An agent's expected utility depends only on mean and variance Consider an agent with the expected utility function $U(L) = \sum_{s=1}^{S}\pi_s U(Y_s)$ over the lottery $L = (Y_s, \pi_s)$ where $\pi_s$ is the probability of state $s$, $Y_s$ are state $s$ payoffs, and $U(y_s) = -\frac{1}{2}(\alpha - Y_s)^2$ for $Y_s &lt; \alpha$ is the utility index
  over payoffs. Show that this agent's expected utility depends upon only
  the mean and variance of the state-contingent payoffs.


I do not really understand what the question is asking of me to show. Any suggestions or comments are greatly appreciated. Specifically, is the question asking me to find $$E[U(s_s)]$$ and $$Var[U(Y_s)]$$ if so how do we do that when we don't really have any defined distribution for $Y_s$? Also what does even mean that the mean the agent's expected utility depends upon only the mean and variance of the state-contingent payoffs. Does not make sense to me, I do not have much of an economics background as a graduate student in Applied Mathematics.
In order to understand this problem, I will work through the generic case. Say that a user had generalized quadratic (Bernoulli) utility, similar to your problem:

$$u(x) = \beta x^2 + \gamma x$$

and suppose that there is a distribution for the outcome of $x$, denoted $F(x)$. Thus, utility over this distribution is equal to

$$\begin{align}
\int u(x) \text{d}F(x) &amp; = \int (\beta x^2 + \gamma x) \text{d}F(x) \\
&amp; = \beta \int x^2 \text{d}F(x) + \gamma \int x \text{d}F(x) \\
&amp; = \beta \int x^2 \text{d}F(x) + \left[ - \beta \left(\int x \text{d}F(x)\right)^2 + \beta \left(\int x \text{d}F(x)\right)^2 \right] + \gamma \int x \text{d}F(x) \\
&amp; = \left[\beta \int x^2 \text{d}F(x) - \beta \left(\int x \text{d}F(x)\right)^2 \right] + \beta \left(\int x \text{d}F(x)\right)^2  + \gamma \int x \text{d}F(x) \\
&amp; = \beta\cdot(\text{variance of F(x)}) + \beta\cdot(\text{mean of F(x)})^2 + \gamma \cdot (\text{mean of F(x)})
\end{align}$$

So utility is determined by the mean and variance of the payoff distribution. If you understand the above work, then it should be easy to do the specific case you have given us. Try it for yourself.



Considering your specific case:

$$U(Y_s) = -\frac{1}{2}(\alpha - Y_s)^2$$

implies

$$U(L) = \sum_{s=1}^{S}\pi_s (-\frac{1}{2}(\alpha - Y_s)^2)$$

You can technically work with either finding $\int U(L) \text{d}L$ or $\int U(Y_s) \text{d}L$. Best of luck with your work.

Edit: Working in the discrete case will mean using expected values may prove helpful.