Averie Ferguson

2022-04-03

Does the logistic function uniquely satisfy these three conditions?
Given
$r\left(t\right)=\frac{f\left(t\right)}{1-F\left(t\right)}$ {Eq. 1}
where
$f\left(t\right)=\frac{dF}{dt}$ {Eq. 2}
and the conditions:
$\underset{t\to \mathrm{\infty }}{lim}r\left(t\right)=1$ {Eq. 3}
$\underset{t\to \mathrm{\infty }}{lim}F\left(t\right)=1$ {Eq. 4}
$\underset{t\to \mathrm{\infty }}{lim}f\left(t\right)=1-F\left(t\right)$ {Eq. 5}
I can think of just one function F satisfying these three conditions--the logistic function:
$F\left(t\right)=\frac{1}{1+{e}^{-t}}$ {Eq. 6}
(which can also be expressed $F\left(t\right)=r\left(t\right)$)
Is this is the only function satisfying these conditions? If so, is there a way to prove it?

Boehm98wy

$r\left(t\right)=\frac{{F}^{\prime }\left(t\right)}{1-F\left(t\right)}$
$r\left(t\right),dt=\frac{dF}{1-F}$
$\int r\left(t\right),dt=-\mathrm{log}\left(1-F\left(t\right)\right)+C$
$1-\mathrm{exp}\left(-\int r\left(t\right),dt+C\right)=F\left(t\right).$
So you can put as many things in the role of F as will fit in the role of r, i.e. things satisfying your Eq. 3.

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