Proof that the complementary CDF equals the expected value for non-negative RVs

$$F_X(x)=P(X\leq x)$$ $$f_x(x)=\frac{dF_X(x)}{x}$$ $$s_X(x)=\int_x^\infty f_X(t)dt$$

$X$ is almost surely* non-negative and continuous

*The difference between almost surely and surely is this: a sure event will always happen, anything else is impossible. E.g. if $X \in [0,1]$, then the event $0\leq X \leq 1$ is a sure event. But the event $0 < X \leq 1$ is an almost sure event. It is almost certain to happen, with a vanishingly small probability of not happening.

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

Let's start from the definition of $E[X]$: $$E[X] = \int_{-\infty}^{\infty}xf_X(x)dx = \int_{0}^{\infty}xf_X(x)dx$$

As $f_X(x)=0\; \forall\; x < 0$.

$$\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$$

What do we choose to be $u$ and $dv/dx$? As always, we choose whatever simplifies the integral - differentiating $x$ gives a simpler result than integrating $x$. Integrating $f_X(x)$ gives $F_X(x)$ which is actually part of the result, so this seems like the way to go!

Set $u=x$ and $v'=f_X(x)$. Now we need the integral of $f_X(x)$, but integration is only defined up to a constant, so let's take $v(x)=F_X(x)-1$. If you're new to calculus, this might not seem allowed - surely this will change the result? The answer is it won't - it will just change the appearance. Look again at the left hand side of the formula for integration by parts and say we added a constant $C$ to $v$: $$u(v+C) - \int (v+C)\frac{du}{dx}dx$$ $$=uv - \int v\frac{du}{dx}dx + uC - C\int \frac{du}{dx}dx$$ $$=uv - \int v\frac{du}{dx}dx + uC - uC $$ $$= uv - \int v\frac{du}{dx}dx$$ By the definition of integration.

Now we have: $$E[X] = [x(F_X(x)-1)]_0^\infty + \int_0^\infty(1-F_X(x))dx$$ We're nearly there, we just need to prove that the $uv$ term $[x(F_X(x)-1)]_0^\infty = 0$. A classic method to prove something equals 0 is to prove it must be $\geq 0$ and it also must be $\leq 0$.

Now as $F_X(x)=P(X\leq x)$ then $1 - F_X(x) = P(X>x)=\int_x^\infty f_X(t)dt$. Define $s_X(x)=\int_x^\infty f_X(t)dt$ for convenience (also known as the survival function).

Back to that pesky $uv$ term: $$-[xs_X(x)]_0^\infty = xs_X(x)|_{x=0} - xs_X(x)|_{x\rightarrow\infty}$$ $$= - xs_X(x)|_{x\rightarrow\infty}$$ As $xs_X(x)|_{x=0} = 0$. Let's think about what happens to this term as $x\to\infty$: the $x$ term gets large linearly, but the $s_X(x)$ term goes to 0 as the range of the integral is squashed. Which one wins?

Now $x\geq 0$ and $s_X(x)\geq 0$ as $f_X(x)\geq 0$, all by or obvious from definition. $$\therefore -xs_X(x)|_{x\rightarrow\infty} \leq 0$$

Now comes the cool bit - pay attention: $$-xs_X(x)|_{x\rightarrow\infty} = -\lim_{x\to \infty}x\int_x^\infty f_X(t)dt$$ $$\geq -\lim_{x\to \infty}\int_x^\infty t f_X(t)dt$$ The last step just needs you to realise that when you move $x$ to inside the integral and change it to $t$, it is always larger than $x$ itself since the integral is going from $x$ to $\infty$, so the smallest value of $t$ is $x$ and the largest is $\infty$ at the limit of the integral.

Now the finishing touch: $$-\lim_{x\to \infty}\int_x^\infty t f_X(t)dt = 0$$ Why? Because as you take the limit of the integral as $x \to \infty$, you are squeezing the range of the integral to nothing, it becomes: $$-\int_\infty^\infty t f_X(t)dt$$ so $$0 \leq -xs_X(x)|_{x\rightarrow\infty} \leq 0$$ $$\therefore -xs_X(x)|_{x\rightarrow\infty}=0$$

$$E[X] = -[xs_X(x)]_0^\infty + \int_0^\infty(1-F_X(x))dx$$ $$=\int_0^\infty(1-F_X(x))dx$$ </p> $$\tag*{$\blacksquare$}$$