Nathan Kong

Expectation Value Using the Cumulative Distribution Function

We assume that the random variable $X$ is non-negative. We want to show that: $\mathbb{E}[x] = \displaystyle\int_0^\infty \left[1-F(z)\right] \operatorname{d}\!z$
Proof:
$\begin{align} \mathbb{E}[x] &= \displaystyle\int_0^\infty x f(x) \operatorname{d}\!x \\ &= \displaystyle\int_0^\infty \int_0^x \operatorname{d}\!z \, f(x) \operatorname{d}\!x \\ &= \displaystyle\int_0^\infty \left[\int_0^\infty \mathbb{1}(z < x) f(x) \operatorname{d}\!z \right]\operatorname{d}\!x \\ &= \displaystyle\int_0^\infty \left[\int_0^\infty \mathbb{1}(z < x) f(x) \operatorname{d}\!x \right]\operatorname{d}\!z \,\,\,\,\text{(Fubini's theorem)}\\ &= \displaystyle\int_0^\infty \left[\int_z^\infty f(x) \operatorname{d}\!x \right]\operatorname{d}\!z\\ &= \displaystyle\int_0^\infty \left[1-F(z)\right] \operatorname{d}\!z \end{align}$
where $\mathbb{1}(\cdot)$ is the indicator function.

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