Let $X_1, X_2, X_3$ be i.i.d exponential random variables with mean $1$. What is $\operatorname{Pr}(X_1 < X_2 < X_3)$? 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?
 A: 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 $Pr(X_1 < X_2 < X_3) = 1/6$.
A: This can also be written as 
$P(X_1<X_2<X_3) = \int_{0}^{\infty}\int_{x_1}^{\infty}\int_{x_2}^{\infty} e^{-{(x_1+x_2+x_3)}}dx_3dx_2dx_1= \frac{1}{6}$
A: 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 $\frac16$.
