Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 6 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.

share|cite|improve this answer
"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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.