<p>While it is true that a function has the expected utility form if and only if it is linear (in probabilities), it is not the case that <em>any</em> linear function can represent a preference that satisfies the vNM axioms. The expected utility theorem simply says that when a preference satisfies the vNM axioms, <strong>there exists</strong> a linear utility function that represents it. The theorem does not say, in particular, that <em>all</em> linear utility function represents a preference that satisfies the axioms. </p>

<p>Also, as @Giskard mentioned, utility representation of a vNM-consistent preference does not have to take the expected utility form. For instance, if <span class="math-container">$()$</span> is a linear expected utility function representing a vNM-consistent preference, then <span class="math-container">$()=[()]^2$</span> is another utility function representing the same preference, except that it is a non-linear (or non-vNM) expected utility function.</p>

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2019-05-16T16:43:52.883Z
You cost me my green check mark :'( but alas, it was fair competition.
Giskard
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2019-05-16T19:30:09.830Z
@Giskard: Sorry :P I wasn't expecting a second mover advantage when you nudge me to turn my comments into an answer.
Herr K.
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2019-05-16T16:02:30.367Z
2019-05-16T16:02:30.367Z
Herr K.
<p>It would suffice to show that <span class="math-container">$U$</span> is linear. But is <span class="math-container">$U$</span> necessarily linear if it satisfies the vNM axioms?</p>

<p>Hint: No.</p>

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2019-05-15T20:05:13.383Z
My reading is that he defines "Ex" as the expectation operator. That is, what you write is equivalent to what he means and to what he expresses in his notation.
Bayesian
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2019-05-15T20:06:50.863Z
@Bayesian That makes sense, thx.
Giskard
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2019-05-15T20:56:37.593Z
Can you provide me an example? As I understand, von Neumann-Morgenstern's theorem is about this - it says that my set of implications actually is in the form $\iff$ for all implications.
Marcelo Gelati
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2019-05-15T19:57:02.563Z
2019-05-15T20:12:23.580Z
2019-05-15T20:12:23.580Z
Giskard
<p>I'm having trouble with a question from Ariel Rubinstein's book, Lecture Notes in Microeconomic Theory. It's the problem 2 from Problem Set 7. Here's the question:</p>

<blockquote>
  <p>Show that the utility function <span class="math-container">$u(L) = \mathbb E(L) - (\mathbb E(L))^2 - var(L)$</span> is consistent with vNM assumptions. </p>
</blockquote>

<p>Where <span class="math-container">$\mathbb E(L)$</span> and <span class="math-container">$var(L)$</span> are the expected value and the variance of the lotteries, respectively.</p>

<p>So here's what I thought: we know the following set of implications
<span class="math-container">$$\text{Function is linear} \implies \text{Has the expected utility form} \implies \succsim \text{satisfies vNM assumptions.}$$</span></p>

<p>So, it suffices to show that <span class="math-container">$u(L)$</span> is linear. As we know that <span class="math-container">$var(L) = \mathbb E(L^2) - (\mathbb E(L))^2$</span>, let's rewrite our utility function.</p>

<p><span class="math-container">$$u(L) = \mathbb E(L) - (\mathbb E(L))^2 - (\mathbb E(L^2) - (\mathbb E(L))^2) = \mathbb E(L) - \mathbb E(L^2).$$</span></p>

<p>Take two lotteries, <span class="math-container">$L$</span>, <span class="math-container">$M$</span>. We should have that</p>

<p><span class="math-container">$$U(\alpha L + (1 - \alpha)M) = \alpha U(L) + (1-\alpha ) U(M) \qquad \alpha \in [0,1].$$</span></p>

<p>So, <span class="math-container">$$U(\alpha L + (1 - \alpha)M) = \mathbb E(\alpha L + (1 - \alpha)M) - \mathbb E((\alpha L + (1 - \alpha)M)^2) = \alpha \mathbb E(L) + (1-\alpha)\mathbb E(M) - \alpha^2 \mathbb E(L^2) - 2\alpha (1 - \alpha)\mathbb E(LM) - (1 - \alpha)^2 \mathbb E(M^2).$$</span></p>

<p>But the above isn't equal to <span class="math-container">$$\alpha U(L) + (1-\alpha ) U(M) = \alpha (\mathbb E(L) - (\mathbb E(L))^2) + (1 - \alpha)(\mathbb E(M) - (\mathbb E(M))^2).$$</span></p>

<p>Can you guys help me to see where I got it wrong?</p>

CC BY-SA 4.0
2019-05-15T22:16:54.273Z
While it is true that a function has the expected utility form if and only if it is linear (in probabilities), it is not the case that *any* linear function can represent a preference that satisfies the vNM axioms. The expected utility theorem simply says that when a preference satisfies the vNM axioms, **there exists** a linear utility function that represents it. The theorem does not say, in particular, that *all* linear utility function represents a preference that satisfies the axioms.
Herr K.
CC BY-SA 4.0
2019-05-15T22:31:48.953Z
Also, as @Giskard mentioned, utility representation of a vNM-consistent preference does not have to take the expected utility form. For instance, if $U(L)$ is a linear expected utility function representing a vNM-consistent preference, then $V(L)=[U(L)]^2$ is another utility function representing the same preference, except that it is a non-linear (or non-vNM) expected utility function.
Herr K.
CC BY-SA 4.0
2019-05-16T05:26:02.843Z
@HerrK. Please post answers as answers.
Giskard
<microeconomics><expected-utility><proof>
Expected Utility with expected value and variance
Marcelo Gelati
# Expected Utility with expected value and variance

