Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider real random variable $X \in \mathbb{R}$. I know that if I consider r.v. $U = F_X(X)$ where $F_X(x)$ is $X$'s CDF, we get a uniform r.v. in $[0,1]$. So the following holds:

$$U \sim \text{Unif}(0,1)$$

How can this be proved?

share|improve this question
3  
First figure out the possible values of $U=F_X(X)$. Then ask yourself "What is the probability that $U \leq a$?" and try and figure out what are the possible values of $X$ that will make $F_X(X)$ have value less than $a$. If $F_X(X) \leq a$ for all $X \in B_a$ where $B_a$ is a set that is determined by your choice of $a$, then $$P\{U \leq a\} = P\{F_X(X)\leq a\} = P\{X \in B_a\}.$$ So now all you got to do is figure out $P\{X \in B_a\}$. Go on, try it for yourself. Take $a=0.5$ for starters and see if you can get the answer. Then try $a=0.4$, $a=0.6$ etc till you see a pattern emerging. –  Dilip Sarwate Mar 31 '13 at 22:30

1 Answer 1

First of all, $X$ needs to be a continuous random variable, as if it was discrete, then any its function, in particular $F_X(X)$, would be discrete.

Let $t\in [0,1]$, and $(\Omega,P)$ be the probability space. We have $$\begin{align} P(U<t)&=P(\{\omega\,\mid\,F_X(X\omega)<t\}) \\ &=P(\{\omega\,\mid\,P(X<X\omega)\,<t\}) \end{align}$$ So, let $A:=\{\omega\,\mid\,P(X<X\omega)\,<t\}$, and we want to prove that its measure (probability) is exactly $t$. For this, observe that $$A=\bigcup_{a\in\Bbb R\,:\,P(X<a)<t} (X\le a) \,.$$ (For $\subseteq$ consider $a:=X\omega$, and the inclusion $\supseteq$ goes as follows: say $\omega\in (X\le a)$, that is, $X\omega\le a$, then $P(X<X\omega)\le P(X<a)<t$.)

Finally, by continuity of $X$, i.e. of $F_X$, we get a $z\in\Bbb R$ such that $F_X(z)=t$ (take the minimal among them), then we can conlcude $A=(X<z)$.

And thus, indeed $$P(U<t)=P(A)=t\,.$$

share|improve this answer
    
Following Dilip Sarwate's thought in the comment, probably it would be enough to consider only the supremum $z$ of those $a\in\Bbb R$ values for which $P(X<a)<t$, and prove directly that $A=(X<z)$.. –  Berci Mar 31 '13 at 23:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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