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Let's difine a set $U=\{u_1, u_2,\dotsc, u_n\}$, with cardinality $|U|=n$, and a collection of sets of permutations or $n$-tuples of length $n$ each: $O=\{o_1, o_2, \dotsc, o_k\}$ with cardinality $|O|=k$. For instance for an element $o_1 \in O =\{u_4, u_1, u_7, \dotsc, u_n \}$.

Then I need to define an index set of positions of each element in $U$, in all $O$ orders but I'm not sure how. Example: for one element in $u_1 \in U$, the position of $u_1$ in $o_1$ is $p_1=2$, the position of $u_1$ in $o_2$ is $p_2=3$, the position of $u_1$ in $o_3$ is $p_3=3$, and so on until the last element $o_k \in O$. This will yield a index set of $u_1$ of size $k$, an example: $R_{u{_1}}=\{p_1, p_2, p_3, \dots, p_k\}=\{2,3,3, \dots, p_k\}$.

Finally I want to formulate with good notation an average $\sum R_{u_1}=(p_1 + p_2 + p_3 + \dots p_k)/k $, of this index set for all $u \in U$.

How do I express this procedure with a correct notation?

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  • $\begingroup$ This is confusing, but let me try to understand. You have a set with $n$ elements, $U$. You are considering all $n!$ possible orderings of this set, $O$. For each element $u\in U$, you want to find its average index in the order, $\frac1{n!}\sum_{o\in O}(\text{index of $u$ in $o$})$. Correct? $\endgroup$ – Mike Earnest Feb 7 at 17:51
  • $\begingroup$ No, this is not what I mean. $O$ is not all the permutations $n!$, is only 'some' permutations, I think is clearer if I say, k-tuples of length $n$?. For each element in $U$, I want to get the average of positions in all sets $o_k \in O$. An example is, $U=\{u_1, u_2, u_3 \}$; $O=\{o_1, o_2, \dots, o_n \}$ are n-tuples of length $3$, Let's define only 2, $o_1=\{u_2, u_1, u_3 \}$ and $o_1=\{u_1, u_3, u_2 \}$. Then the index function that I'm looking should map elements of $U$, for instance $u_1$, in $O$, e.g., $R_{u_{1}}=\{2,1\} $, $R_{u_{2}}=\{1,3\} $ and $R_{u_{3}}=\{3,2\} $. $\endgroup$ – Mario GS Feb 7 at 18:16
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For each permutation $o_i\in O$ consider a standard permutation $\sigma_i$ of the set $\{1,\dots,n\}$ such that $o_i=\{u_{\sigma_i(1)}, u_{\sigma_i(2)},\dots, u_{\sigma_i(n)}\}$. For instance, if $o_1 \in O =\{u_4, u_1, u_7, \dotsc, u_n \}$ then $\sigma_1=\{4, 1, 7, \dotsc, n\}$. Then a position $p_i$ of an element $u_j\in U$ in $o_i$ satisfies $u_{\sigma_i(p_i)}=u_j$, that is $j=\sigma_i(p_i)$ and $p_i=\sigma^{-1}_i(j)$, so $\sum R_{u_j}=\tfrac 1k\sum \sigma_i^{-1}(j)$.

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