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.


1 Answer 1


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}

  • $\begingroup$ Thanks for the answer. How did you get that $P(U \le F(c)) = F(c)$? $\endgroup$ Oct 1, 2019 at 22:04
  • 1
    $\begingroup$ @The Pointer Since $U$ is uniform $P(U<a)=a$ for $a \in(0,1)$ $\endgroup$ Oct 1, 2019 at 22:05
  • $\begingroup$ Ahh, of course. Thank you very much for the clarification. $\endgroup$ Oct 1, 2019 at 22:07

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