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EDITED with insights from the comment below.

Consider a decision maker who has to choose an action among $\mathcal{Y}\equiv \{1,2,...,L\}$. The payoff from choosing action $y\in \mathcal{Y}$ depends on the state of the world, $V$, with support $\mathcal{V}$. Specifically, choosing action $y\in \mathcal{Y}$ leads the payoff $u(y,v)$, where $u:\mathcal{Y}\times \mathcal{V}\rightarrow \mathbb{R}$.

Suppose the decision maker has complete information about the realisation of $V$ drawn by nature.

A (mixed) strategy of this choice problem is a probability kernel, $\mathcal{P}_{Y|V}\equiv \{P_{Y}(\cdot| v)\in \Delta(\mathcal{Y}): v\in \mathcal{V}\}$, collecting probability distributions of $Y$ conditional on every realisations $v$ of $V$.

Hence, $\mathcal{P}_{Y|V}$ is an optimal strategy of the choice problem above if $\forall v\in \mathcal{V}$ such that $P_{Y}(y|v)>0$, and $\forall \tilde{y}\neq y$ $$ \begin{aligned} u(y, v) \geq u(\tilde{y},v). \\ \end{aligned} $$

Let $\mathcal{Q}^*$ be the collection of all optimal strategies of the choice problem above, that is $$ \mathcal{Q}^*\equiv \Big\{\mathcal{P}_{Y|V}: \forall v\in \mathcal{V}, \forall y \in \mathcal{Y}\\ \hspace{6cm}\underbrace{P_{Y}(y|v)>0 \Rightarrow u(y, v) \geq u(\tilde{y},v)\text{ } \forall \tilde{y}\neq y}_{\text{This is not a linear constraint because of the form "IF ... THEN ..."}}\Big\} $$

Question 1) The definition of $\mathcal{Q}^*$ just given seems to highlight that $\mathcal{Q}^*$ is not a convex set. This is because it is defined by a constraint of the type "IF ... THEN ...", which is not linear.

Is this comment correct?

Question 2) Consider a payoff function $u(1,v)=u(L,v)>u(y,v)$ $\forall y \neq 1,L$ and $\forall v \in \mathcal{V}$. Consider the following strategies $$ 1) \mathcal{P}_{Y|V}\text{ s.t. } P_{Y}(1|v)=1 \text{ and }P_{Y}(y|v)=0 \text{ }\forall y\neq 1, \forall v \in \mathcal{V} $$ $$ 2) \tilde{\mathcal{P}}_{Y|V}\text{ s.t. } \tilde{P}_{Y}(L|v)=1 \text{ and }\tilde{P}_{Y}(y|v)=0 \text{ }\forall y\neq L, \forall v \in \mathcal{V} $$ $$ 3) \mathcal{P}^*_{Y|V;\alpha}\text{ s.t. } P^*_{Y}(1|v;\alpha)=\alpha P_Y(1|v) \text{, } P^*_{Y}(L|v;\alpha)=(1-\alpha) \tilde{P}_Y(L|v) \text{, and }P^*_{Y}(y|v;\alpha)=0 \text{ }\forall y\neq 1,L, \forall v \in \mathcal{V}, \forall \alpha \in (0,1) $$ I believe that the set $$ \mathcal{B}\equiv \{\mathcal{P}_{Y|V}, \tilde{\mathcal{P}}_{Y|V}, \mathcal{P}^*_{Y|V;\alpha} \text{ }\forall \alpha\in (0,1)\} $$ is convex. Indeed it seems to me that $\mathcal{B}$ is the convex hull of $\{\mathcal{P}_{Y|V}, \tilde{\mathcal{P}}_{Y|V}\}$.

Correct?

What is the relation between $\mathcal{Q}^*$ and $\mathcal{B}$?

I think that $\mathcal{B}\subseteq \mathcal{Q}^*$. This is because for each element in $\mathcal{B}$, the "IF ... THEN ..." condition defining $ \mathcal{Q}^*$ is satisfied.

Does $\mathcal{Q}^*\subseteq \mathcal{B}$ too? If my assertion is question 1) is correct, then it should be $\mathcal{Q}^*\supset \mathcal{B}$ because otherwise $\mathcal{Q}^*$ would be convex. But here I'm lost: which element of $\mathcal{Q}^*$ does not belong to $\mathcal{B}$?

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  • $\begingroup$ Regarding your first question, we know that if a set is described by linear inequalities only, then it is a convex set. Your's is not only described by linear inequalities. However, you cannot conclude it is not a convex set. $\endgroup$ – Regio Aug 24 at 21:23
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The set $Q^*$ is the set of probability distributions over the maximizers of $u$ for each value of $v$. So for a fixed $v$ all the values of $Y$ that have positive probability must give the same utility. That is, if $P_Y(y|v)>0$ and $P_Y(y'|v)>0$ then $u(y,v)=u(y',v)\geq u(\tilde y,v)$ for all $\tilde y\neq y, y'$. Therefore, convex combinations of elements in $Q^*$ must also be collections of probability distributions $P_{Y|V}$ supported over the same elements in $Y$. That is, the convex combination will also be supported over maximizers of $u$ for each $v$.

In conclusion, even though the set of maximizers is not guaranteed to be convex, the set $Q^*$, of distributions, is a convex set.

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  • $\begingroup$ Thanks. Hence, $\mathcal{Q}^*=\mathcal{B}$ and it is convex. Correct? $\endgroup$ – user3285148 Aug 26 at 12:58
  • $\begingroup$ Yes, for the particular $u$ described in your second question, those two sets are the same, I think. $\endgroup$ – Regio Aug 27 at 14:31

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