# Expected Utility with expected value and variance

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?

• 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. May 15, 2019 at 22:16
• 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. May 15, 2019 at 22:31
• Thank you! Now I get it. May 15, 2019 at 22:34
• @Giskard: Thanks. Just did. May 16, 2019 at 16:02

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.

• You cost me my green check mark :'( but alas, it was fair competition. May 16, 2019 at 16:43
• @Giskard: Sorry :P I wasn't expecting a second mover advantage when you nudge me to turn my comments into an answer. May 16, 2019 at 19:30

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

Hint: No.

• 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. May 15, 2019 at 20:05
• @Bayesian That makes sense, thx. May 15, 2019 at 20:06
• 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. May 15, 2019 at 20:56