Timeline for How do I convert a uniform value in [0,1) to a standard normal (Gaussian) distribution value?

9 events

when toggle format	what		by	license	comment
Oct 22, 2022 at 0:36	history	edited	KConrad	CC BY-SA 4.0	edited body
Oct 20, 2021 at 0:24	history	edited	KConrad	CC BY-SA 4.0	deleted 90 characters in body
S Jul 25, 2015 at 14:22	history	suggested	Amir Sagiv	CC BY-SA 3.0	changed R to math-type mathbb R
Jul 25, 2015 at 14:12	comment	added	Amir Sagiv		I'm not sure that in this case it is so expensive, as efficient methods and precalculated values are known for the inverse-error function.
Jul 25, 2015 at 14:04	review	Suggested edits
S Jul 25, 2015 at 14:22
Jun 19, 2010 at 0:29	comment	added	KConrad		Michael, I did see that, but (a) I'm not a probabilist and that's my excuse for not knowing what "inverse transform method" meant when I first saw it (once I looked at it later I understood it immediately, of course) and (b) the answer which said this inverse transform method is not so efficient was posted after mine, chronologically.
Jun 18, 2010 at 23:58	comment	added	Michael Hardy		KConrad, someone already gave this answer above (by linkning to a Wikipedia article) and someone else pointed out that it's computationally expensive.
Jun 17, 2010 at 23:04	comment	added	KConrad		Gerry's answer suggests a possibly more practical method: take a large number of samples from a uniform distribution on $(0,1)$ and use the central limit theorem, which explains how a standard normal distribution is a limit of a normalized average of independent identically distributed random variables.
Jun 17, 2010 at 22:05	history	answered	KConrad	CC BY-SA 2.5