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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.

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    $\begingroup$ It is not clear to me what part of the question you do not understand. The problem could be one of economics, mathematics or English. Please elaborate. $\endgroup$ – Giskard Jan 14 '17 at 7:44
  • $\begingroup$ @denesp I am not sure how to get the mean and variance. $\endgroup$ – Wolfy Jan 14 '17 at 17:18
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\begin{eqnarray*} \displaystyle U(L) & = &\sum_{s=1}^{S}\pi_s U(Y_s) = \sum_{s=1}^{S} \left(-\frac{1}{2}\pi_s(\alpha - Y_s)^2\right) = -\frac{1}{2}\sum_{s=1}^{S} \left(\pi_s(\alpha^2 + Y_s^2-2\alpha Y_s)\right) \\ &=& -\frac{1}{2}\left(\alpha^2\sum_{s=1}^{S} \pi_s + \sum_{s=1}^{S} \pi_sY_s^2-2\alpha \sum_{s=1}^{S} \pi_sY_s\right) = -\frac{1}{2}\left(\alpha^2 + \mathbb{E}(L^2)-2\alpha \mathbb{E}(L)\right) \\ &=& -\frac{1}{2}\left(\alpha^2 + \mathbb{E}(L^2) - (\mathbb{E}(L))^2 + (\mathbb{E}(L))^2 -2\alpha \mathbb{E}(L)\right) \\ &=& -\frac{1}{2}\left(\alpha^2 + \mathbb{V}(L) + (\mathbb{E}(L))^2 -2\alpha \mathbb{E}(L)\right) \end{eqnarray*}

So utility from the lottery only depends on expected value - $\mathbb{E}(L)$ and variance - $\mathbb{V}(L)$ of the state-contingent payoffs.

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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

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

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.

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  • $\begingroup$ so, coming back to this. What am I trying to find $E[U(L]$ and $Var[U(L)]$? $\endgroup$ – Wolfy Jan 14 '17 at 23:52
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I think this may not always be true as stated

You say "$U(y_s) = -\frac{1}{2}(\alpha - Y_s)^2$ for $Y_s < \alpha$" so presumably $U(y_s) = 0$ for $y_s \ge \alpha$ to avoid utilities decreasing as large prizes increase

Suppose as an example $\alpha=50$ and there are two different lotteries:

  • In the first, the possible outcomes are a payoff of $0$ with probability $\frac12$ or a payoff of $30$ with probability $\frac12$

    • This has a mean of $ 0\times \frac12 + 30\times \frac12=15$ and a variance of $225$
    • The expected utility is $ -\frac{1}{2}(50 - 0)^2\times \frac12 -\frac{1}{2}(50 - 30)^2\times \frac12 = -725$

  • In the second, the possible outcomes are a payoff of $10$ with probability $\frac9{10}$ or a payoff of $60$ with probability $\frac1{10}$

    • This also has a mean of $10\times \frac9{10} + 60\times \frac1{10}=15$ and a variance of $225$
    • The expected utility is $ -\frac{1}{2}(50 - 10)^2\times \frac9{10} +0\times \frac1{10} = -720$ which is different
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    $\begingroup$ It seems suspicious that the counterexample does not work without your assumption "$U(y_s) = 0$ for $y_s \ge \alpha$". And the assumption is not referred to in any way in the question. $\endgroup$ – Giskard Jan 14 '17 at 12:22
  • $\begingroup$ @denesp: But $\alpha$ is mentioned in the question (it is implicitly the potential prize which maximises utility) and it would be strange if larger prizes than $\alpha$ could lead to more negative utilities than $\alpha$ would (as implicitly assumed in Kitsune Cavalry's answer). If you removed my assumption then the second lottery would have a lower utility (of $-725$) than a lottery with a payoff of $10$ with probability $\frac9{10}$ or a payoff of $50$ (rather than $60$) with probability $\frac1{10}$ (utility of $-720$) $\endgroup$ – Henry Jan 14 '17 at 12:42
  • $\begingroup$ It is a standard assumption that well-behaved utility functions are monotonic $\endgroup$ – Henry Jan 14 '17 at 12:47
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    $\begingroup$ @denesp In a lottery, if a prize is too big, you can always refuse or give away the excess. Here this is suggested by the statement $U(y_s) = -\frac{1}{2}(\alpha - Y_s)^2$ for $Y_s < \alpha$ $\endgroup$ – Henry Jan 14 '17 at 15:15
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    $\begingroup$ This last one is a good point. Still, given the problem one might argue that you are supposed to assume that for all $s$ $Y_s < \alpha$. I am not saying this assumption is better than yours. But given the nature of the exercise it seems more in line with the intent of the author. $\endgroup$ – Giskard Jan 14 '17 at 17:54

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