<p>Your problem can be simply expressed as 
<span class="math-container">$$\arg\max_{y\in\mathcal{Y}}\sum_{v\in\mathcal{V}}u_i(y,v)P_V(v)$$</span>
Note that the subscript in <span class="math-container">$u_i$</span> is un-necessary. Furthermore, this is not a game so "mixed strategies" are only a solution if the maximum is not unique, but you should not worry about them  (more on that later). </p>

<p>First, note that the maximization problem is equivalent to the inequality you present. Presenting it as a maximization problem is more akin to how you solve a single agent problem, and circumvents the need to specify the quantifiers. However, strictly speaking,  it only characterizes "pure strategies", (a better term would be deterministic actions), but this is, really, without loss of generality. A probabilistic action is justified only when the maximizer is not unique, in which case the <span class="math-container">$\arg\max$</span> operator will already be giving you a set, and any mixture between the maximizers in that set would be optimal. </p>

<p>Besides being more clear, another benefit from presenting it as a maximization problem is that you can use standard theorems to guarantee existence and uniqueness, for example, if <span class="math-container">$\mathcal{Y}$</span> is finite then <span class="math-container">$\sum u(y,v)P_V(v)$</span> is continuous in <span class="math-container">$y$</span> (using the discrete metric) and Weierstrass ensures the existence of a solution. In general  you need a <span class="math-container">$\mathcal{Y}$</span> to be compact and <span class="math-container">$\sum u(y,v)P_V(v)$</span> to be  continuous in <span class="math-container">$y$</span>. Uniqueness is a bit more tricky, but if <span class="math-container">$\sum u(y,v)P_V(v)$</span> is strictly concave with respect to <span class="math-container">$y$</span>, then uniqueness is guarantied.  </p>

<p>In your particular example given the finiteness of the action space, existence is trivial, but not uniqueness. Knowing that <span class="math-container">$P_V$</span> is normal is not enough since we don't know anything about the shape of <span class="math-container">$u(\cdot)$</span>. </p>

CC BY-SA 4.0
2019-08-22T12:05:15.713Z
Could you help me with this question here https://economics.stackexchange.com/questions/30601/is-the-set-of-optimal-strategies-convex-in-a-single-agent-decision-choice-proble? I think it is related to what you were pointing out in your answer here, but I want to understand it more formally.
Star
CC BY-SA 4.0
2019-06-26T10:22:43.217Z
Thanks. Regarding the last point, in my example I have $u(y,v)\equiv v_y$ where $v\equiv (v_1,v_2,v_3)$. Is that enough for uniqueness?
Star
CC BY-SA 4.0
2019-06-26T12:04:24.327Z
What do you mean by "and circumvents the need to specify the quantifiers"?
Star
CC BY-SA 4.0
2019-06-26T17:12:34.910Z
No, it is not enough. I think I now understand your utility. What you are assuming is that there are three random variables and I need to choose one of them. its realization wil be my payoff. The optimal decision will be to choose the random variable with the highest expected payoff given the prior. Suppose that the prior is such that the expected values of each of them are  $E(v_1)=5, E(v_2)=3, E(v_3)=5$ then the solution is not unique. It is pretty easy to think of examples where two or more random variables have the same mean (it should be clear that normality does not help here).
Regio
CC BY-SA 4.0
2019-06-26T17:14:36.473Z
the only thing I mean my "circumvents..." is that you don't have to specify, that the solution is a measure  $P_Y$ such that for all $y$ with $P_Y(y)>0$ and for all  $\tilde y\neq y$ ...". I.e. the only thing I mean to say is that the maximization is more clear.
Regio
CC BY-SA 4.0
2019-06-25T23:00:23.457Z
2019-06-25T23:00:23.457Z
Regio
<p>Consider the following single-agent choice problem under uncertainty. </p>

<p>Let <span class="math-container">$V$</span> be the state of the world with support <span class="math-container">$\mathcal{V}$</span> and probability distribution <span class="math-container">$P_V\in \Delta(\mathcal{v})$</span>. First, let nature draw a realisation <span class="math-container">$v$</span> of <span class="math-container">$V$</span> from <span class="math-container">$P_V$</span>. 
Then, let the decision maker choose an action <span class="math-container">$y\in \mathcal{Y}$</span>, with <span class="math-container">$\mathcal{Y}$</span> finite, without observing <span class="math-container">$v$</span>. Upon the decision has been made, the decision maker gets a payoff <span class="math-container">$u(y,v)$</span>.</p>

<p>For example, suppose that <span class="math-container">$\mathcal{Y}\equiv \{1,2,3\}$</span>. <span class="math-container">$V$</span> is a <span class="math-container">$3\times 1$</span> random vector, <span class="math-container">$V\equiv (V_1,V_2,V_3)$</span>. <span class="math-container">$P_V$</span> is the 3-variate standard normal distribution. <span class="math-container">$u(y,v)\equiv v_y$</span>.  </p>

<p><strong>What is the definition of an optimal strategy for the decision maker in this setting?</strong> </p>

<p>I'm thinking about using "a sort of" Bayesian Nash equilibrium for a single-agent setting, i.e., an optimal strategy is <span class="math-container">$P_Y\in \Delta(\mathcal{Y})$</span> such that, <span class="math-container">$\forall y\in \mathcal{Y}$</span> such that <span class="math-container">$P_Y(y)&gt;0$</span> and <span class="math-container">$\forall \tilde{y}\neq y$</span>, 
we have that 
<span class="math-container">$$
\sum_{v\in \mathcal{V}} u_i(y,v)P_V(v)\geq \sum_{v\in \mathcal{V}} u_i(\tilde{y},v)P_V(v)
$$</span>
that is, in my example, 
<span class="math-container">$$
\sum_{v\in \mathcal{V}} (v_y -v_{\tilde{y}}) P_V(v)\geq 0
$$</span></p>

