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Let $(X,\leq)$ be a poset, $X=\{x_1,x_2,\dots,x_n\}$. Preference matrix $\textbf{P}=[p_{ij}]$ (which is known and fixed), satisfies $p_{ij}=1-p_{ji},p_{ii}=\frac{1}{2}$, and

$$\forall i \neq j, x_i \leq x_j \iff p_{ij} < \frac{1}{2},$$

$$ \forall i \neq j, x_i \| x_j \;(x_i \text{ can't compare with } x_j) \Longleftrightarrow p_{ij}=\frac{1}{2}.$$

Define sigmod function $f(x)=\frac{1}{1+e^{-x}}$, and
$$ g(a_1,a_2,\dots,a_n) = \sum_{i,j}|f(a_i-a_j)-p_{ij}|. $$

I'd like to know a good lower bound for the function $g$. I'm sure this lower bound is greater than $0$. Can anyone give me some hint?

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  • $\begingroup$ What do you mean by "a good lower bound"? What properties of the bound will make it good enough for you? $\endgroup$ Commented Jun 2, 2024 at 17:46
  • $\begingroup$ I think a non-trivial lower bound is sufficient , possibly related to chains and antichains. $\endgroup$ Commented Jun 3, 2024 at 3:24
  • $\begingroup$ What if $X=\{1,\dots,n\}$, $a_i=i$ for all $i$, and $p_{ij}=f(i-j)$ for all $i,j$? Then $g(a_1,\dots,a_n)=0$. $\endgroup$ Commented Jun 3, 2024 at 12:51
  • $\begingroup$ OK. This is a special case of a totally ordered set; while I'd like to consider the partially ordered set that is not totally ordered. $\endgroup$ Commented Jun 3, 2024 at 14:48
  • $\begingroup$ Saying $p_{ii} = \frac{1}{2}$ is redundant. $\endgroup$ Commented May 1 at 12:46

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Under the conditions specified in your post, the best lower bound on $g(a_1,\dots,a_n)$ is $\sum'_{i,j}|f(a_i-a_j)-1/2|$, where $\sum'_{i,j}$ denotes the summation over all pairs $(i,j)$ with $x_i ||x_j$. Indeed, if for some $n$-tuple $(a_1,\dots,a_n)$ with $a_1<\cdots<a_n$ and all pairs $(i,j)$ with $i\ne j$ and $x_i\le x_j$ we have $p_{ij}=f(a_i,a_j)$, then your conditions on the $p_{ij}$'s hold, whereas $g(a_1,\dots,a_n)=\sum'_{i,j}|f(a_i-a_j)-1/2|$, so that the lower bound is attained.

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  • $\begingroup$ I think there may not exist $(a_1,...,a_n)$ such that $p_{ij}=f(a_i,a_j)$ for all $x_i \leq x_j$. Because the number of equations is greater than the number of $a_i$( i.e. $n$). $\endgroup$ Commented Jun 3, 2024 at 15:44
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    $\begingroup$ @MixiAndrew : Of course, the only bound on $g(a_1,\dots,a_n)$ that is exact for all feasible $p_{ij}$'s and all $a_i$'s is the tautological bound, that is, $g(a_1,\dots,a_n)$ itself. Any non-tautological bound will be exact only under a certain, rather narrow condition. In this answer, this condition for the exactness is that the $p_{ij}$'s and $a_i$'s are matching, in the sense that $p_{ij}=f(a_i-a_j)$ for $(i,j)$ such that .... However, the mentioned lower bound holds for all $p_{ij}$'s and all $a_i$'s satisfying your conditions. $\endgroup$ Commented Jun 3, 2024 at 16:41
  • $\begingroup$ Thank you for your attention to this problem. I don't fully understand your response. I should include a constraint in my question: $p_{ij}$'s are known and fixed. $\endgroup$ Commented Jun 4, 2024 at 1:59

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