4
$\begingroup$

Consider the following theoreom: If a random variable $X$ has CDF $F,$ then $F(X)\sim U[0,1]$ where $F(.)$ is the c.d.f of $X$.

The demonstration is very straightforward:

$\mathbb{P}\left(F\left(X\right)\leq f\right)=\mathbb{P}\left(X\leq F^{-1}\left(f\right)\right)=F\left(F^{-1}\left(f\right)\right)=f$.

for $F(.)$ invertible.

Is there any intuition behind this result? Why is, for instance the probability that $F(X)\leq 0.1=0.1$?

Many thanks!

$\endgroup$
1

2 Answers 2

5
$\begingroup$

The intuition is that if, for any $p \in (0,1)$ and any $x$ such that $F(x)=\mathbb P(X \le x) =p$, you have $\mathbb P(F(X) \le F(x))=p$ and $\mathbb P(F(X) \le p)=p$ which happens iff $F(X) \sim \mathcal U(0,1)$.

In words, that is saying that if the CDF of $X$ is a function of $x$, then the probability of the function being less than or equal to a particular value $p$ must be the probability $X$ is less than or equal to a value of $x$ that gives that $p$ in the CDF, which is of course $p$ itself. That only happens when the distribution is uniform on the unit interval.

Clearly you need the CDF to be continuous for this to work, since there needs to be an $x$ which gives precisely that $p$ with $F(x)=p$.

$\endgroup$
1
$\begingroup$

Think of probability density function of a standard uniform distribution, it is a rectangle. A flat line between zero and one is the probability destiny equal for all values on the $x$-axis. The area under the curve is one. The cumulative distribution function tells us what is $\Pr(X\le x)$, it starts at 0 with probability equal to 0 and needs to end at 1 with probability equal to 1. If the distribution was discrete, to calculate the intermediate values you would take cumulative sum over probabilities for $x$’s, since they are all equal, each jump in the cumulative sum would be the same. With continuous values you don’t sum the probabilities but take an integral, but still for each $x+\epsilon$ the jump would be the same, so the cumulative distribution function would be a line joining (0,0) and (1,1).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.