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The logistic function is the inverse of the natural logit function
The standard logistic function looks like (equation_1)
$$ {\displaystyle {\begin{aligned} f(x)&={\frac {1}{1+e^{-x}}}={\frac {e^{x}}{e^{x}+1}}={\frac {1}{2}}+{\frac {1}{2}}\tanh({\frac {x}{2}})\\ \end{aligned}}} $$
the natural logit function looks like (equation_2)
$$logit(p) = \log\left(\dfrac{p}{1-p}\right)$$
how to justify equation_1 is the inverse of equation_2?
$\endgroup$ 12 Answers
$\begingroup$let $y = logit(x) = log \dfrac{x}{1-x}$
$e^y = \dfrac{x}{1-x}$
$1 + e^y = \dfrac{1-x}{1-x} + \dfrac{x}{1-x} = \dfrac{1}{1-x}$
$\dfrac{1}{1 + e^y} = 1 - x$
$x = 1 - \dfrac{1}{1 + e^y} = \dfrac{e^y}{1 + e^y}$
$\endgroup$ $\begingroup$just calculate $f\big(\operatorname{logit}(p)\big)=p$ and $\operatorname{logit}\big(f(x)\big)=x$;$$f\big(\operatorname{logit}(p)\big) = \frac{1}{1+e^{-\log\left(\frac{p}{1-p}\right)}}=\frac{1}{1+\frac{1-p}{p}}=\frac{1}{\frac{1}{p}}=p$$and$$\operatorname{logit}\big(f(x)\big)=\log\left(\frac{\frac{1}{1+e^{-x}}}{1-\frac{1}{1+e^{-x}}}\right)=\log\left(\frac{1}{1+e^{-x}-1}\right)=\log\left(e^x\right)=x.$$
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