Prove relation between expected utility and utility of the mean Assume $U(x)$ is a concave, monotonically increasing function. If we know that 

$$U\left[E(p)\right] &gt; U\left[E(q)\right]$$

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

$$E(U_p) &gt; E(U_q)$$
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 $\dfrac{d U(x)}{d x}&gt;0$. 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:

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

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

Assuming that $U(x_p) &gt; U(x_q)$:

$$E[U(x_p)] = U(x_p + \pi) &gt; U(x_p) &gt; U(x_q) $$

However, $E[U(x_q)] &gt; U(x_q)$. 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 $U(x_p) &gt; U(x_q)$ implies that:

$$ E[U(x_p)] &gt; E[U(x_q)] $$

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