# Prove relation between expected utility and utility of the mean

Assume $U(x)$ is a concave, monotonically increasing function. If we know that

$$U\left[E(p)\right] > U\left[E(q)\right]$$

where $p$ and $q$ are probability distributions, how can we prove that

$$E(U_p) > E(U_q)$$

• What is the shape of $u(x)$? Concave? Convex? – Herr K. Jul 17 '17 at 18:03
• @HerrK. assume it concave, monotonically increasing – Mino Jul 17 '17 at 21:36
• This question has been cross-posted here. – luchonacho Jul 28 '17 at 7:29

The conclusion you want to prove is not always true. I will give you a graphical and mathematical answer.

Graphical analysis

Consider the example in the image below:

There is one lottery, with expected payoffs $z_1$ and $z_2$, and average $E(z)$. The expected value of this lottery in terms of utility is $U[E(z)]$.

Now, assume a second lottery, with higher $z_1$ and smaller $z_2$, such that the expected value is the same, i.e. $E(z)$. The expected value of this lottery is the same, $U[E(z)]$. Yet, because of lower variance in the second lottery, the point E in the second lottery is closer to D. (In the extreme case of no uncertainty, D and E coincide). In other words, $E[u(z)]$ in lottery one is lower than $E[u(z)]$ in lottery two, even if they pay the same on average.

The contradiction of your statement follows immediately. Take the second lottery just introduced, and decrease $z_1$ by a small $\epsilon$. Now, the expected income of the lottery is marginally lower than $E(z)$, which is the opposite relation to the one in your question. There clearly is a small $\epsilon$ such that the relation between $E[u(z)]$ in the two lotteries remains as before.

The proof follows the same logic for the case of risk loving individuals (where uncertainty is preferred to certainty and the utility function is convex).

Actually, the only case where your statement does hold is for risk neutrality. This can trivially be seen in the graph too.

Mathematical analysis

First, we know that $\dfrac{d U(x)}{d x}>0$. Then, it is true that:

$$U(x_p + \pi)=E[U(x_p)]$$

$$U(x_q + \phi)=E[U(x_q)]$$

where the term inside the parenthesis in the left-hand side are the certainty equivalent. The sign of $\pi$ and $\phi$ depend on the nature of the utility function. There are three cases:

• Risk aversion

If utility is concave (i.e. agent is risk averse), $\pi$ and $\rho$ are positive. Therefore:

$$U(x_p) < U(x_p + \pi) = E[U(x_p)]$$

$$U(x_q) < U(x_q + \phi) = E[U(x_q)]$$

Assuming that $U(x_p) > U(x_q)$:

$$E[U(x_p)] = U(x_p + \pi) > U(x_p) > U(x_q)$$

However, $E[U(x_q)] > U(x_q)$. It is evident that nothing more can be said regarding the relation between $E[U(x_p)]$ and $E[U(x_q)]$. The result you want to prove does not necessarily hold in this case.

Risk acceptance

The proof is exactly the same, but reversing the signs.

Risk neutrality

$\pi$ and $\phi$ are zero.

$$U(x_p) = E[U(x_p)]$$

$$U(x_q) = E[U(x_q)]$$

Assuming $U(x_p) > U(x_q)$ implies that:

$$E[U(x_p)] > E[U(x_q)]$$

Thus, in this case the conclusion you want to prove is true.

• where does your conterexample fail under this extra assumption? math.stackexchange.com/questions/2360958/… – Mino Jul 26 '17 at 14:15
• @Mino You are not supposed to cross-post questions. In any case, I think it fails in that first-order stochastic dominance (FSD) implies two lotteries over the same $z$ cannot have the same mean, because the probabilities must be different (they accumulate faster over $z$ in one than in another case, whereas in my example the probabilities do not accumulate faster in one, but they reverse over $z$, violating FSD). Hence, my graphical example is not pertinent for such case. – luchonacho Jul 28 '17 at 7:37
• I got two different timings I didn't know what to do. I think that the two answers are very good together, in the sense that they are not a duplicate and they complement each other for a good understanding (from my side, at least). any suggestion is welcome – Mino Jul 28 '17 at 8:22
• @Mino I would leave them. As they stand, they are not exactly the same. In the other you assume FSD, in this one you don't. The answers seems to be different in either case. If they help you understand the problem, then it might be ok. – luchonacho Jul 28 '17 at 8:32