# EX proof

I have a problem understanding the following proof:
EX = integral from 0 to infinity of 1 - F(x) dx
Say this integral can be:
EX = integral from 0 to infinity of P(X > x), then
EX = integral from 0 to infinity of integral from x to infinity of f(y)dydx
= integral from 0 to infinity of integral from 0 to y of dxf(y)dy = EX

Can somebody explain me how and why the last integral is included?
Thanks.

## Answers and Replies

EnumaElish
Science Advisor
Homework Helper
jetoso said:
EX = integral from 0 to infinity of integral from x to infinity of f(y)dydx
= integral from 0 to infinity of integral from 0 to y of dxf(y)dy = EX
$$\int_0^\infty \left(\int_x^\infty f(y)dy\right) dx \overset{?}{=} \int_0^\infty \left(\int_0^y dx\right) f(y)dy$$

Think of this as integrating the area of the unit square. You could first integrate along the y axis then integrate that integral along the x axis. Alternatively you could first integrate over x then integrate that over y.

Or suppose you are to integrate function f over the area that lies below the y = x line on the unit square. You can either integrate f from 0 to x on the y axis then integrate that from 0 to 1 on the x axis. Alternatively you could integrate f from 0 to y on the x axis then integrate that from 0 to 1 on the y axis.

This isn't an exact or perfect explanation but I hope that it will be useful to some degree.

statdad
Homework Helper
The integral

$$\int_0^\infty \left(1-F(x)\right) \, dx$$

can be rewritten with this step.

$$\int_0^\infty \left(1-F(x)\right) \, dx = \int_0^\infty \left(\int_x^\infty f(y)\dy\right) \,dx$$

The region over which we are integrating is the portion of the first quadrant that is on and to the right of line $$y = x$$

This can also be calculated by reversing the order of integration. In this case the inner integral goes from $$x=0$$ to $$x = y$$. Here is the work all in a single location.

\begin{align*} \int_0^\infty \left(1-F(x)\right) \, dx & = \int_0^\infty \left(\int_x^\infty f(y) \,dy\right) \,dx\\ & = \int_0^\infty \left(\int_0^y \, dy \right) f(x) \,dx = \int_0^\infty y f(y) \,dy \end{align*}

If you remember that the variable names are simply placeholders in this work you realize that the final integral is simply the expected value of $$X$$.