Assuming F is the CDF of R.V. X and is strictly increasing, the following is true: F(X) ~ Unif(0,1).

I was able to prove this as follows,

Let Y = F(X), Y is an R.V. with sample space Y = [0,1]. Mathematically,

Y = F(X) = P(X $\leq$ X),

P(Y $\leq$ y) = P(F(X) $\leq$ y) = P(X $\leq$ $F^{-1}$(y)) = F($F^{-1}$(y)) = y

I am confused by this result. Intuitively, the probability of the event (X <= X) would be 1 (once X is realized), right? Even before going with the proof, I was guessing the pdf of Y would be 1 at Y = 1 and 0 everywhere else. Can someone please help understand the gap in my understanding?

Side question: while thinking about the universality of uniform distribution and working through the above proof, I was trying to come up with an intuition for the inverse of a CDF. But, the inverse is not making a lot of sense to me. If someone has interpretations for this, I would really appreciate it.

  • $\begingroup$ Your $F'$ should be $F^{-1}$ $\endgroup$
    – jlammy
    Commented Feb 4, 2021 at 15:17
  • $\begingroup$ yes, I didn't realize using F' is confusing, changed $\endgroup$ Commented Feb 4, 2021 at 16:28
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    $\begingroup$ The problem is with notation $F(X)$. Note that $F$ is a function with domain $\mathbb R$, hence it need to take value from $\mathbb R$ and not a random variable itself, so it isn't true that $F(X) = \mathbb P(X \le X) = 1$. You have to evaluate $F(X)$ pointwise, that is, it should be understood as: given $p \in \Omega$, then $X(p)$ is some value in $\mathbb R$ and now it makes sense to tell something about $F(X(p)) = \mathbb P(X \le X(p)) = \mathbb P(\{\omega \in \Omega : X(\omega) \le X(p)\})$ and we understood $F(X)$ in that sense (in fact, for any function $f$, $f(X)$ is pointwise defined) $\endgroup$ Commented Feb 10, 2021 at 13:07
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    $\begingroup$ As for your side question, try google quantile function, it stands for "generalized inverse of CDF" (if inverse exists, then it's exactly the inverse). Maybe wiki page or some other source will help you with intuition $\endgroup$ Commented Feb 10, 2021 at 13:09

1 Answer 1


Lets remember if we have a probability space $(\Omega,\mathcal{F},\mathcal{P})$ i.e a (set, sigma algebra, probability measure), and $X$ is a (say real valued) random variable w.r.t to this guy, then $X:\Omega \to \mathbb{R}$, and so how you have defined $Y$ we similarly have $Y:\Omega \to \mathbb{R}$.

Now your proof is fine (as long as you state for $y\in[0,1]$) so I will address your concerns. Firstly you saying the pdf of $Y$ should put all the mass into $1$, i.e because $X\leq X$ is always true $Y$ should always take the value $1$. The problem here is your reading of $Y=F(X)$, this is a random variable so once we give it an $\tilde{\omega} \in \Omega$ we get

$$ Y(\tilde{\omega})=\mathbb{P}\Big(X\leq X (\tilde{\omega})\Big)=\mathbb{P}\Big(\Big\{ \omega\in \Omega ~:~X(\omega)\leq X (\tilde{\omega})\Big\}\Big), $$

so you can see where your confusion came from.

The uniform random variable is fundamental for instance in programming / simulations, take a general random variable if you have the inverse of its CDF you can numerically draw from its distribution using the uniform random variable.


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