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Given

$r(t)=\frac{f(t)}{1-F(t)} \tag{Eq. 1}$ where $f(t)=\frac{dF}{dt} \tag{Eq. 2}$

and the conditions:

$\lim_{t\rightarrow \infty} r(t)=1 \tag{Eq. 3}$ $\lim_{t\rightarrow \infty} F(t)=1 \tag{Eq. 4}$ $\lim_{t\rightarrow \infty} f(t)=1-F(t) \tag{Eq. 5}$

I can think of just one function $F$ satisfying these three conditions--the logistic function:

$F(t)=\frac{1}{1+e^{-t}} \tag{Eq. 6}$

(which can also be expressed $F(t)=r(t)$)

Is this is the only function satisfying these conditions? If so, is there a way to prove it?

1 Answers 1

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$ r(t) = \frac{F'(t)}{1-F(t)} $ $ r(t)\,dt = \frac{dF}{1-F} $ $ \int r(t)\,dt = -\log(1-F(t)) + C $ $ 1-\exp\left(-\int r(t)\,dt+C\right)=F(t). $

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 $\mathrm{Eq.}\ 3$.