<p>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.</p>
<p>But it need not be as good in every single state, which is what the formalism of the question seems to ask for.</p>
<p>Suppose there are two actions, <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, and three equally likely states, <span class="math-container">$0$</span>, <span class="math-container">$1$</span>, and <span class="math-container">$2$</span>. The payoff function is given by <span class="math-container">$u(a,0)=u(a,1)=u(a,2)=1$</span>, <span class="math-container">$u(b,0)=u(b,1)=-1$</span>, and <span class="math-container">$u(b,2)=4$</span>. With no information, playing <span class="math-container">$a$</span> is clearly optimal. Suppose now the decision-maker only learns whether the state is <span class="math-container">$0$</span> or not. In state <span class="math-container">$0$</span>, the decision-maker will still play <span class="math-container">$a$</span>. But in the remaining states, the decision maker will play <span class="math-container">$b$</span>. Consequently, the payoff received in state <span class="math-container">$1$</span> will be worse.</p>

CC BY-SA 4.0
2022-10-06T09:34:07.600Z
Thanks, it makes sense. Is this an obvious result or do I need a formal proof to claim it?
Star
CC BY-SA 4.0
2022-10-06T11:22:29.607Z
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.
Michael Greinecker
CC BY-SA 4.0
2022-10-05T21:07:22.277Z
2022-10-05T21:07:22.277Z
Michael Greinecker
<p><strong>Model</strong></p>
<p>Consider a <strong>single-agent decision problem with uncertainty.</strong></p>
<p>A decision maker (DM) has to choose action <span class="math-container">$y\in \mathcal{Y}$</span> possibly without being fully aware of the state of the world. <span class="math-container">$\mathcal{Y}$</span> is a finite set.
The state of the world is a random variable <span class="math-container">$V$</span> with support <span class="math-container">$\mathcal{V}$</span>.
When the DM chooses action <span class="math-container">$y\in \mathcal{Y}$</span> and the state of the world is <span class="math-container">$v\in \mathcal{V}$</span>, she receives the payoff <span class="math-container">$u(y,v)$</span>.
Let <span class="math-container">$P_V\in \Delta(\mathcal{V})$</span> be the DM's prior about the state of the world <span class="math-container">$V$</span>.</p>
<p>The DM also processes some signal <span class="math-container">$T$</span> with support <span class="math-container">$\mathcal{T}$</span> and  distribution <span class="math-container">$P_{T|V}$</span> conditional on <span class="math-container">$V$</span>  to refine his prior and get a posterior on <span class="math-container">$V$</span>, denoted by <span class="math-container">$P_{V|T}$</span>, via the Bayes rule.</p>
<p>A strategy for the DM is a distribution of actions conditional on the signal, which we denote by <span class="math-container">$P_{Y|T}$</span>. Such a strategy is optimal if it maximises his expected payoff, where the expectation is computed using the posterior  <span class="math-container">$P_{V|T}$</span>.</p>
<p>Hereafter, we call <span class="math-container">$S\equiv (\mathcal{T}, P_{T|V})$</span> as the DM's information structure.</p>
<hr />
<p><strong>Question</strong></p>
<p>In the worst-case scenario, the signal is uninformative about <span class="math-container">$V$</span> (null information structure). In this scenario, the DM with assigned state <span class="math-container">$v$</span> will choose based on the prior <span class="math-container">$P_V$</span> and get the utility
<span class="math-container">$$
\bar{u}(v)\equiv u\Big(\text{argmax}_{y\in \mathcal{Y}} \int_\mathcal{V} u(y,x) dP_V(x), v\Big).
$$</span>
Can we show that <span class="math-container">$\bar{u}(v)$</span> is the <strong>lowest</strong> utility that the DM can attain across every possible information structures? In other words, take any information structure that is <strong>at least as informative as the null information structure</strong>; suppose the DM gets some signal <span class="math-container">$t$</span> from such information structure; does it hold that<br />
<span class="math-container">$$
u\Big(\text{argmax}_{y\in \mathcal{Y}} \int_\mathcal{V} u(y,x) dP_{V|T}(x|t), v\Big)\geq \bar{u}(v)\quad ?
$$</span></p>

<decision-theory><uncertainty>
Lower bound for the utility in a decision problem with uncertainty
Star
# Lower bound for the utility in a decision problem with uncertainty

(Asked by: username_0 on 2022-10-03)

**Model**

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.

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**Question**

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  
$$
u\Big(\text{argmax}\_{y\in \mathcal{Y}} \int\_\mathcal{V} u(y,x) dP\_{V|T}(x|t), v\Big)\geq \bar{u}(v)\quad ?
$$

**Tags:** decision-theory, uncertainty


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## 1 answers

### Answer by username_1 on 2022-10-05. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks, it makes sense. Is this an obvious result or do I need a formal proof to claim it?  
> **username_1**: 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.  

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