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I've been going back over my notes from Stats class and came across the Probability Integral Transform. From my limited understanding, the basic idea is that from a cdf in terms of one variable, can be transformed into another cdf in terms of different variable:

  • i.e. from $F_x(x)$ to --> $F_y(y)$

Is this understanding correct? What is the purpose behind this? Finally, is there a general procedure in performing the transformation?

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1 Answer 1

up vote 5 down vote accepted

Your understanding looks basically correct to me.

As far as purpose, I've seen it used mostly to generate random variables from continuous distributions. For instance, if $X$ has a $U(0,1)$ distribution, then $F_X(x) = x$. Thus the requirement $F_X(x) = F_Y(y)$ in the probability integral transform reduces to $x = F_Y(y)$ or $y = F_Y^{-1}(x)$. Since $y$ is an observation from the probability distribution $Y$, this means that we can generate observations from the distribution $Y$ by generating $U(0,1)$ random variables (which most software programs can do easily) and applying the $F_Y^{-1}$ transformation.

For example, suppose you want to generate instances of an exponential$(\lambda)$ random variable. The cdf is $$F(y) = \int_0^y \lambda e^{-\lambda t} dt = 1 - e^{-\lambda y}.$$ Solving for $y$, we have $$F(y) - 1 = - e^{-\lambda y} \Rightarrow -\lambda y = \ln (1- F(y)) \Rightarrow y = F^{-1}(x) = -\ln(1-x)/\lambda.$$

Thus if $x$ is an observation from a $U(0,1)$ distribution, then $y = -\ln(1-x)/\lambda$ is an observation from an exponential$(\lambda)$ distribution. Moreover, $x$ having a $U(0,1)$ distribution is equivalent to $1-x$ having a $U(0,1)$ distribution, so we often express the transformation as $y = -\ln x/\lambda$.

As far as a general procedure for performing the transformation, what I've done here with the uniform and exponential distributions should give you a guide. Unfortunately, though, there aren't that many commonly-used distributions for which the cdf can be inverted analytically.

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"Unfortunately, though, there aren't that many commonly-used distributions for which the cdf can be inverted analytically." - indeed. :( –  J. M. Apr 20 '11 at 0:05

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