Given a pdf: $$f(x)=\tau x \exp\left(\frac{-\tau x^2}{2}\right); \quad x, \tau > 0$$

So I found the corresponsing cdf: $$F(x)=1 - \exp\left(\frac{-\tau x^2}{2}\right)$$

Then I got given a value for tau: $\tau=0.2$ and I derived the inverse function $F^{-1}_X(u).$: $$F^{-1}_X(u)=\sqrt{\color{red}{\frac{2}{\tau}}10 \log\left(\frac{1}{1-u}\right)}; \quad u\sim U[0,1]$$

Now from what I have been told, $F^{-1} \sim F$ but I just can't see it. I may be drawing wrong conclusions but $$0< F<1$$ and $$-\infty<F^{-1}<\infty$$

Is what I did correct?

EDIT: My lecture notes: enter image description here


1 Answer 1


The statement is

If $u \sim U[0,1]$ then $F_X^{-1}(u)\sim f$

To show you this I sample a uniform distribution and calculate

$$ x = F^{-1}_X(u) = \sqrt{-\frac{2}{\tau}\ln(1 - u)} $$

Then I make a histogram and overplot the distribution

$$ f_X(x) = \tau x e^{-\tau x^2/2} $$

enter image description here

  • $\begingroup$ Could you look at the lecture notes I included and tell me what's going on there? It made more sense for me that the inverse must follow the cdf, but I was looking at my notes which used capital F And thank you for answering of course! $\endgroup$ Jan 30, 2018 at 15:24
  • $\begingroup$ @KuderaSebastian Could be a typo $\endgroup$
    – caverac
    Jan 30, 2018 at 15:27

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