<p>Yes, an optimal strategy always exists for any information structure. </p>

<p>Proof: 1st- the set <span class="math-container">$\mathcal{Y}$</span> is finite (thus compact). 2nd- for each state of the world, <span class="math-container">$v$</span>, <span class="math-container">$u(y,v)$</span> is continuous in <span class="math-container">$y$</span> using the discrete metric. 3rd- Given the linearity of integrals, for each <span class="math-container">$t\in\mathcal{T}$</span> we have that <span class="math-container">$\int_{\mathcal{V}}u(y,u)P_{T|v}(v)P_V(dv)$</span> is also continuous in <span class="math-container">$y$</span>. Therefore Weierstrass theorem ensures the existence of a minimum and a maximum. </p>

<p>Let <span class="math-container">$y^*_t$</span> be one such maximizer when the signal is <span class="math-container">$t$</span> (note it can be different for each signal realization, and that the maximizer might not be unique. In the latter case, I arbitrarily choose one for simplicity). Finally, define <span class="math-container">$P_{Y|T}(y|t)=\mathbb{1}\{y=y^*_t\}$</span>. I.e. to play <span class="math-container">$y^*_t$</span> with probability one when the signal is <span class="math-container">$t$</span>. </p>

<p>In this setup, existence might only fail if <span class="math-container">$\mathcal{Y}$</span> was not compact, or if <span class="math-container">$u(y,v)$</span> was not continuous with respect to <span class="math-container">$y$</span>. Note I said might, even if these two conditions are not present, an optimal might still exist. </p>

CC BY-SA 4.0
2019-06-27T20:31:21.923Z
2019-06-27T20:31:21.923Z
Regio
<p>Consider a decision maker choosing an action, <span class="math-container">$y$</span>, from the finiteset <span class="math-container">$\mathcal{Y}$</span>, possibly without having complete information about the state of the world. </p>

<p>More precisely, let <span class="math-container">$V$</span> be a <em>continuously</em> distributed random variable (or vector) representing the state of the world, with support, <span class="math-container">$\mathcal{V}$</span>. Let <span class="math-container">$P_{V}\in \Delta(\mathcal{V})$</span> be the probability distribution of <span class="math-container">$V$</span>.</p>

<p>Nature draws a realisation, <span class="math-container">$v$</span>, of <span class="math-container">$V$</span> from <span class="math-container">$P_{V}$</span>. The decision maker is not aware of <span class="math-container">$v$</span>. However, she can refine her prior, <span class="math-container">$P_{V}$</span>, upon reception of a private signal which may be informative about the state of the world. In particular, let <span class="math-container">$T$</span> be a random variable (or vector) representing the private signal received the decision maker, with support <span class="math-container">$\mathcal{T}$</span>. Let <span class="math-container">$P_{T|v}\in \Delta(\mathcal{T})$</span> be the probability distribution of <span class="math-container">$T$</span> conditional on <span class="math-container">$v$</span>.<br>
Nature draws a realisation, <span class="math-container">$t$</span>, of <span class="math-container">$T$</span> from <span class="math-container">$P_{T|v}$</span>. The decision maker observes <span class="math-container">$t$</span>, uses the Bayes rule to update <span class="math-container">$P_{V}$</span> with the posterior <span class="math-container">$P_{V|t}\in \Delta(\mathcal{V})$</span>, and chooses an action, <span class="math-container">$y\in \mathcal{Y}$</span>. Finally, the decision maker receives a payoff, <span class="math-container">$u(y, v)$</span>.</p>

<p>Let us now define an optimal strategy of the decision maker. </p>

<p>A (mixed) strategy is <span class="math-container">$P_{Y|T}\equiv \{P_{Y|t}\in \Delta(\mathcal{Y}): t\in \mathcal{T}\}$</span>,  collecting probability distributions of  <span class="math-container">$Y$</span> conditional on every realisations <span class="math-container">$t$</span> of <span class="math-container">$T$</span>. </p>

<p>A  strategy <span class="math-container">$P_{Y|T}$</span> is optimal if it allows the decision maker to maximise her expected payoff:  <span class="math-container">$\forall t\in \mathcal{T}$</span>,  <span class="math-container">$\forall y\in \mathcal{Y}$</span> such that <span class="math-container">$P_{Y|t}(y)&gt;0$</span>, and  <span class="math-container">$\forall \tilde{y}\in \mathcal{Y}\setminus \{y\}$</span>
<span class="math-container">$$
\int_{ \mathcal{V}} u(y,v) P_{T|v}(t)  P_{V}(\text{d}v) \geq \int_{ \mathcal{V}} u(\tilde{y}, v) P_{T|v}(t) P_{V}(\text{d} v;\theta_v). $$</span></p>

