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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.

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3 Answers 3

up vote 7 down vote accepted

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

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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


Does that help?

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The solution below is much, much more elegant. I'm ashamed! –  Sebastien May 11 '11 at 5:13

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

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