(Asked by: username_0 on 2019-05-15)

I'm having trouble with a question from Ariel Rubinstein's book, Lecture Notes in Microeconomic Theory. It's the problem 2 from Problem Set 7. Here's the question:

> Show that the utility function $u(L) = \mathbb E(L) - (\mathbb E(L))^2 - var(L)$ is consistent with vNM assumptions.

Where $\mathbb E(L)$ and $var(L)$ are the expected value and the variance of the lotteries, respectively.

So here's what I thought: we know the following set of implications
$$\text{Function is linear} \implies \text{Has the expected utility form} \implies \succsim \text{satisfies vNM assumptions.}$$

So, it suffices to show that $u(L)$ is linear. As we know that $var(L) = \mathbb E(L^2) - (\mathbb E(L))^2$, let's rewrite our utility function.

$$u(L) = \mathbb E(L) - (\mathbb E(L))^2 - (\mathbb E(L^2) - (\mathbb E(L))^2) = \mathbb E(L) - \mathbb E(L^2).$$

Take two lotteries, $L$, $M$. We should have that

$$U(\alpha L + (1 - \alpha)M) = \alpha U(L) + (1-\alpha ) U(M) \qquad \alpha \in [0,1].$$

So, $$U(\alpha L + (1 - \alpha)M) = \mathbb E(\alpha L + (1 - \alpha)M) - \mathbb E((\alpha L + (1 - \alpha)M)^2) = \alpha \mathbb E(L) + (1-\alpha)\mathbb E(M) - \alpha^2 \mathbb E(L^2) - 2\alpha (1 - \alpha)\mathbb E(LM) - (1 - \alpha)^2 \mathbb E(M^2).$$

But the above isn't equal to $$\alpha U(L) + (1-\alpha ) U(M) = \alpha (\mathbb E(L) - (\mathbb E(L))^2) + (1 - \alpha)(\mathbb E(M) - (\mathbb E(M))^2).$$

Can you guys help me to see where I got it wrong?

**Tags:** microeconomics, expected-utility, proof
> **Comments:**
> **username_1**: While it is true that a function has the expected utility form if and only if it is linear (in probabilities), it is not the case that *any* linear function can represent a preference that satisfies the vNM axioms. The expected utility theorem simply says that when a preference satisfies the vNM axioms, **there exists** a linear utility function that represents it. The theorem does not say, in particular, that *all* linear utility function represents a preference that satisfies the axioms.  
> **username_1**: Also, as @Giskard mentioned, utility representation of a vNM-consistent preference does not have to take the expected utility form. For instance, if $U(L)$ is a linear expected utility function representing a vNM-consistent preference, then $V(L)=[U(L)]^2$ is another utility function representing the same preference, except that it is a non-linear (or non-vNM) expected utility function.  
> **username_2**: @HerrK. Please post answers as answers.  

---

## 2 answers

### Answer by username_1 on 2019-05-16. Score: 2 | Accepted answer

While it is true that a function has the expected utility form if and only if it is linear (in probabilities), it is not the case that *any* linear function can represent a preference that satisfies the vNM axioms. The expected utility theorem simply says that when a preference satisfies the vNM axioms, **there exists** a linear utility function that represents it. The theorem does not say, in particular, that *all* linear utility function represents a preference that satisfies the axioms.

Also, as @Giskard mentioned, utility representation of a vNM-consistent preference does not have to take the expected utility form. For instance, if $()$ is a linear expected utility function representing a vNM-consistent preference, then $()=[()]^2$ is another utility function representing the same preference, except that it is a non-linear (or non-vNM) expected utility function.

> **Comments:**
> **username_2**: You cost me my green check mark :'( but alas, it was fair competition.  
> **username_1**: @Giskard: Sorry :P I wasn't expecting a second mover advantage when you nudge me to turn my comments into an answer.  

---

### Answer by username_2 on 2019-05-15. Score: 2

It would suffice to show that $U$ is linear. But is $U$ necessarily linear if it satisfies the vNM axioms?

Hint: No.

> **Comments:**
> **username_3**: My reading is that he defines "Ex" as the expectation operator. That is, what you write is equivalent to what he means and to what he expresses in his notation.  
> **username_2**: @Bayesian That makes sense, thx.  
> **username_0**: Can you provide me an example? As I understand, von Neumann-Morgenstern's theorem is about this - it says that my set of implications actually is in the form $\iff$ for all implications.  

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