Consider a single-agent decision problem with uncertainty.

A decision maker (DM) has to choose action $y\in \mathcal{Y}$ possibly without being fully aware of the state of the world. $\mathcal{Y}$ is a finite set. The state of the world is a random variable $V$ with support $\mathcal{V}$. When the DM chooses action $y\in \mathcal{Y}$ and the state of the world is $v\in \mathcal{V}$, she receives the payoff $u(y,v)$. Let $P_V\in \Delta(\mathcal{V})$ be the DM's prior about the state of the world $V$.

The DM also processes some signal $T$ with support $\mathcal{T}$ and distribution $P_{T|V}$ conditional on $V$ to refine his prior and get a posterior on $V$, denoted by $P_{V|T}$, via the Bayes rule.

A strategy for the DM is a distribution of actions conditional on the signal, which we denote by $P_{Y|T}$. Such a strategy is optimal if it maximises his expected payoff, where the expectation is computed using the posterior $P_{V|T}$.

Hereafter, we call $S\equiv (\mathcal{T}, P_{T|V})$ as the DM's information structure.


In the worst-case scenario, the signal is uninformative about $V$ (null information structure). In this scenario, the DM with assigned state $v$ will choose based on the prior $P_V$ and get the utility $$ \bar{u}(v)\equiv u\Big(\text{argmax}_{y\in \mathcal{Y}} \int_\mathcal{V} u(y,x) dP_V(x), v\Big). $$ Can we show that $\bar{u}(v)$ is the lowest utility that the DM can attain across every possible information structures? In other words, take any information structure that is at least as informative as the null information structure; suppose the DM gets some signal $t$ from such information structure; does it hold that
$$ u\Big(\text{argmax}_{y\in \mathcal{Y}} \int_\mathcal{V} u(y,x) dP_{V|T}(x|t), v\Big)\geq \bar{u}(v)\quad ? $$


1 Answer 1


Trivially, the decision maker can ignore all additional information and simply play the same action they would play without any information. The optimal strategy must be then at least as good, on average.

But it need not be as good in every single state, which is what the formalism of the question seems to ask for.

Suppose there are two actions, $a$ and $b$, and three equally likely states, $0$, $1$, and $2$. The payoff function is given by $u(a,0)=u(a,1)=u(a,2)=1$, $u(b,0)=u(b,1)=-1$, and $u(b,2)=4$. With no information, playing $a$ is clearly optimal. Suppose now the decision-maker only learns whether the state is $0$ or not. In state $0$, the decision-maker will still play $a$. But in the remaining states, the decision maker will play $b$. Consequently, the payoff received in state $1$ will be worse.

  • $\begingroup$ Thanks, it makes sense. Is this an obvious result or do I need a formal proof to claim it? $\endgroup$
    – Star
    Oct 6, 2022 at 9:34
  • $\begingroup$ The proof is obvious once the problems is clearly formulated. Here you need to use strategies, functions from signals to actions to define the problem. Your formalization leaves them out. $\endgroup$ Oct 6, 2022 at 11:22

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