<hr>

<p><strong>Question:</strong> does an optimal strategy always exists f<strong>or any information structure</strong> <span class="math-container">$S\equiv (\mathcal{T}, P_{T|V})$</span>? Could you make an example on non-existence, if any?</p>

<decision-theory><bayesian-game><uncertainty>
Existence of optimal strategy in a choice problem with uncertainty and information structure
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# Existence of optimal strategy in a choice problem with uncertainty and information structure

(Asked by: username_0 on 2019-06-27)

Consider a decision maker choosing an action, $y$, from the finiteset $\mathcal{Y}$, possibly without having complete information about the state of the world.

More precisely, let $V$ be a *continuously* distributed random variable (or vector) representing the state of the world, with support, $\mathcal{V}$. Let $P\_{V}\in \Delta(\mathcal{V})$ be the probability distribution of $V$.

Nature draws a realisation, $v$, of $V$ from $P\_{V}$. The decision maker is not aware of $v$. However, she can refine her prior, $P\_{V}$, upon reception of a private signal which may be informative about the state of the world. In particular, let $T$ be a random variable (or vector) representing the private signal received the decision maker, with support $\mathcal{T}$. Let $P\_{T|v}\in \Delta(\mathcal{T})$ be the probability distribution of $T$ conditional on $v$.  
Nature draws a realisation, $t$, of $T$ from $P\_{T|v}$. The decision maker observes $t$, uses the Bayes rule to update $P\_{V}$ with the posterior $P\_{V|t}\in \Delta(\mathcal{V})$, and chooses an action, $y\in \mathcal{Y}$. Finally, the decision maker receives a payoff, $u(y, v)$.

Let us now define an optimal strategy of the decision maker.

A (mixed) strategy is $P\_{Y|T}\equiv \{P\_{Y|t}\in \Delta(\mathcal{Y}): t\in \mathcal{T}\}$, collecting probability distributions of $Y$ conditional on every realisations $t$ of $T$.

A strategy $P\_{Y|T}$ is optimal if it allows the decision maker to maximise her expected payoff: $\forall t\in \mathcal{T}$, $\forall y\in \mathcal{Y}$ such that $P\_{Y|t}(y)>0$, and $\forall \tilde{y}\in \mathcal{Y}\setminus \{y\}$
$$
\int\_{ \mathcal{V}} u(y,v) P\_{T|v}(t) P\_{V}(\text{d}v) \geq \int\_{ \mathcal{V}} u(\tilde{y}, v) P\_{T|v}(t) P\_{V}(\text{d} v;\theta\_v). $$

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**Question:** does an optimal strategy always exists f**or any information structure** $S\equiv (\mathcal{T}, P\_{T|V})$? Could you make an example on non-existence, if any?

**Tags:** decision-theory, bayesian-game, uncertainty

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## 1 answers

### Answer by username_1 on 2019-06-27. Score: 1 | Accepted answer

Yes, an optimal strategy always exists for any information structure.

Proof: 1st- the set $\mathcal{Y}$ is finite (thus compact). 2nd- for each state of the world, $v$, $u(y,v)$ is continuous in $y$ using the discrete metric. 3rd- Given the linearity of integrals, for each $t\in\mathcal{T}$ we have that $\int\_{\mathcal{V}}u(y,u)P\_{T|v}(v)P\_V(dv)$ is also continuous in $y$. Therefore Weierstrass theorem ensures the existence of a minimum and a maximum.

Let $y^\*\_t$ be one such maximizer when the signal is $t$ (note it can be different for each signal realization, and that the maximizer might not be unique. In the latter case, I arbitrarily choose one for simplicity). Finally, define $P\_{Y|T}(y|t)=\mathbb{1}\{y=y^\*\_t\}$. I.e. to play $y^\*\_t$ with probability one when the signal is $t$.

In this setup, existence might only fail if $\mathcal{Y}$ was not compact, or if $u(y,v)$ was not continuous with respect to $y$. Note I said might, even if these two conditions are not present, an optimal might still exist.


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