Inverse transform sampling: Difference between revisions

Revision as of 16:38, 26 February 2021

Revision as of 23:28, 22 March 2021

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[[File:Inverse transform sampling.png|thumbnail|360px|right|Inverse transform sampling for normal distribution]]

We are randomly choosing a proportion of the area under the curve and returning the number in the ___domain such that exactly this proportion of the area occurs to the left of that number. Intuitively, we are unlikely to choose a number in the far end of tails because there is very little area in them which would require choosing a number very close to zero or one.

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The [[probability integral transform]] states that if <math>X</math> is a [[continuous random variable]] with [[cumulative distribution function]] <math>F_X</math>, then the random variable <math>Y=F_X(X)</math> has a [[uniform distribution (continuous)|uniform distribution]] on [0,&nbsp;1]. The inverse probability integral transform is just the inverse of this: specifically, if <math>Y</math> has a uniform distribution on [0,&nbsp;1] and if <math>X</math> has a cumulative distribution <math>F_X</math>, then the random variable <math>F_X^{-1}(Y)</math> has the same distribution as <math>X</math> .

[[File:InverseFunc.png|thumb|360px|Graph of the inversion technique from <math>x</math> to <math>F(x)</math>. On the bottom right we see the regular function and in the top left its inversion.]]

== Intuition ==

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==The method==

[[File:Generalized inversion method.svg|thumb|300px360px|Schematic of the inverse transform sampling. The inverse function of <math>y=F_X(x)</math> can be defined by <math>F_X^{-1}(y)=\mathrm{inf}\{x| F_X(x)\geq y\}</math>.]]

[[File:Inverse Transform Sampling Example.gif|thumb|360px|right|An animation of how inverse transform sampling generates normally distributed random values from uniformly distributed random values]]

The problem that the inverse transform sampling method solves is as follows: