Timeline for Kahn Pseudo Normal distribution

7 events

when toggle format	what		by	license	comment
May 10, 2017 at 14:06	comment	added	Glen_b		I nowhere say that $\log(1/x-1)$ is the inverse cdf (it isn't), However, I do say the method corresponds to the inverse cdf method. Note that $\log(1/v-1) = \log((1-v)/v)= -\log(v/(1-v))$, Keep in mind that the standard logistic is symmetric about $0$. Alternatively (but equivalently), note that if $u=1-v$ $\log(1/v-1)=\log(u/(1-u))$, and if $V$ is uniform on $(0,1)$ then so is $U=1-V$.
May 10, 2017 at 13:58	comment	added	plasmacel		In some sources I see $\log(x/(1-x))$ as the inverse CDF of the logistic distribution. This differs from $\log(1/x-1)=-\log(x/(1-x))$ in a sign. Which is the correct one?
May 10, 2017 at 12:44	history	edited	Glen_b	CC BY-SA 3.0	added 168 characters in body
May 10, 2017 at 12:36	vote	accept	plasmacel
May 10, 2017 at 8:46	history	edited	Glen_b	CC BY-SA 3.0	added 77 characters in body
May 10, 2017 at 7:57	history	edited	Glen_b	CC BY-SA 3.0	added 668 characters in body
May 10, 2017 at 5:37	history	answered	Glen_b	CC BY-SA 3.0