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

Hint: No.