<p>The set <span class="math-container">$Q^*$</span> is the set of probability distributions over the maximizers of <span class="math-container">$u$</span> for each value of <span class="math-container">$v$</span>. So for a fixed <span class="math-container">$v$</span> all the values of <span class="math-container">$Y$</span> that have positive probability must give the same utility. That is, if <span class="math-container">$P_Y(y|v)&gt;0$</span> and <span class="math-container">$P_Y(y'|v)&gt;0$</span> then <span class="math-container">$u(y,v)=u(y',v)\geq u(\tilde y,v)$</span> for all <span class="math-container">$\tilde y\neq y, y'$</span>. Therefore, convex combinations of elements in <span class="math-container">$Q^*$</span> must also be collections of probability distributions <span class="math-container">$P_{Y|V}$</span> supported over the same elements in <span class="math-container">$Y$</span>. That is, the convex combination will also be supported over maximizers of <span class="math-container">$u$</span> for each <span class="math-container">$v$</span>. </p>

<p>In conclusion, even though the set of maximizers is not guaranteed to be convex, the set <span class="math-container">$Q^*$</span>, of distributions, is a convex set. </p>

CC BY-SA 4.0
2019-08-26T12:58:32.893Z
Thanks. Hence, $\mathcal{Q}^*=\mathcal{B}$ and it is convex. Correct?
Star
CC BY-SA 4.0
2019-08-27T14:31:34.863Z
Yes, for the particular $u$ described in your second question, those two sets are the same, I think.
Regio
CC BY-SA 4.0
2019-08-24T20:11:05.933Z
2019-08-24T20:11:05.933Z
Regio
<p><strong>EDITED with insights from the comment below.</strong></p>

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

<p>Suppose the decision maker has complete information about the realisation of <span class="math-container">$V$</span> drawn by nature. </p>

<p>A (mixed) strategy of this choice problem is a probability kernel, <span class="math-container">$\mathcal{P}_{Y|V}\equiv \{P_{Y}(\cdot| v)\in \Delta(\mathcal{Y}): v\in \mathcal{V}\}$</span>,  collecting probability distributions of  <span class="math-container">$Y$</span> conditional on every realisations <span class="math-container">$v$</span> of <span class="math-container">$V$</span>. </p>

<p>Hence, <span class="math-container">$\mathcal{P}_{Y|V}$</span> is an optimal strategy of the choice problem  above if  <span class="math-container">$\forall v\in \mathcal{V}$</span> such that <span class="math-container">$P_{Y}(y|v)&gt;0$</span>, and  <span class="math-container">$\forall \tilde{y}\neq y$</span>
<span class="math-container">$$
\begin{aligned}
u(y, v)  \geq u(\tilde{y},v). \\
\end{aligned}
$$</span></p>

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

<p><strong>Question 1)</strong> The definition of <span class="math-container">$\mathcal{Q}^*$</span> just given seems to highlight that <span class="math-container">$\mathcal{Q}^*$</span> is not a convex set. This is because it is defined by a constraint of the type "IF ... THEN ...", which is not linear. </p>

<p>Is this comment correct?</p>

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

<p>Correct? </p>

<p>What is the relation between <span class="math-container">$\mathcal{Q}^*$</span> and <span class="math-container">$\mathcal{B}$</span>? </p>

<p>I think that <span class="math-container">$\mathcal{B}\subseteq \mathcal{Q}^*$</span>. This is because for each element in <span class="math-container">$\mathcal{B}$</span>, the "IF ... THEN ..." condition defining <span class="math-container">$ \mathcal{Q}^*$</span> is satisfied. </p>

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

CC BY-SA 4.0
2019-08-24T21:23:48.573Z
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.
Regio
<game-theory><decision-theory><probability><choice-theory>
Is the set of optimal strategies convex in a single-agent decision choice problem?
Star
# Is the set of optimal strategies convex in a single-agent decision choice problem?

(Asked by: username_0 on 2019-08-21)

**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 $\mathcal{B}$?

**Tags:** game-theory, decision-theory, probability, choice-theory

> **Comments:**
> **username_1**: 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.  

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## 1 answers

### Answer by username_1 on 2019-08-24. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks. Hence, $\mathcal{Q}^*=\mathcal{B}$ and it is convex. Correct?  
> **username_1**: Yes, for the particular $u$ described in your second question, those two sets are the same, I think.  

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