If $ F(x) = P[X\le x] $ is continuous in x, show that $ Y=F(X) $ is measurable and that $Y$ has a uniform distribution $ P[Y\le y] = y, \; 0\le y \le 1 $

My first question is about notation. What does $ F(X) $ mean? I cant make sense of $F(X) = P[X \le X] $.

Also how do you show $Y$ is measurable?

For the last part if $F^{-1}$ exists then we get that $ P(Y \leq y ) = P(F(X) \leq y) = P(X \leq F^{-1}(y)) = F(F^{-1}(y)) = y $. Which would show that $Y$ has a uniform distribution on $[0,1]$. But how do I know that $F^{-1}$ exists. Wouldn't $F$ need to be strictly increasing for the inverse to exist? But we only know that it is non decreasing.


$F(x) = P(X\leq x)$ is the CDF of $X$ and is a regular old function (well, a nondecreasing right continuous function whose limit as you tend to $-\infty$ is zero and to $\infty$ is one). Borel Measurability follows from monotonicity (see this question, for example).

$F(X)$ is the random variable you get when you take the function $F(x)$ and plug $X$ into it. For example, if you have $X \sim Exp(1)$, $F(x) = \begin{cases} 1 - e^{-x} & x \geq 0 \\ 0 & o.w. \end{cases}$. Then, $F(X)$ is the random variable given by \begin{cases} 1 - e^{-X} & X \geq 0 \\ 0 & o.w. \end{cases}.

As for the second part of your question, you can define the generalized inverse $F^{-1}(y) = \inf\{ x: F(x) \geq y\}$, which is well defined and measurable for $y \in(0,1)$.

  • $\begingroup$ Would this work to prove $Y$ is measurable. $X$ being a random variable is a measurable function and $ X:(\Omega , \mathcal B) \rightarrow (\mathbb R , \mathcal B (\mathbb R)) $ and $F$ is also a measurable function and $ F:(\mathbb R , \mathcal B (\mathbb R)) \rightarrow ([0,1] , \mathcal B (\mathbb [0,1]) $ so the composition $F(X)$ is measurable. $\endgroup$ – alpastor Mar 10 '16 at 5:13
  • $\begingroup$ See your text -- there should be a proof that borel measurable functions of a RV are also RV's. $\endgroup$ – Batman Mar 10 '16 at 6:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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