# probability

• Jun 19th 2010, 12:07 PM
Sampras
probability
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.
• Jun 20th 2010, 06:09 AM
theodds
Try integration by parts, noting that $\displaystyle P(X > \lambda) = 1 - F(\lambda)$ where F is the CDF. Also, I think you are missing an assumption on the support of $\displaystyle X$.
• Jun 20th 2010, 03:03 PM
jamix
Quote:

Originally Posted by Sampras
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.

That statement isn't true unless $\displaystyle \lambda$ is a nonnegative random variable.

Did you really mean to prove the following:

$\displaystyle E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda - \int_{0}^{\infty} P(X \leq -\lambda) \ d \lambda$
• Jun 21st 2010, 04:21 PM
Sampras
$\displaystyle X$ 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?

• Jun 21st 2010, 04:27 PM
Sampras
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: $\displaystyle P(X \geq \lambda) \leq \frac{E(X)}{\lambda}$?
• Jun 22nd 2010, 09:55 AM
jamix
Quote:

Originally Posted by Sampras
$\displaystyle X$ 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

$\displaystyle E(X) = \lambda P(X \leq \lambda)- P(X \leq \lambda)$

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.
• Jun 22nd 2010, 01:58 PM
awkward
Quote:

Originally Posted by Sampras
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.

Hint:

$\displaystyle \int_0^\infty P(X > \lambda) \, d\lambda = \int_0^\infty \int_\lambda^\infty f(x) \,dx \, d\lambda$

Now interchange the order of integration in the double integral.