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Let X be a continuous random variable with a continuous distribution function $F$ that is strictly increasing where it is not $0$ or $1$. Show that the random variable $Y = F(X)$ is uniformly distributed on the interval $[0, 1]$.

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closed as too localized by cardinal, Benjamin Lim, The Chaz 2.0, Did, Zev Chonoles May 8 '12 at 2:39

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Homework is not a stand alone tag on this site. Please use another subject tag. What have you tried BTW? –  user21436 Apr 19 '12 at 0:06
Related:meta.math.stackexchange.com/questions/4001/… –  Henry Apr 19 '12 at 0:10
You should be able to do this, if you've understood the previous responses to your earlier questions. Find the distribution function of $Y$ first, then differentiate. Note that $F_X$ has an inverse function $F^{-1}$ on a suitably restricted domain. –  David Mitra Apr 19 '12 at 0:24
@DavidMitra : To speak of differentiation is to make the matter too complicated. That isn't needed here. –  Michael Hardy Apr 19 '12 at 0:34
Related meta thread: meta.math.stackexchange.com/questions/4001/… –  cardinal Apr 19 '12 at 23:56

1 Answer 1

up vote 7 down vote accepted

For $0\le y\le 1$, $$\Pr(Y\le y)= \Pr(F(X)\le y) = \Pr(X\le F^{-1}(y))=F(F^{-1}(y))=y.$$ So you have $\Pr(Y\le y)=y$. Now think about what "uniform distribution" means.

Now here's something for you to think about: It said $F$ is strictly increasing where its value is neither $0$ nor $1$. Think about how that was used above. What is it that I wrote above that wouldn't work if $F$ were not strictly increasing?

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