Expectation Integral Identity

Question

it's known that for any random variable X:

$$ \mathbb{E}[X] = \int_{0}^{\infty} P(X>y) dy - \int_{-\infty}^{0} P(X<y) dy $$

I'm trying to prove this doing the following:

$$ \mathbb{E}[X] = \int_{-\infty}^{\infty} x f_X(x) dx = \int_{-\infty}^{\infty} \int_{0}^{x} dy f_X(x) dx = \int_{-\infty}^{\infty} \int_{0}^{x} f_X(x) dydx = \int_{-\infty}^{\infty} \int_{x}^{\infty} f_X(x) dxdy = \\ =\int_{-\infty}^{\infty} P(X \geq x)dy = \int_{0}^{\infty} P(X \geq x)dy + \int_{-\infty}^{0} P(X \geq x)dy $$

what am I missing?

Answer 1

Thanks for the suggestion/correction of kccu:

$$ \mathbb{E}[X] = \int_{-\infty}^{\infty} x f_X(x) dx = \int_{0}^{\infty} x f_X(x) dx + \int_{-\infty}^{0} x f_X(x) dx = \\ \int_{0}^{\infty} \int_{0}^{x} f_X(x) dydx - \int_{-\infty}^{0} \int_{x}^{0} f_X(x) dydx = \int_{0}^{\infty} \int_{y}^{\infty} f_X(x) dxdy - \int_{-\infty}^{0} \int_{-\infty}^{y} f_X(x) dxdy = \\ \int_{0}^{\infty} P(X>x) dx - \int_{-\infty}^{0} P(X<x) dx $$