Why isn't $F_{X}^{-1}(U) \sim F_{X}(x) $?

Question

In the following passage I am having difficulty understanding we don't have $F_{X}^{-1}(U) \sim F_{X}(x)$? I also want to ask why is the uniform distribution so involved in simulating random variables?

Answer 1

$F_X(x)$ is not a random variable, so we can't say that it is "distributed equivalently" to a random variable. In fact, $F_X^{-1}(U) \sim X$ means that the random variable $Y = F_X^{-1}(U)$, the result of applying the inverse of $X$'s c.d.f. to the uniform distribution $U$, is distributed equivalently to $X$. That is, the c.d.f. of $F_X^{-1}(U)$ is $F_X(x)$.

The uniform distribution comes up so often in simulating random variables exactly because of the given theorem. For any random variable, if you can invert its c.d.f. and sample from a uniform distribution, the you can sample from your original random variable. It is generally possible to sample from uniform distributions, so this can often be done. This technique is called inverse transform sampling.

As an example, the exponential distribution has c.d.f. $$F_X(x) = 1 - e^{-x/\beta}.$$ The inverse is then $F_X^{-1}(x) = -\beta \ln(1 - x)$. It follows that, if $U$ is a uniform random variable, then $Y = - \beta \ln(1 - U)$ is exponentially distributed with parameter $\beta$. The Python numeric library NumPy actually samples from the exponential distribution in this way, as can be seen in their random sampling source code.