Probability computation involving order statistics Let $X_1$, $X_2$.. $X_n$ be iid uniform random variables i.e.   $X_i \sim U(0,1)$. We know that the order statistics, $X_{(i)}$ is beta distributed $X_{(k)} \sim B(k,n+1-k)$.
Also let $Y_1$, $Y_2$.. $Y_n$ be another set of uniform random variables i.e.   $Y_i \sim U(0,1)$. The order statistics $Y_{(i)}$ are also beta distributed $Y_{(k)} \sim B(k,n+1-k)$.
I'm interested in  finding the following probability,
$\Pr(X_{(1)} < Y_{(2)}, X_{(2)} < Y_{(3)})= ?$ 
The problem here is that the events $E_1 \equiv \{X_{(1)} < Y_{(2)}\}$ and $E_2 \equiv \{X_{(2)} < Y_{(3)}\}$ etc are not independent, and I'm unable to come up with ways to make them independent so as to make use of the known marginal densities.
Is there any simple way to solve this problem?  Any pointers to a solution are welcome.
 A: One solution is to use joint distribution for two order statistics:
$$
   f_{X_{1:n}, X_{2:n}}\left(x_1, x_2\right) = n(n-1) \left(1-x_2\right)^{n-2} \left[ 0 < x_1 < x_2 <1\right]
$$
$$
   f_{Y_{2:n}, Y_{3:n}}\left(y_2, y_3\right) = n(n-1)(n-2) y_2 \left(1-y_3\right)^{n-3} \left[ 0 < y_2 < y_3 <1\right]
$$
To evaluate the probability use
$$ \begin{eqnarray}
    \Pr\left(X_{1:n}\lt Y_{2:n},X_{2:n}\lt Y_{3:n}\right) &=& \mathbb{E}\left( \Pr\left(X_{1:n}\lt Y_{2:n},X_{2:n}\lt Y_{3:n} \mid Y_{2:n},Y_{3:n}\right) \right) \\ &=& \mathbb{E}\left(  F_{X_{1:n},X_{2:n}}\left(Y_{2:n},Y_{3:n}\right)\right)
\end{eqnarray}
$$
The probability $F_{X_{1:n},X_{2:n}}(y_2,y_3)$ is computed as follows, assuming $0<y_2<y_3<1$
$$\begin{eqnarray}
   F_{X_{1:n},X_{2:n}}(y_2,y_3) &=& n(n-1) \int_0^{y_3} \mathrm{d}x_2 \int_{0}^{\min(y_2,x_1)} \mathrm{d}x_1 \cdot  \left(1-x_2\right)^{n-2} \\&=& 1 - \left(1-y_2\right)^{n} - n y_2 \left(1-y_3\right)^{n-1}
\end{eqnarray}$$
Now
$$
 \begin{eqnarray}
    \Pr\left(X_{1:n}\lt Y_{2:n},X_{2:n}\lt Y_{3:n}\right) &=&  1 - \mathbb{E}\left(  \left(1-Y_{2:n}\right)^{n} \right) - n \mathbb{E}\left( Y_{2:n} \left(1-Y_{3:n}\right)^{n-1}\right) \\
  &=& 1 - \mathbb{E}\left(  Y_{n-1:n}^{n} \right) - n \mathbb{E}\left( Y_{2:n} \left(1-Y_{3:n}\right)^{n-1}\right)
\end{eqnarray}
$$
The latter integral is easy to evaluate:
$$
\begin{eqnarray}
  n \mathbb{E}\left( Y_{2:n} \left(1-Y_{3:n}\right)^{n-1}\right) &=& \int_0^{1} \mathrm{d}y_3 \int_0^{y_3} \mathrm{d} y_2 \cdot n^2(n-1)(n-2) y_2^2 (1-y_3)^{2n-4}  \\
  &=& \frac{n^2 (n-1)(n-2)}{ \frac{1}{2} (2n)(2n-1)(2n-2)(2n-3)} \int_0^{1} \mathrm{d}y_3 \int_0^{y_3} f_{Z_{3:2n},Z_{4:2n}}\left(y_2,y_3\right) \\
  &=& \frac{n(n-2)}{2 (2n-1)(2n-3)}
\end{eqnarray}
$$
The remaining integral is evaluated using singe statistics density function:
$$
  \mathbb{E}\left(  Y_{n-1:n}^{n} \right) = \int_0^1 x^n \cdot n(n-1) x^{n-2} (1-x) \mathrm{d} x = \frac{n(n-1)}{2n(2n-1)} \int_0^1 f_{Y_{2n-1:2n}}(x) \mathrm{d}x =  \frac{n-1}{2(2n-1)} 
$$
Thus
$$
   \Pr\left(X_{1:n}\lt Y_{2:n},X_{2:n}\lt Y_{3:n}\right) = 1-\frac{n-1}{2(2n-1)}-\frac{n(n-2)}{2 (2n-1)(2n-3)}=\frac{(5n-4)(n-1)-1}{2(2n-1)(2n-3)}
$$
