calculus - Prove that $E(X) = int_{0}^{infty} P(X>x),dx = int_{0}^{infty} (1-F_X(x)),dx$.

Let $X$ be a continuous non-negative random variable (i.e. $R_x$ has only non-negative values). Prove that $$E(X) = \int_{0}^{\infty} P(X>x)\,dx = \int_{0}^{\infty} (1-F_X(x))\,dx$$ where $F_X(x)$ is the CDF for $X$. Using this result, find $E(X)$ for an exponential ($\lambda$) random variable.

I know that by definition, $F_X(x) = P(X \leq x)$ and so $1 - F_X(x) = P(X>x)$

The solution is:
$$\int_{0}^{\infty} \int_{x}^{\infty} f(y)\,dy dx

= \int_{0}^{\infty} \int_{0}^{y} f(y)\,dy dx
= \int_{0}^{\infty} yf(y) dy.$$

I'm really confused as to where the double integral came from. I'm also rusty on multivariate calc, so I'm confused about the swapping of $x$ and $\infty$ to $0$ and $y$.

Any help would be greatly appreciated!

Answer

Observe that for a continuous random variable, (well absolutely continuous to be rigorous):

$$\mathsf P(X> x) = \int_x^\infty f_X(y)\operatorname d y$$

Then taking the definite integral (if we can):

$$\int_0^\infty \mathsf P(X> x)\operatorname d x = \int_0^\infty \int_x^\infty f_X(y)\operatorname d y\operatorname d x$$

To swap the order of integration we use Tonelli's Theorem, since a probability density is strictly non-negative.

Observe that we are integrating over the domain where $0< x< \infty$ and $x< y< \infty$, which is to say $0

$$\begin{align}\int_0^\infty \mathsf P(X> x)\operatorname d x = & ~ \iint_{0< x< y< \infty} f_X(y)\operatorname d (x,y)

\\[1ex] = & ~ \int_0^\infty \int_0^y f_X(y)\operatorname d x\operatorname d y\end{align}$$

Then since $\int_0^y f_X(y)\operatorname d x = f_X(y) \int_0^y 1\operatorname d x = y~f_X(y)$ we have:

$$\begin{align}\int_0^\infty \mathsf P(X> x)\operatorname d x = & ~ \int_0^\infty y ~ f_X(y)\operatorname d y \\[1ex] = & ~ \mathsf E(X \mid X\geq 0)~\mathsf P(X\geq 0) \\[1ex] = & ~ \mathsf E(X) & \textsf{when $X$ is strictly positive} \end{align}$$

$\mathcal {QED}$