the predictor is +1. And then turn that comparison into log odds?

• Why don't you just do power analysis for the logistic regression? I was not aware you had to run any analysis to get at statistical power which is inherent in the data and the method. I don't think it is possible to go from a linear regression coefficient to a logistic one, they represent totally different things. Beyond that the model assumptions are very different, logic regression is inherently non-linear for example. – user54285 Apr 16 at 23:40

Amemiya (1981, p. 1488) shows that for slopes, $$\beta_{Logit} \approx 4 \cdot \beta_{OLS}$$. He shows that these comparisons work quite well for slope parameters if 0.1 ≤ p ≤ 0.9. This pins down the logit slope to 2 in your example.
$$\frac{1}{N} \sum_i^N \hat p= \frac{1}{N} \sum_i^N \Lambda(\alpha + \beta \cdot x_i).$$
Here $$\Lambda()$$ is the inverse logit. If you happen to know the average success rate in your OLS sample and are willing to make some assumptions about $$x_i$$s in your new experiment, you can solve that for $$\alpha$$. Or perhaps there is a better way to get that from the log odds.