# Can we maximize the log of the odds instead of log of the probability?

Cross Validated Asked on September 9, 2020

As above. This question relates to all optimization formulations in statistics.

Yes, because if you have two different probabilities $$p_1$$ and $$p_2$$ with $$p_1 < p_2$$, the following are all equivalent: begin{align} p_1 & < p_2 \ - p_1 & > - p_2 \ 1 - p_1 & > 1 - p_2 \ frac{1}{1 - p_1} & < frac{1}{1 - p_2} end{align} because applying an increasing function does not reverse the inequality (like $$f(x) = x + 1$$ from line 2 to 3) while applying a decreasing function (like $$g(x) = -x$$ from lines 1 to 2 and $$h(x) = 1/x$$ for positive values of $$x$$ from lines 3 to 4) does reverse the inequality. Then multiplying the first and last lines gives you $$frac{p_1}{1 - p_1} < frac{p_2}{1 - p_2}.$$ These are also equivalent to $$log(p_1) < log(p_2)$$ and $$log left( frac{p_1}{1 - p_1} right) < log left( frac{p_2}{1 - p_2} right)$$ because $$log$$ is an increasing function.