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.

Let $X,Y,X$ be random variables defined on the same probability space $(\Omega,F,P)$. Suppose that $X,Y,Z$ are independent and identically distributed and the common distribution is continuous. Prove that $\displaystyle P\{X<Y<Z\}=\frac{1}{3!}.$

My try: well I've done this in a particular case, and as the problem says, it's true when $X, Y, Z$ have exponential distributions with parameter $\lambda$. There it is easy because the problem reduces to calculating the integral $\displaystyle \int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}\lambda^{3}e^{-\lambda(x+y+z)}dz dy dx.$ But I don't know how to do it in general. Any suggestions? Thanks beforehand.

share|improve this question
add comment

3 Answers

up vote 7 down vote accepted

Hint: ${\rm P}(X < Y < Z) = {\rm P}(X < Z < Y) = \ldots$.

share|improve this answer
add comment

You should replace the integrand by the joint probability distribution $p_{XYZ}(x, y, z)$. Now $X$, $Y$ and $Z$ are independent, so

$p_{XYZ}(x, y, z) = p_X(x)p_Y(y)p_Z(z)$

besides they are identically distributed, so $p_X = p_Y = p_Z = p$. So I think you are back to almost the same calculations as you did in your particular case, the only property needed being

$\int_{-\infty}^{+\infty}p(x)dx=1$

Does that help?

share|improve this answer
    
The solution below is much, much more elegant. I'm ashamed! –  Sebastien May 11 '11 at 5:13
add comment

Symmetry rules the world, my friends... See this answer.

share|improve this answer
add comment

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.