Solving for the quantile of a probability density function.

Question

What is the quantile, p, from the density $e^{-x}(1+e^{-x})^{-2}$?

I believe I am on the right path to the solution, but I am stuck part way through.

I figure that the CDF is almost certainly $(1+e^{-x})^{-1}$.

So to start, then we must have $(1+e^{-x})^{-1} = p$, which I have managed to rearrange as $-ln((1-p)/p) = x$. The concern that I have is that the question is multiple choice, and none of the solutions match the solution I have arrived at.

So there must be some crucial insight that I have overlooked in the process of the trying to resolve this, that or the CDF I have arrived at is incorrect.

For reference, the choices are:

$p/(1-p)$

$(1-p)/p$

$ln((1-p)/p)$, which is the solution that most closely resembles mine were it not for the fact that is its negative.

$ln(p/(1-p))$

$1/(1+e^{x})$

Answer 1

By setting $a = (1-p)/p$ in the general formula $\ln(a^{-1})=-\ln(a)$, you have

$$x = -\ln((1-p)/p) = \ln\left(\left(\frac{1-p}p\right)^{-1}\right) = \ln\left(\frac{p}{1-p}\right)$$

Answer 2

First you have to define your pdf completely, that is $$f(x)=e^{-x}(1+e^{-x})^{-2}$$ for $-\infty < x <\infty $. You have to find $a$ such that $$\int_{-\infty}^{a} e^{-x}(1+e^{-x})^{-2} \, dx = p.$$ Thus $$ \frac{e^a}{e^a+1}=p.$$ Can you solve for $a$? \begin{align*} \frac{d}{da}\left( \frac{e^a}{e^a+1} \right) &=\frac{\left(\frac{d}{da}e^a\right)(e^a+1)-\left[\frac{d}{da}(e^a+1)\right]e^a}{(e^a+1)^2}\\ &=\frac {e^a(e^a+1)-e^a\cdot e^a}{(e^a+1)^2}\\ &=\frac{e^a}{(e^a+1)^2}\\ &=\frac{e^a}{e^{2a}(1+e^{-a})^2}\\ \end{align*}