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We consider a function $f(s_{1:p}, a_{1:p})$, where $p>1$ is an integer, $s_{1:p}$ denotes $(s_1,\ldots,s_p)^\top \in R^p$, and $a_{1:p}$ denotes $(a_1,\ldots,a_p)^\top \in R^p$.

Question: What is the necessary condition such that $$a^*_{1:p}(s_{1:p}) \triangleq \arg\max_{a_{1:p}} f(s_{1:p}, a_{1:p})$$ admits the factorization $$a_i^*(s_{1:p}) \equiv a_i^* (s_{n(i)})~\text{for all }i\in\{1, \ldots, p\},$$ where $n(i) \subsetneq \{1, \ldots, p\}$ with $|n(i)|<p$?

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  • $\begingroup$ It seems that the choice $n(i) \equiv \{1,\ldots,p\}$ is always possible? But maybe I do not understand the definition of $a_i^*(s_{n(i)})$. $\endgroup$
    – gerw
    Commented Oct 14, 2021 at 11:08
  • $\begingroup$ @gerw Sorry for the confusion. I have added the requirement that $|n(i)| < p$. Here $a_i^*(s_{n(i)})$ means that the $i$-th coordinate of the argmax only depends on a subset $n(i)$ of the coordinates of $s_{1:p}$. $\endgroup$
    – Minkov
    Commented Oct 14, 2021 at 23:29
  • $\begingroup$ The notation $a^*_i(s_{n(i)})$ still seems unwieldy and unclear to me — can you give a non-trivial example or two? Also, what conditions on $f$ do you assume to ensure that this and the $\arg \max$’s are unique? $\endgroup$
    – user44143
    Commented Oct 15, 2021 at 16:32

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