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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?

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    $\begingroup$ 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. $\endgroup$ – Herr K. May 15 at 22:16
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    $\begingroup$ 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. $\endgroup$ – Herr K. May 15 at 22:31
  • $\begingroup$ Thank you! Now I get it. $\endgroup$ – Marcelo Gelati May 15 at 22:34
  • $\begingroup$ @HerrK. Please post answers as answers. $\endgroup$ – Giskard May 16 at 5:26
  • $\begingroup$ @Giskard: Thanks. Just did. $\endgroup$ – Herr K. May 16 at 16:02
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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 $𝑈(𝐿)$ 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.

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  • $\begingroup$ You cost me my green check mark :'( but alas, it was fair competition. $\endgroup$ – Giskard May 16 at 16:43
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    $\begingroup$ @Giskard: Sorry :P I wasn't expecting a second mover advantage when you nudge me to turn my comments into an answer. $\endgroup$ – Herr K. May 16 at 19:30
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It would suffice to show that $U$ is linear. But is $U$ necessarily linear if it satisfies the vNM axioms?

Hint: No.

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  • $\begingroup$ 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. $\endgroup$ – Bayesian May 15 at 20:05
  • $\begingroup$ @Bayesian That makes sense, thx. $\endgroup$ – Giskard May 15 at 20:06
  • $\begingroup$ 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. $\endgroup$ – Marcelo Gelati May 15 at 20:56

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