# Equivalence of Logistic regression to Gaussian naive bayes

I was revisiting the differences between logistic regression and Naive Bayes, and had a conceptual question. A logistic regression classifier makes intuitive sense to me as a classifier that directly estimates $$P(y \mid x)$$ without making any assumptions about how the data is distributed (except that the conditional distribution $$P(y \mid x)$$ is Bernoulli). To my understanding, however, logistic regression is exactly equal to a Naive Bayes classifier with an assumed Gaussian distribution and constant variance. So, I was wondering, why is it equal if Naive Bayes is making the explicit assumption of a Gaussian distribution?

Thanks.

• Your assertion would benefit from a numerical illustration, as it is not immediately clear what you mean. The logistic model $p(x)=\frac{1}{1+\exp(-(\beta_0 +\beta_1x))}$ is less Gaussian in nature than the probit model $p(x)= \Phi ( \gamma_0 +\gamma_1x)$ though easier to explain in terms of log-odds Commented Mar 25, 2023 at 10:52

Logistic regression assumes a particular functional form for the conditional probabilities $$P(c|\mathbf x)$$ ($$c$$ being the class index).
Under the Naive Bayes assumptions with some further restrictions ($$x_i|c$$ are independent, gaussian distributed with the same variance), the conditional probabilities $$P(c|\mathbf x)$$ happen to have the same functional form. So you can view Logistic regression as a generalized model, of which this specific instance of Naive Bayes is a special case. Of course this can also viewed as one of the motivations for choosing this particular functional form.