# 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 $Y$

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.

• From your own description of the problem, $U$ is a continuous random variable that is uniformly distributed over the interval $[0,1]$. Can you clarify what it is about $U$ that you don't understand?
– Ceph
Nov 4, 2019 at 20:08
• I don't understand how you can apply $F_X^{-1}$ to $U$. Is it $F_X^{-1}(F_U(x))$? Nov 4, 2019 at 20:11
• $U$ is a random variable, taking values between 0 and 1. $F_X$ is a function whose range is the interval $[0,1]$. So, for any real number $z\in[0,1]$, $F_X^{-1}(z)$ gives you whatever real number $w$ is such that $F_X(w)=z$. So $F_X^{-1}(U)$ defines a random variable $W$, such that $F_X(W)=U$.
– Ceph
Nov 4, 2019 at 20:15
• Since $X$ is a continuous random variable $F_X^{-1}$ exists giving unique values except possibly on a subset of $[0,1]$ of measure $0$ Nov 4, 2019 at 20:25
• @Ceph I'm not sure what I'm supposed to do then to find $F_Y(y)$. Nov 4, 2019 at 20:34

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

• I dont understand how you get $F_X^{-1}(U)\sim Exp(1)$. Nov 5, 2019 at 2:44
• Illustrated by simulation above. See Wikipedia for discussion. Also search this site for inverse CDF transform simulation including this Q&A. Nov 5, 2019 at 3:15

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

$$Y \sim X$$