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We know that the cumulative distribution function (CDF) follows the $U[0,1]$ distribution. What is the distribution of (1-CDF)? Is it also follows the $U[0,1]$ ? (I believe it's true for the normal distribution, how about in general)

Thank you in advance.

whuber
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score324
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    Would you perhaps be asking about the distribution of $1-F_X(X)$ where $F_X$ is the CDF of a *continuous* random variable $X$? Assuming so, your question is about the distribution of $1-U$ where $U$ is a uniform random variable. What steps have you taken to determine the answer? – whuber Sep 18 '18 at 22:10
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    Let $X\sim U[0,1]$, then let $Y = 1-X$. Now we can find the distribution of $Y$ using $f(y) = f(g^{-1}(y))|J|$. Thus, $1- F(X)$ also follows the $U[0,1]$ right? – score324 Sep 18 '18 at 22:37
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    A much easier way to approach this is to compute $\Pr(1 - U \le u)$ for any $0 \le u \le 1:$ that, by definition, is the distribution function of $1-U$ (at least for arguments $u$ between $0$ and $1,$ but it should be obvious these are the only ones that matter). Use the fact that $F_U(u)=u.$ – whuber Sep 18 '18 at 23:02

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