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I came across a problem where X is a non-negative RV with CDF $F_X(x)$, and I have to show that $$E[min(X,1)] = P(X\geq U),$$ being U a uniformly distributed variable from 0 to 1, independent of X. In this question they show that $$E[min(X,1)] = \int_0^1 P(X\geq x)dx,$$ but I didn't grasp how this probability can "get out" of the integral for a uniformly distributed U.

Emílio Dolgener Cantú
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    Try writing out the integral explicitly. First we can assume that $X \in [0,1]$. Then, $P(X \geq U)$ is an integral over a subset of $[0,1] \times [0,1]$, and the pdf is just $p(x)$. (Nitpick: You forgot to mention that $U$ is independent from $X$.) – Elle Najt Mar 27 '21 at 18:26
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    $P(X \ge U) = \int_0^1 \int_u^\infty f_X(x) \ dx du = \int_0^1 [1 - F_X(u)] \ du = \int_0^1 P(X \ge u) \ du$. As Lorenzo points out, this requires $X,U$ to be independent. – Gregory Mar 27 '21 at 18:26
  • Oops, indeed it is stated in the problem that they are i.i.d. Edited the question to add this information. – Emílio Dolgener Cantú Mar 27 '21 at 18:37

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If we assume that the uniform variable is between $k/n$ and $(k+1)/n,$ then the probability that your variable is bigger approximately equals $P(X\geq (k+1)/n).$ Now, the probability that $X>U$ is the sum of the disjoint events $U_k,$ so the Riemann sum converges exactly to the integral on your RHS.

Igor Rivin
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