# Relating the CDF to the probability density

1. Aug 30, 2014

### bonfire09

1. The problem statement, all variables and given/known data
If $X$ is any random variable defined on $[0,\infty]$ with continuous CDF $F_X(t)$. Prove that $E(X)=\int_1^\infty (1-F_X(t)) dt$.
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2. Relevant equations

3. The attempt at a solution
I am not sure how to go about this. I think double integration can be used to prove it, but I don't want to go down that path. Is there another which I can prove this. Thanks.

Last edited by a moderator: Sep 1, 2014
2. Aug 30, 2014

### haruspex

Are you sure about those limits? I don't see how E(X) can be independent of the behaviour of F between 0 and 1.

3. Aug 30, 2014

### bonfire09

Well it says that $X$ is a random variable with range $[0, \infty]$.

4. Aug 30, 2014

### Ray Vickson

Somebody has asked you to prove a false result. The correct version is
$$EX = \int_0^{\infty} [1 - F_X(x)] \,dx .$$

Exactly how you should prove it depends on what you know about Lebesgue-Stieltjes integrals, or whether you even need to use them. The proof is easiest if you assume that $F_X(\cdot)$ is absolutely continuous, in which case there is a density $f_X(\cdot)$ on $(0,\infty)$ giving $F_X(x) = \int_0^x f_X(t) \, dt$ for all $x > 0$. If $F_X$ is not absolutely continuous, it can be continuous but not have a density function; then the proof is harder.

I suggest you try it first for the case where a density function exists. Think integration by parts.

Last edited: Aug 30, 2014
5. Aug 30, 2014

### pasmith

I agree; it should be $E(X) = \int_0^\infty 1 - F_X(t)\,dt$.

This is an application of integration by parts. You want to calculate $$E(X) = \int_0^\infty xF_X'(x) \,dx.$$ Note that $$F_X'(x) = -\frac{d}{dx}(1 - F_X(x)).$$