Compute the expected value of a random variable $Y$ that has a distribution function defined by parts

Question

Let $X$ be a nonnegative random variable with continuous distribution $F$. And let $Y$ be a random variable with CDF \begin{equation} G(y) = \begin{cases} 1- \alpha\int_{y}^{\infty}(1- F(x) )dx, &y>0 \\ 0, &y \leq 0, \end{cases} \end{equation} where $\alpha = \frac{1}{E(X)}$.

Here is my attempt:

$$ E(Y) =\int_{0}^{\infty} [1-G(y)]dy = \int_{0}^{\infty} [1-\{ 1-\alpha \int_{y}^{\infty}(1- F(x)) dx \} ]dy = \alpha \int_{0}^{\infty} \int_{y}^{\infty}(1- F(x) )dxdy. $$

But at this point I'm stuck, I do not know how to proceed. Do you know how compute $E(Y)$ using this formula? or Do you know other way to compute $E(Y)$?

Answer 1

By interchanging the order of integration, your last integral becomes $$\int_0^\infty (1-F(x))\int_0^x \, dy \, dx = \int_0^\infty x(1-F(x)) \, dx = \frac{1}{2} E[X^2]$$ where the last equality is shown here.