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The text I'm working in has this cdf function

$$F(x)=1-\left(1+5 e^{-x}\right)^{-.2}, \quad-\infty<x<\infty$$

For the inverse of the cdf, it then sets $u=F(x)$ and solves for $u$. After that it provides this for the inverse cdf without showing any steps.

$$F^{-1}(u)=\log \left\{.2\left[(1-u)^{-5}-1\right]\right\}, \quad 0<u<1$$

I am trying to work through and solve this on my own and I am getting stuck at what should be the last step. I'm coming up with the inverse function they provide, but it's equal to $-x$, not $x$. Is there some obvious way to get to just $x$ from where I'm leaving off, or am I making some error working my way to it?

\begin{align} u &= 1-\left(1+5 e^{-x}\right)^{-.2} \\ 1-u &= \left(1+5 e^{-x}\right)^{-.2} \\ (1-u)^{-5} &= 1+5 e^{-x} \\ (1-u)^{-5} - 1 &= 5 e^{-x} \\ .2[(1-u)^{-5} - 1] &= e^{-x} \\ \log \left\{.2\left[(1-u)^{-5}-1\right]\right\}&=-x \\ \end{align}

At this point $F^{-1}(u)$ should equal $x$, but I'm stuck with $-x$. Thank you for any assistance!

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Your derivation is correct. There is a typo in the question statement. It should be $5e^x$ and not $5e^{-x}$. Namely, in the last line of your attempt, there's indeed no negative sign for the lone $x$ on the right hand side.

For a plot, see Wolfram Alpha. You can see what happens when you try to change the correct 5Exp[x] to the WRONG 5Exp[-x].

Basically, you can check that $F(x) \to 0$ when $x \to -\infty$ and $F(x) \to 1$ when $x \to \infty$. Here to make this even clearer, the correct CDF is $$F(x) = 1-\frac1{(1+5 e^{x})^{1/5}}$$ as $x \to \infty$ the denominator blows up and makes the whole fraction vanish, resulting in the leading $1$ (probability one) as needed.

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  • $\begingroup$ Thank you for your help! $\endgroup$ Feb 14, 2020 at 2:14

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