Let $X_1, X_2, X_3$ be i.i.d exponential random variables with mean $1$. What is $\operatorname{Pr}(X_1 < X_2 < X_3)$?

Question

Ive been working on this question and just want to know if i'm on the right track of if im completely off.

The join pdf for the order statistic is

$f_{x_{(1)}x_{(2)}x_{(3)}}(y_1,y_2,y_3)$ = $3!e^{-(y_1 +y_2 + y_3)} $

by integrating, i get the densities for $x_{(1)}$ and $x_{(2)}$, and for $x_{(2)}$ and $x_{(3)}$

$f_{x_{(1)}x_{(2)}}(y_1,y_2)$ = $6(e^{-(y_1 + y_2)} - e^{-(y_1 + 2y_2)})$

$f_{x_{(2)}x_{(3)}}(y_1,y_2)$ = $6(e^{-(2y_2 + y_3)} - e^{-(y_2 + y_3)})$

We need to fine $Pr(X_1 < X_2 < X_3)$

$Pr(X_1 < X_2 < X_3)$ = $Pr(X_2 - X_3) - Pr(X_1 < X_2)$

where

$Pr(X_2 < X_3)$ = $\int_{0}^\infty \int_{0}^{x_{3}} 6(e^{-(2y_2 + y_3)} - e^{-(y_2 + y_3)}) $

$Pr(X_1 < X_2)$ = $\int_{0}^\infty \int_{0}^{x_{2}} 6(e^{-(y_1 + y_2)} - e^{-(y_1 + 2y_2)}) $

Thus

$Pr(X_1 < X_2 < X_3)$ = $\int_{0}^\infty \int_{0}^{x_{3}} 6(e^{-(2y_2 + y_3)} - e^{-(y_2 + y_3)}) $ - $\int_{0}^\infty \int_{0}^{x_{2}} 6(e^{-(y_1 + y_2)} - e^{-(y_1 + 2y_2)}) $

This is where im at so far. Is this correct?

Answer 1

Probably computing order statistics is much too hard.

Since exponential distributions are continuous, we have $Pr(X_1 = X_2) = Pr(X_1 = X_3) = Pr(X_2=X_3) = 0$. Thus we see that there are six different orders for $X_1, X_2, X_3$, all equally likely, exhaustive, and mutually exclusive. So $Pr(X_1 < X_2 < X_3) = 1/6$.

Answer 2

This can also be written as

$P(X_1<X_2<X_3) = \int_{0}^{\infty}\int_{x_1}^{\infty}\int_{x_2}^{\infty} e^{-{(x_1+x_2+x_3)}}dx_3dx_2dx_1= \frac{1}{6}$

Answer 3

Because the $X_i$ are i.i.d, the six vector-valued random variables $(X_\sigma(1), X_\sigma(2), X_\sigma(3))$ for each permutation $\sigma$ are all identically distributed. Let $f(x,y,z)$ be the function equal to $1$ if $x<y<z$ and $0$ otherwise. Then it follows from what precedes that the variables $f(X_\sigma(1), X_\sigma(2), X_\sigma(3))$ are also all identically distributed.

Therefore the events $f(X_\sigma(1), X_\sigma(2), X_\sigma(3))=1$ for each $\sigma$ all have the same probability. Since they're disjoint and cover the entire space (minus a null set), each one is of probability $\frac16$.