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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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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