It is well known that for a random variable $X$ with a continuous distribution function $F(x)$, $F(X)$ is uniformly distributed on $[0,1]$.

Can we now prove the same for a conditional distribution function. So we have two real valued, integrable random variables $X,Y$ ($Y$ can be $\Omega \to \mathbb R^d, d \in \mathbb N$) and $F(x,y) = \mathbb E[ \mathbb I \{X \in (-\infty, x) \}| Y = y]$. How can I show that $F(X,Y)$ is uniformly distributed on $[0,1]$.

I would appreciate a measure theoretical proof.


1 Answer 1


We need to find $P(F(X,Y) \le t), t \in [0,1]$.

$$P(F(X,Y) \le t)= \int \mathbb I \{F(X,Y) \le t \} dP \overset{*}{=} \int \mathbb I \{F(x,y) \le t \} dP^{X,Y}$$

I think for * above we must assume that $F$ is continuous.

There is a theorem saying that there is a version of $\mathbb E[ \mathbb I \{X \in B| Y = y]$, we call it $P^{X|Y}(y,B)$, which defines the Markov kernel for $P^{X,Y}$, in other words $P^{X,Y}=P^{X|Y} \otimes P^Y$ and $P^{X|Y}(y,B)$ is a probability mass $\forall y$.

$$\int \mathbb I \{F(x,y) \le t \} dP^{X,Y} = \int \mathbb I \{F(x,y) \le t \} dP^{X|Y} \otimes P^Y = \\ \int \int \mathbb I \{F(x,y) \le t \} P^{X|Y}(y,dx) P^Y(dy)= \int P^{X|Y}(\{F(x,y) \le t \},y) P^Y(dy) = \\ \int t P^Y(dy)=t$$.

  • $\begingroup$ Could please somebody vote it up or down, so I know that I have (not) managed to answer my own question. $\endgroup$
    – zesy
    Aug 7, 2015 at 17:28

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