# Let $X$ be an r.v. with CDF $F$. Then $F(X) \sim \text{Unif}(0,1)$?

I recently encountered a theorem called universality of the uniform:

Let $$F$$ be a CDF which is a continuous function and strictly increasing on the support of the distribution. This ensures that the inverse function $$F^{−1}$$ exists, as a function from $$(0, 1)$$ to $$\mathbb{R}$$. We then have the following results:

• Let $$U \sim \text{Unif}(0,1)$$ and $$X = F^{−1}(U)$$. Then $$X$$ is an r.v. with CDF $$F$$.

• Let $$X$$ be an r.v. with CDF $$F$$. Then $$F(X) \sim \text{Unif}(0,1)$$.

The second result is clear to me. However, I'm uncomfortable with the first result; specifically, the idea that $$X$$ is then an r.v. with CDF $$F$$. It was originally stated that $$F$$ is a CDF. If the inverse function $$F^{-1}$$ exists, and $$X = F^{-1}(U)$$, then how is it the case that the CDF of $$X$$ is $$F$$? I'm not seeing this.

I would appreciate it if people could please take the time to clarify this.

Given $$U\sim Unif(0,1)$$, $$X=F^{-1}(U)$$, to show that $$X$$ has CDF $$F$$, we check \begin{align} P(X\le c)&=P(F^{-1}(U)\le c)\\&=P(U\le F(c))\\&=F(c) \end{align}
• Thanks for the answer. How did you get that $P(U \le F(c)) = F(c)$? Oct 1, 2019 at 22:04
• @The Pointer Since $U$ is uniform $P(U<a)=a$ for $a \in(0,1)$ Oct 1, 2019 at 22:05