let $X=\min(x_{11},x_{12},…,x_{1n},x_{21},x_{22},…,x_{2n})$，such as，$X=x_{1k}$
and
$Y=\min(x_{21},x_{22},…,x_{2,k-1},x_{2,k+1},…,x_{2n})$
and $\forall x_{ij}$ is i.i.d, uniform random variable on [0,1].
so we can get X's cdf which is $F_X(x)=1-(1-x)^{2n}$ and its pdf $f_X(x)=2n(1-x)^{2n-1}$。then $E(X)=1/(2n+1)$
similarly，$F_Y(y)=1-(1-y)^{n-1}$，$f_Y(y)=(n-1)(1-y)^{n-2}$。then $E(Y)=1/n$.
but obviously $Y＞=X$ so i want to know how to get $E(Y|Y＞=X)$ under these conditions.
And the real question is that given a circle road and its length is 1,and car’s direction must be clockwise.Assume that there are n persons and 2 cars uniformly distributed on this circle.let S(n,2) be the expected average optimal matching distance,and can we get the closed form of S(n,2).
I found that if we calculate the distance matrix,and just find the minimum value in the matrix,then delete the corresponding row and column,then find the minimum Value among the remaining elements ,and the sum of these two operations is the optimal matching distance. That’s why I raised this question.
Now I realize there is another issue. Assuming there are two cars with coordinates $x_1$ and $x_2$,and two people with coordinates $y_1$ and $y_2$,because the cars can only move clockwise, the distance between a car $x$ and a person $y$ depends on whether $x<=y$. If $x<=y$,the distance is $y−x$;otherwise, it is $1+y−x$.
This allows us to obtain a $2×2$ distance matrix. Initially, I thought that the four elements in the distance matrix were all independent, but now I am not so sure.
Best Answer
Let us first clearly define the random variables. We have
$$X=\min (A_1,\dots, A_n, B_1,\dots, B_n)=\min (A_\zeta, B_\zeta)$$
where $\zeta$ is a random variable taking values in $\{1,\dots,n\}$. Then, we can define
$$ Y= \begin{cases} \min (B_1,\dots,B_{\zeta-1}, B_{\zeta+1},\dots, B_n)& \quad X=A_\zeta \\ \min (A_1,\dots,A_{\zeta-1}, A_{\zeta+1},\dots, A_n)& \quad X=B_\zeta. \end{cases} $$
As each car can move only clockwise, each of $A_i, B_i, i=1,\dots,n$ is uniformly distributed over $(0,1)$. When, each car is allowed to move in both directions, each of $\, A_i, B_i, i=1,\dots,n \,$ follows the distribution of $Z=\min (U, 1-U)$ with $U \sim \text{Uniform}(0,1)$, where we can show $Z \sim \text{Uniform} \left (0,\frac{1}{2} \right)$.
We always have $Y \ge X$, so
$$\mathbb E (Y | Y \ge X)= \mathbb E (Y).$$
Because of the symmetry, we can see that
$$\mathbb E (Y) = \sum_{i=1}^{n} \frac{1}{2n} \mathbb E \big (Y \big | X=A_i \big )+\sum_{i=1}^{n} \frac{1}{2n} \mathbb E \big (Y \big | X=B_i \big )= \mathbb E \big (Y \big | X=A_1).$$
The expectation $\mathbb E \big (Y \big | X=A_1)$ can be computed as
$$\mathbb E (Y | X=A_1) =\int_0^1 f_{A_1}(a) \mathbb E \bigg (Y \big | X=A_1, A_1=a \bigg) \text{d}t \, \text{d}a=\\ \int_0^1 1 \times \mathbb E (\min (B_2,\dots, B_n) \big | A_2,\dots, A_n, B_1,\dots, B_n > a) \text{d}a=\\ \int_0^1 \int_a^1 \mathbb P (\min (B_2,\dots, B_n)>t \big | A_2,\dots, A_n, B_1,\dots, B_n > a) \text{d}t \, \text{d}a= \\ \int_0^1 \int_a^1 \frac{(1-t)^{n-2}(1-a)^{n}}{(1-a)^{2n-1}} \text{d}t \, \text{d}a =\frac{1}{n-1}.$$