Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I stumbled upon an expression in an article of statistics for an $n$-th moment with $X$ being a random variable over $[0, \infty)$.

$$\mathbb{E} X^{n} = \int^{\infty}_{0} nz^{n-1}\; \text{Pr}(X > z) \; \text{dz}$$

Could someone enlighten me on why the above is true? It indeed works for the exponential distribution.

share|improve this question
2  
If you are familiar with the formula $$E[Y] = \int_0^{\infty} P\{Y > y\}\,\mathrm dy$$ for a nonnegative random variable, set $Y = X^n$ and do a change of variable $z = y^{1/n}$ in $$E[Y] = \int_0^{\infty} P\{Y > y\}\,\mathrm dy = \int_0^{\infty} P\{X^n > y\}\,\mathrm dy = \int_0^{\infty} P\{X > y^{1/n}\}\,\mathrm dy.$$ –  Dilip Sarwate Jun 14 '12 at 11:45
    
Thank you! Knowing the formula surely helps! –  johnny Jun 14 '12 at 12:14
1  
You are welcome. For the intuition behind the formula as well as a formal proof via Tonelli/Fubini as you asked of @StefanHansen, see the answers to Intuition behind using complementary CDF to compute expectation for nonnegative random variables –  Dilip Sarwate Jun 14 '12 at 12:30
    
The article I'm reading is treating a continuously distributed random variable, so that's probably why they won't make a difference. But your point is of course very valid. –  johnny Jun 14 '12 at 12:42
1  
The functions $P(X>z)$ and $P(X\geq z)$ can only differ at a countable set of $z$ values, and so it doesn't matter which of these you integrate. –  Byron Schmuland Jun 14 '12 at 13:12

1 Answer 1

up vote 2 down vote accepted

First, use Tonelli's theorem to conclude that (write the probability as an integral and interchange the two integrals) $$ E[X^n]=\int_0^\infty P\left(X^n> z\right)\, \mathrm{d} z. $$ Now write $P\left(X^n> z\right)=P\big(X> z^{1/n}\big)$ and use change of variables with $t=z^{1/n}$.

share|improve this answer
    
Would you add a line or two to show how you apply Tonelli's theorem to get the formula for expectation? I can't see it immediately. I'll accept the answer then. Thanks! –  johnny Jun 14 '12 at 12:13
    
It's alright. See @DilipSarwate comment for a link to the proof of that relation. –  johnny Jun 14 '12 at 12:32
    
@DilipSarwate: Of course, I edited it now. Thanks for the heads up. –  Stefan Hansen Jun 14 '12 at 13:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.