# What exactly is the Probability Integral Transform?

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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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 $y = -\ln x/\lambda$.