Expected value of a distribution function

Here is an interesting probability problem: Suppose {X} is a random variable with absolutely continuous distribution function {F}. What is {\mathbb E[F(X)]}?
Since {F} is absolutely continuous, {F} is differentiable a.e. and {X} has a density, i.e. there is a nonnegative function {f} such that

\displaystyle F(x) = \int_{-\infty}^t f(t)\mathsf dt

for each {x\in\mathbb R}, where {F'=f} a.e. By the law of the unconscious statistician,

\displaystyle \mathbb E[F(X)] = \int_\mathbb R F(x)\mathsf dF(x) = \int_{-\infty}^\infty F(x)f(x)\mathsf dx.

Replacing {F} with the integral of its density and applying Tonelli’s theorem, we have

\displaystyle \begin{aligned} \mathbb E[F(X)] &= \int_{-\infty}^\infty \int_{-\infty}^t f(x)\mathsf dx\; f(t)\mathsf dt\\ &= \int_{-\infty}^\infty \int_x^\infty f(t)\mathsf dt \;f(x)\mathsf dx\\ &= \int_{-\infty}^\infty (1-F(x))f(x)\mathsf dx. \end{aligned}

Adding the two equations, we get

\displaystyle 2\mathbb E[F(X)] = \int_{-\infty}^\infty f(x)\mathsf dx = 1,

and hence

\displaystyle \mathbb E[F(X)] = \frac12.

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