As per my understanding of logistic regression, a log of odds of the desired value of “y” should be in linear relation with the log (x). Does that mean that independent variables should have exponential distribution, which is again a non-normal distribution?
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$\begingroup$ Thanks, Ronaldo. I thought because of ln(p/1-p)=b0 + b1 ln(e^(X)). Thus, ln(p/1-p)= b0 + b1.X. $\endgroup$– Payal BhatiaCommented Aug 23, 2019 at 5:12
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$\begingroup$ @rolando2: do you want to post your comment(s) as an answer? Better to have a short answer than no answer at all. Anyone who has a better answer can post it. $\endgroup$– kjetil b halvorsen ♦Commented Nov 23, 2020 at 1:34
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No, the logistic model is actually simpler than you are describing. The log of odds of the predicted value of “y” should be in linear relation simply with "x". And "x" might follow any of a variety of distributions.
