Has any one got any idea about this problem? I found my formula is too complicated to get a closed form.
Let $P_{2n}$ be the set of all $(2n)!$ permutations of $\{1,2,3,···,2n\}$. For any $\sigma = (a_1,a_2,···,a_{2n})$ in $P_{2n}$, a pair of positions $(i,j)$ such that $i < j$ is called an inversion in $\sigma$ if $a_i > a_j$. For example, in the permutation $(a_1,a_2,a_3,a_4) = (2,4,1,3)$, $(2,4)$ is an inversion as $a_2 = 4 > a_4 = 3$; in fact, in this case there are exactly $3$ inversions $(1, 3)$, $(2, 3)$, $(2, 4)$.
For any $\sigma = (a_1,a_2,···,a_{2n})$ in $P_{2n}$, let $f(\sigma)$ be the permutation obtained from $\sigma$ by sorting the sublist of odd positions. That is, $f(\sigma) =(b_1,b_2,···,b_{2n}) \in P_{2n}$, where $b_{2k} = a_{2k}$ for $1\le k\le n$, and $b_1 < b_3 < b_5 < ··· < b_{2n-1} $ is the sorted list of $a_1,a_3,···,a_{2n−1}$. For example, for $\sigma = (3,8,2,5,6,7,1,4)$, $f(\sigma) = (1,8,2,5,3,7,6,4)$.
For a random $\sigma $ uniformly chosen from $P_{2n}$, let $I_n$ be the random variable corresponding to the number of inversions in the permutation $f(\sigma)$. Determine $E(I_n)$ and $\text{Var}(I_n)$.