# $X$~UNI[0,1], $F$ a distribution function, then $Y:= F^{-1}(X)$ has distribution $F$. Looking for intuition in this technical result

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.

• 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. Commented Sep 29, 2022 at 11:35

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

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~$$[\alpha, \beta]$$.
• 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. Commented Sep 29, 2022 at 16:26