<p>It is actually easy to find <span class="math-container">$v$</span> such that the operator that does the job is not <span class="math-container">$E(v(\cdot))$</span> but instead <span class="math-container">$E(v^2(\cdot))$</span>, just take <span class="math-container">$v=\sqrt{u}$</span> with the <span class="math-container">$u$</span> defined by the theorem. This means that <span class="math-container">$E$</span> could be replaced by any <span class="math-container">$L^p$</span> norm for <span class="math-container">$p&lt;\infty$</span>.</p>

CC BY-SA 4.0
2024-01-31T22:33:14.717Z
You do not change operators in what you write. You change the function on which the same operator, the expected value, operates.
Alecos Papadopoulos
CC BY-SA 4.0
2023-12-29T14:40:08.220Z
2023-12-29T14:40:08.220Z
Marco Disce
<p>The <a href="https://en.m.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem" rel="nofollow noreferrer">Von Neumann–Morgenstern utility theorem</a> states that</p>
<blockquote>
<p>For any VNM-rational agent (i.e. satisfying axioms 1–4), there exists a function <span class="math-container">$u$</span> which assigns to each outcome <span class="math-container">$A$</span> a real number <span class="math-container">$u(A)$</span> such that for any two lotteries <span class="math-container">$L\prec M$</span> iff <span class="math-container">$E(u(L)) &lt; E(u(M))$</span></p>
</blockquote>
<p>I would like to know whether a similar result can be derived if we replace the operator &quot;expected value&quot; <span class="math-container">$E(\cdot)$</span> with some other continuous operator on the space of distribution, such as <a href="https://en.wikipedia.org/wiki/Lp_space" rel="nofollow noreferrer">the <span class="math-container">$L^p$</span> norm</a>.</p>
<p>For example: is it true that</p>
<blockquote>
<p>For any VNM-rational agent (i.e. satisfying axioms 1–4), there exists a function <span class="math-container">$u$</span> which assigns to each outcome <span class="math-container">$A$</span> a real number <span class="math-container">$u(A)$</span> such that for any two lotteries <span class="math-container">$L\prec M$</span> iff <span class="math-container">$\|u(L)\|_p &lt;\|u(M)\|_p$</span>?</p>
</blockquote>
<p>Or is it true for some other operator that is not <span class="math-container">$E(\cdot)$</span>? Or is the expected value the &quot;special&quot; unique operator with this property?</p>

<utility><expected-utility>
How special is the "expected value" operator in Von Neumann–Morgenstern utility theorem?
Marco Disce
# How special is the "expected value" operator in Von Neumann–Morgenstern utility theorem?

(Asked by: username_0 on 2023-12-29)

The [Von Neumann–Morgenstern utility theorem](https://en.m.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem) states that

> For any VNM-rational agent (i.e. satisfying axioms 1–4), there exists a function $u$ which assigns to each outcome $A$ a real number $u(A)$ such that for any two lotteries $L\prec M$ iff $E(u(L)) < E(u(M))$

I would like to know whether a similar result can be derived if we replace the operator "expected value" $E(\cdot)$ with some other continuous operator on the space of distribution, such as [the $L^p$ norm](https://en.wikipedia.org/wiki/Lp_space).

For example: is it true that

> For any VNM-rational agent (i.e. satisfying axioms 1–4), there exists a function $u$ which assigns to each outcome $A$ a real number $u(A)$ such that for any two lotteries $L\prec M$ iff $\|u(L)\|\_p <\|u(M)\|\_p$?

Or is it true for some other operator that is not $E(\cdot)$? Or is the expected value the "special" unique operator with this property?

**Tags:** utility, expected-utility


---

## 1 answers

### Answer by username_0 on 2023-12-29. Score: -1

It is actually easy to find $v$ such that the operator that does the job is not $E(v(\cdot))$ but instead $E(v^2(\cdot))$, just take $v=\sqrt{u}$ with the $u$ defined by the theorem. This means that $E$ could be replaced by any $L^p$ norm for $p<\infty$.

> **Comments:**
> **username_1**: You do not change operators in what you write. You change the function on which the same operator, the expected value, operates.  

---

