Agent's expected utility depends only on mean and variance

Question

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 $Y_s < \alpha$ 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.

Answer 1

Let me clarify the question.

First, writing $U(L)=\sum_{s=1}^{S}\pi_{s}U(Y_{s})$ does not make sense. The two $U$ functions are different objects. Correct way is $U(L)=\sum_{s=1}^{S}\pi_{s}u(Y_{s})$, were $U$ is expected utility (of lottery $L$) and $u$ is Bernoulli utility (of payoff $Y_{s}$). $U$ is a function that lives on space of lotteries. $u$ is a function that lives on space of real numbers. It is the Bernoulli utility that is quadratic, that is, $u(x)=-\frac{1}{2}(\alpha-x)^{2}$.

Second, a lottery is $L=(Y_{s},\pi_{s})_{s=1}^{S}$. That is, for each state $s\in\{1,\ldots,S\}$, it gives probability of $s$ occurring, $\pi_{s}$, and (state-contingent) payoff/return from the lottery, $Y_{s}$, in $s$. Namely, $Y_{s}$ is a number (e.g. amount of money you win in state $s$), not a random variable. Lottery $L$ is a random variable. One that gives realization $Y_{s}$ with probability $\pi_{s}$ and has discrete support. (In view of the convention to denote r.v.s by capital letters, this notation is unfortunate. Let me switch to $y_{s}$ then.)

Since $L$ is r.v., $\sum_{s=1}^{S}\pi_{s}=1$. Let me denote by $\mu_{L}=\sum_{s=1}^{S}\pi_{s}y_{s}$ mean (return) of the lottery $L$. Let me denote by $\sigma^{2}_{L}=\sum_{s=1}^{S}\pi_{s}(y_{s}-\mu_{L})^{2}=\left(\sum_{s=1}^{S}\pi_{s}y_{s}^{2}\right)-\mu_{L}^{2}$ variance (of returns) of the lottery (the last equality is standard algebra, in fact the equation is discrete version of $\mathbb{E}[x^{2}]-\mathbb{E}[x]^{2}$).

So now we have $U(L)=\sum_{s=1}^{S}\pi_{s}u(y_{s})$, $u(y_{s})=-\frac{1}{2}(\alpha-y_{s})^{2}$, $L=(y_{s},\pi_{s})_{s=1}^{S}$. We want to show $U(L)$ depends only on mean and variance of the lottery. We have \begin{equation*}\begin{aligned} U(L)&=-\tfrac{1}{2}\sum_{s=1}^{S}\pi_{s}(\alpha^{2}-2\alpha y_{s}+y_{s}^{2})\\ &=-\tfrac{1}{2}\left(\alpha^{2}-2\alpha\sum_{s=1}^{S}\pi_{s}y_{s}+\sum_{s=1}^{S}\pi_{s}y_{s}^{2}\right)\\ &=-\tfrac{1}{2}\left(\alpha^{2}-2\alpha\mu_{L}+\sigma^{2}_{L}+\mu_{L}^{2}\right)\\ &=-\tfrac{1}{2}\left(\sigma_{L}^{2}+(\alpha-\mu_{L})^{2}\right) \end{aligned}\end{equation*} which is what we wanted to show.

Answer 2

Hint:

First obtain the mean and variance of the state-contingent payoffs, which are random variables that with probability $\pi_s$ take value $Y_s$ (you win the lottery!) and with probability $1-\pi_s$ take value $0$. Call these means and variances $\mu_s$ and var$_s$.

Now try to rewrite $E[U(\cdot)]$ using only the quantities $\mu_s$ and var$_s$ in the formula, and in particular avoiding any reference $U(\cdot)$ or $\pi_s$. That's what the question is about.

Note: if the problem is from a textbook, add a reference so other people can find it!