How do you show $\displaystyle E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda $? We know that $\displaystyle E(X) = \int_{0}^{\infty} \lambda \ p(\lambda) \ d \lambda $. In other words, $\displaystyle P(X > \lambda) $ controls all the moments.
How do you show $\displaystyle E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda $? We know that $\displaystyle E(X) = \int_{0}^{\infty} \lambda \ p(\lambda) \ d \lambda $. In other words, $\displaystyle P(X > \lambda) $ controls all the moments.
$\displaystyle X $ is non-negative.
So we know that $\displaystyle E(X) = \int_{0}^{\infty} \lambda p(\lambda) \ d \lambda $. Let $\displaystyle u = \lambda $ and $\displaystyle dv = p(\lambda) \ d \lambda $. Then $\displaystyle du = 1 $ and $\displaystyle v = P(X \leq \lambda) $. So
$\displaystyle E(X) = \lambda P(X \leq \lambda)- P(X \leq \lambda) $
$\displaystyle = \lambda P(X \leq \lambda)- E(X) = \int_{0}^{\infty} P(X \leq \lambda) \ d \lambda $
$\displaystyle = \lambda P(X \leq \lambda)- E(X) = \int_{0}^{\infty} 1-P(X > \lambda) \ d \lambda $
$\displaystyle = \lambda P(X \leq \lambda)- E(X) = \lambda- \int_{0}^{\infty} P(X > \lambda) \ d \lambda $
So we have:
$\displaystyle \int_{0}^{\infty} P(X > \lambda) \ d \lambda = E(X)-\lambda [\underbrace{P(X \leq \lambda)-1}_{0}] $
and so the result follows. Is this correct?
Or more generally, one can show that $\displaystyle E(X^p) = p \int_{0}^{\infty} \lambda^{p-1} P(X > \lambda) \ d \lambda $ using induction on $\displaystyle p $? One could use Markov's inequality: $\displaystyle P(X \geq \lambda) \leq \frac{E(X)}{\lambda} $?
Unfortunately integrating by parts won't help much here because were evaluating a definite integral. Your first line would actually be this:
$\displaystyle E(X) = \infty - \int_{0}^{\infty} P(X \leq \lambda) $
which is obviously no good.
I don't know the answer off the top of my head. I'll post back if I have any useful hints.