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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?

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1 Answer 1

up vote 1 down vote accepted

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

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 $\mathrm{Eq.}\ 3$.

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