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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.

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$F(x) = P(X\leq x)$ 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 $y \in(0,1)$.

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  • $\begingroup$ 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

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