Let $X$ be a continuous random variable with cdf, $F_X(x)$ and let $Y=F_X^{-1}(U)$ where $U$ is a continuous uniform from zero to one. Find cdf of $Y$

Question

Let $X$ be a continuous random variable with cdf, $F_X(x)$ and let $Y=F_X^{-1}(U)$.The distribution function is strictly increasing on a single interval (which could be infinite, so that the inverse function $F_X^{-1}(y)$ is defined in the natural way.$U$ is a continuous uniform random variable on the interval zero to one. Find cdf of $Y$

So I believe I have the density function of $U$ is $f(x)=1$ for $0\leq x\leq 1$ and $f(x)=0$ otherwise.

But I'm not sure how to interpret $F_X^{-1}(U)$,

I realize its the inverse of $F_X$ but I don't understand what $U$ is.

Answer 1

Suppose that $X \sim \mathsf{Exp}(\lambda = 1)$ so that $F_X(u) = 1 - e^{-u}.$ Then, after a little algebra one has $F_X^{-1}(u) = -\ln(1-u).$

If you let $U \sim \mathsf{Unif}(0,1),$ then $F_X^{-1}(U) \sim \mathsf{Exp}(1).$

This is a method for simulating a random variable $X$ by inverting its CDF (if feasible), and then transforming a standard uniform random variable $U$ according to the inverse CDF of of $X,$ sometimes known as the quantile function of $X.$ [The output of a satisfactory pseudorandom number generator is essentially indistinguishable from independent realizations of a standard uniform random variable.]

Demonstration in R:

set.seed(411)    # for reproducibility
u = runif(10^5)  # generate values from UNIF(0,1)
x = -log(1 - u)  # quantile transform to get EXP(1)
hist(x, prob=T, ylim=c(0,1), br = 30, col="skyblue2")
 curve(dexp(x), add=T, col="red", lwd=2, n = 10001)

The first 5000 realizations of $X$ above pass a Kolmogorv-Smirnov test for $\mathsf{Exp}(1).$ [The test can't accommodate more than 5000 values.]

ks.test(x[1:5000], "pexp", 1)

        One-sample Kolmogorov-Smirnov test

data:  x[1:5000]
D = 0.012407, p-value = 0.4248
alternative hypothesis: two-sided

Answer 2

Let's find the CDF of Y. Loosely defined, the CDF is the probability that the random variable takes on a a value less than or equal to some threshold. Let us use the notation of $F_{Y}(y)$ to represent the CDF. $$CDF_Y = F_{Y}(y)$$ $$= P(Y\le y)$$ $$=P(F_X^{-1}(U) \le y)$$ As the CDF is monotonically increasing, the inequality is preserved when we apply the CDF in the following step $$=P( U \le F_X(y))$$ Now, if we rewrite this inequality, we can see that this equation represents a Uniform CDF $$=F_U(F_X(y))$$ Now, taking the parameters of U from your description, we have U ~ Uniform(0,1) which yields the following interpretation for $F_U$:

$$ F_U(F_X(y))= \begin{cases} 0 &\text{ if } F_X(y) \lt 0 \\ F_X(y) &\text{ if } 0 \le F_X(y) \lt 1 \\ 1 &\text{ if } F_X(y) \ge 1 \end{cases} $$

Because we know $F_X(y)$ is a CDF, and CDFs only take values from $[0,1]$, we can collapse the previous piecewise definition down to

$$F_U(F_X(y)) = F_X(y)$$

Pulling down the initial part of the equation, we have

$$F_Y(y) = F_X(y)$$

and thus

$$Y \sim X $$