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I have been given a question:

Suppose that $F(x)=P(X\leq x)$. Show that if $F$ is continuous and $Y:=F(X)$, then $P(Y\leq y)=y$ for all $y\in [0,1]$.

I have been having difficult even starting to prove anything, since the definiton of $Y$ seems to me somewhat self refrencing. I will be thankful for any clues.

Keen-ameteur
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    $P(Y\leq y)=P(F(X) \leq y)=P(X \leq F^{-1}(y))=F(F^{-1}(y))$, and I guess there's a typo in the way you described the CDF, it should be $F_X(x).$ – User9523 Oct 31 '17 at 19:41
  • The whole thing does not make sense to me. $X$ is a (real?) function on a probability space. So, what does $F(X)$ mean then? And what is $x$ (which doesn't appear on the LHS)? – amsmath Oct 31 '17 at 19:48
  • I have considered using the method stated above, but it does not take into consideration the continuity of $F$. And to the second commentor, you are correct and I have changed it accordingly. – Keen-ameteur Oct 31 '17 at 19:54
  • $F$ s only right-continuous. It is only continuous at points $x$, where $P(X=x) = 0$. – amsmath Oct 31 '17 at 19:57
  • @amsmath True in general, but here continuity of $F$ is one of the hypotheses. I guess $F(X)$ really means $F \circ X$. –  Oct 31 '17 at 20:22
  • @Bungo I don't know how you read, but OP explicitly says "Show that $F$ is continuous". Why should it be a hypothesis so? EDIT: Before editing, OP wrote $F(X) = P(X\le x)$. This confused me. – amsmath Oct 31 '17 at 20:26
  • @Bungo Oh, I am the one that cannot read... Small words... Excuse me. – amsmath Oct 31 '17 at 20:30
  • @Keen-ameteur, I think some of your confusion about "self-referencing" might arise from the fact that in $P(F(x)\leq y)$, the $F$ only represents the CDF of $X$ in the sense that it is the same function, but isn't actually a CDF there in the sense of being a probability rule. E.g. let $f_Y(y)$ be any PDF, we can take a suitable random variable $X$ and ask $P(f_Y(X)\in A)$, but here $f_Y$ is just a particular function, it's not really a PDF in that particular instantiation of it. Of course, its nature as a PDF may help solve the problem. – jdods Oct 31 '17 at 20:39

2 Answers2

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Let $F_Y$ be the cumulative distribution function of $Y=F(X)$, where $X$ is a random variable. Then

$$ F_Y(y)=\mathbf{P}(Y\leq y) = \mathbf{P} (F(X)\leq y) = \mathbf{P}(X\leq F^{-1}(y))=F(F^{-1}(y))=y$$

for $y \in [0,1].$

Observe that a uniform distribution has such a c.d.f.

$\textbf{To OP:}$ Do check out also

Proof of $Y=F_X(X)$ being uniformly distributed on $[0,1]$ for arbitrary continuous $F_X$

and

Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X

Stoner
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  • You might have a problem at $y=1$. Actually you have problems at every point that is not reached by $F$. – amsmath Oct 31 '17 at 20:04
  • @amsmath I believe the answer to that lies in the links I've shared above. I apologize as I can't give a definite answer which I can say is right for the point you have raised. – Stoner Oct 31 '17 at 20:23
  • When I read the question, I missed the word "if" in "if $F$ is continuous". So, I have to apologize. – amsmath Oct 31 '17 at 20:33
  • @amsmath No worries. We do make mistakes from time to time! – Stoner Oct 31 '17 at 20:36
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I first read "Show that $F$ is continuous", so I missed the word "if"... However, I leave my answer here.

Both claims are false if $F$ is not assume d to be continuous. First, $F$ is in general not continuous. It has jump continuities at those points $x$ for which $P(X=x) > 0$. However, $F$ is right-continuous.

Let $x_0$ be a jump discontinuity of $F$ and set $y_0 := \sup_{x<x_0}F(x)$. Then, $F(x_0) = y_0+P(X=x_0)$. If $y\in (y_0,F(x_0))$, then $$ P(Y\le y) = P(F\circ X < F(x_0)) = P(X < x_0) = F(x_0) - P(X=x_0) = y_0 < y. $$ But sure, if you are given in addition that $P(X=x)=0$ for all $x$, then both claims hold true.

amsmath
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