<p>But maybe a <strong>pure</strong> strategy is what people use? </p>

<p>Is <strong>existence and uniqueness</strong> obvious (at least in my example with a normal distribution)? </p>

<p>Could you also provide a reference discussing <strong>definition, existence, multiplicity</strong>?</p>

<decision-theory><bayesian-game><uncertainty>
Optimal strategy in a single-agent choice problem under uncertainty
Star
# Optimal strategy in a single-agent choice problem under uncertainty

(Asked by: username_0 on 2019-06-25)

Consider the following single-agent choice problem under uncertainty.

Let $V$ be the state of the world with support $\mathcal{V}$ and probability distribution $P\_V\in \Delta(\mathcal{v})$. First, let nature draw a realisation $v$ of $V$ from $P\_V$.
Then, let the decision maker choose an action $y\in \mathcal{Y}$, with $\mathcal{Y}$ finite, without observing $v$. Upon the decision has been made, the decision maker gets a payoff $u(y,v)$.

For example, suppose that $\mathcal{Y}\equiv \{1,2,3\}$. $V$ is a $3\times 1$ random vector, $V\equiv (V\_1,V\_2,V\_3)$. $P\_V$ is the 3-variate standard normal distribution. $u(y,v)\equiv v\_y$.

**What is the definition of an optimal strategy for the decision maker in this setting?**

I'm thinking about using "a sort of" Bayesian Nash equilibrium for a single-agent setting, i.e., an optimal strategy is $P\_Y\in \Delta(\mathcal{Y})$ such that, $\forall y\in \mathcal{Y}$ such that $P\_Y(y)>0$ and $\forall \tilde{y}\neq y$,
we have that
$$
\sum\_{v\in \mathcal{V}} u\_i(y,v)P\_V(v)\geq \sum\_{v\in \mathcal{V}} u\_i(\tilde{y},v)P\_V(v)
$$
that is, in my example,
$$
\sum\_{v\in \mathcal{V}} (v\_y -v\_{\tilde{y}}) P\_V(v)\geq 0
$$

But maybe a **pure** strategy is what people use?

Is **existence and uniqueness** obvious (at least in my example with a normal distribution)?

Could you also provide a reference discussing **definition, existence, multiplicity**?

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

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## 1 answers

### Answer by username_1 on 2019-06-25. Score: 1 | Accepted answer

Your problem can be simply expressed as
$$\arg\max\_{y\in\mathcal{Y}}\sum\_{v\in\mathcal{V}}u\_i(y,v)P\_V(v)$$
Note that the subscript in $u\_i$ is un-necessary. Furthermore, this is not a game so "mixed strategies" are only a solution if the maximum is not unique, but you should not worry about them (more on that later).

First, note that the maximization problem is equivalent to the inequality you present. Presenting it as a maximization problem is more akin to how you solve a single agent problem, and circumvents the need to specify the quantifiers. However, strictly speaking, it only characterizes "pure strategies", (a better term would be deterministic actions), but this is, really, without loss of generality. A probabilistic action is justified only when the maximizer is not unique, in which case the $\arg\max$ operator will already be giving you a set, and any mixture between the maximizers in that set would be optimal.

Besides being more clear, another benefit from presenting it as a maximization problem is that you can use standard theorems to guarantee existence and uniqueness, for example, if $\mathcal{Y}$ is finite then $\sum u(y,v)P\_V(v)$ is continuous in $y$ (using the discrete metric) and Weierstrass ensures the existence of a solution. In general you need a $\mathcal{Y}$ to be compact and $\sum u(y,v)P\_V(v)$ to be continuous in $y$. Uniqueness is a bit more tricky, but if $\sum u(y,v)P\_V(v)$ is strictly concave with respect to $y$, then uniqueness is guarantied.

In your particular example given the finiteness of the action space, existence is trivial, but not uniqueness. Knowing that $P\_V$ is normal is not enough since we don't know anything about the shape of $u(\cdot)$.

> **Comments:**
> **username_0**: Thanks. Regarding the last point, in my example I have $u(y,v)\equiv v_y$ where $v\equiv (v_1,v_2,v_3)$. Is that enough for uniqueness?  
> **username_0**: What do you mean by "and circumvents the need to specify the quantifiers"?  
> **username_1**: No, it is not enough. I think I now understand your utility. What you are assuming is that there are three random variables and I need to choose one of them. its realization wil be my payoff. The optimal decision will be to choose the random variable with the highest expected payoff given the prior. Suppose that the prior is such that the expected values of each of them are  $E(v_1)=5, E(v_2)=3, E(v_3)=5$ then the solution is not unique. It is pretty easy to think of examples where two or more random variables have the same mean (it should be clear that normality does not help here).  
> **username_1**: the only thing I mean my "circumvents..." is that you don't have to specify, that the solution is a measure  $P_Y$ such that for all $y$ with $P_Y(y)>0$ and for all  $\tilde y\neq y$ ...". I.e. the only thing I mean to say is that the maximization is more clear.  
> **username_0**: Could you help me with this question here https://economics.stackexchange.com/questions/30601/is-the-set-of-optimal-strategies-convex-in-a-single-agent-decision-choice-proble? I think it is related to what you were pointing out in your answer here, but I want to understand it more formally.  

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