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Let $F: \mathbb{R} \to [0,1]$ be a distribution function:

  • left continuous,
  • $\lim_{k\to-\infty}F(k)=0$,
  • $\lim_{k\to\infty}F(k)=1$,
  • $F$ is monotonic increasing

Then, suppose $X$~UNI[0,1], then $Y = F^{-1}(X)$ has distribution function $F$.

The proof is simple and clear, but I feel like there must be some intuition/explanation behind this technical result. I have already seen the measure theoretic definition of random variables so hit me with it if it helps.

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    $\begingroup$ The statement is equivalent to $F(Y)$ being uniform on $[0,1]$ for any $Y\sim F$. How intuitive that is depends how intuitive you find $\mathbb P\{F(Y)\le x\}=\mathbb P\{Y\le F^{-1}(x)\}=F\circ F^{-1}(x)=x$ which is the CDF of a uniform. $\endgroup$
    – Kurt G.
    Commented Sep 29, 2022 at 11:35

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I think that I may have asked this question in the past as well, but I think I have the answer now. A simple re-writing makes the lemma much clearer.

Suppose a function $F$ satisfies the requirements of a distribution function, and we know that $Y = F(X)$ while $Y$~UNI$[0,1]$. Then, the distribution function of $X$ is $F$.

$$\mathbb{P}(X<x) = \mathbb{P}(F(X)<F(x)) = \mathbb{P}(Y<F(x)) = F(x)$$

The first equality makes sense, because $F$ is monotonic increasing in one variable (invertible). The last equality makes sense because if $F$ is a distribution function and its image is in $[0,1]$.

An example:

Suppose that we have a function $F$:

  • $F(k)=0$, if $k \leq \alpha$,
  • $F(k)=\frac{k - \alpha}{\beta - \alpha}$, if $k \in [\alpha, \beta]$,
  • $F(k)=1$, if $k \geq \beta$.

Obviously, $F$ is the distribution function of UNI~$[\alpha, \beta]$. Now, let $Y$~UNI$[0,1]$, and suppose we found a random variable $X$ such that $Y=F(X)$. Then this means $Y = \frac{X-\alpha}{\beta - \alpha}$, thus $X$ must be distributed like UNI~$[\alpha, \beta]$.

The phrasing in the previous paragraph is kind of weird, so let us do it differently.

  • Restricted to $[\alpha, \beta]$, our $F^{-1}(k)=(\beta - \alpha)k + \alpha$.
  • Let $X:=F^{-1}(Y)=(\beta - \alpha)Y + \alpha$.
  • Thus, we have that $F(X) = F(F^{-1}(Y)) = Y$, which is UNI~$[0,1]$.
  • Thus, $X$ must be distributed like UNI~$[\alpha, \beta]$.
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  • $\begingroup$ this is a self answer, because I got stuck in this part of my notes several times. I don't know why I bothered writing this to be honest. $\endgroup$ Commented Sep 29, 2022 at 16:26

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