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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 $\forall \tilde{y}\in \mathcal{Y}\setminus \{y\}$ $$ \int_{ \mathcal{V}} u(y,v) P_{T|v}(t) P_{V}(\text{d}v) \geq \int_{ \mathcal{V}} u(\tilde{y}, v) P_{T|v}(t) P_{V}(\text{d} v;\theta_v). $$


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

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1 Answer 1

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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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