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Let $X$ be a continuous random variable. Let $F(t)=P(X\le t)$ be the cdf (cumulative distribution function) of $X$. Then the random variable $Y=F(X)$ takes values in the unit interval $[0,1]$. What is the distribution of $Y$? I read from a book that seems to claim $Y$ has a uniform distribution. But I don't see why. $Y$ is too abstract for me to understand.

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This is known as the probability integral transform. For $0<t<1$ we have \begin{align} F_Y(t) &= \mathbb P(Y\leqslant t)\\ &= \mathbb P(F_X(X)\leqslant t)\\ &= \mathbb P(X\leqslant F_X^{-1}(t))\\ &= F_X(F_X^{-1}(t))\\ &= t, \end{align} so that $Y$ is uniformly distributed over $(0,1)$. Note that when $X$ is not continuous the map $F_X^{-1}$ is not a true inverse, and must instead be defined as the quantile function $F_X^{-1}(t) = \inf\{x:F_X(x)>t\}$.

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