Show that $E[x]=\int_0^\infty(1-F_X(x))dx$

Question

Show that $E[x]=\int_0^\infty(1-F_X(x))dx$ The solution is as follows: $$\begin{align} \int_0^\infty P(X>x)dx &=\int_0^\infty \int_x^\infty f_X(y)dydx\\ &=\int_0^\infty \int_0^y dxf_X(y)dy\\ &=\int_0^\infty yf_X(y)dy\\ &=E[X] \end{align}$$

I don't understand step 3 in the solution, specifically while the limit of integration changed from $\infty$ to $y$.

You're missing something, perhaps that $X$ is nonnegative. – Nov 03 '16 at 02:08 — , Nov 03 '16 at 02:08
yes, its nonnegative. How does this help? @ Zachary Selk – MAS Nov 03 '16 at 02:12 — MAS, Nov 03 '16 at 02:12

RRL · Accepted Answer · 2016-11-03T02:22:30.250

Using the indicator function

$$1_{(y \geqslant x)}= \begin{cases}1, \,\,y \geqslant x \\ 0, \,\, y < x \end{cases},$$

we have

$$\int_0^\infty \left(\int_x^\infty f_X(y)dy\right)dx = \int_0^\infty \left(\int_0^\infty f_X(y)1_{(y \geqslant x)}dy\right)dx. $$

Since the integrand is non-negative, we can apply Tonelli's theorem to switch the order of integration and obtain

$$\int_0^\infty \left(\int_x^\infty f_X(y)dy\right)dx = \int_0^\infty \left(\int_0^\infty f_X(y)1_{(y \geqslant x)}dx\right)dy \\ = \int_0^\infty \left(\int_0^y f_X(y)dx\right)dy $$

Show that $E[x]=\int_0^\infty(1-F_X(x))dx$

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