0
$\begingroup$

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.

$\endgroup$

2 Answers 2

Reset to default
1
+100
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
6
  • $\begingroup$ I see, so am I computing $E[U(Y_s)]$ and $Var[U(Y_s)]$? $\endgroup$ Commented Jan 14, 2017 at 3:26
  • $\begingroup$ Ok, then I am completely lost, how can one compute $E[Y_s]$ when $Y_s$ is not explicitly defined? $\endgroup$ Commented Jan 14, 2017 at 3:38
  • $\begingroup$ I probably did not make myself very clear, sorry. You don't "compute it". Just see if you can write $E[U(y_s)]$ (who cares about the variance of the utility function? the question only mentions the expectation) as a function of the sole unknowns $E[y_s]$, $Var[y_s]$, by plugging those values "appropriately" into the expression of $E[U(\cdot)]$. Note that $y_s$ is the random variable, $Y_s$ is the positive value it takes with probability $\pi_s$. Better now? $\endgroup$ Commented Jan 14, 2017 at 3:45
  • $\begingroup$ Seems to me that we have $$E[U(y_s)] = -\frac{1}{2}\left(E[Y_s^2] - 2aE[Y_s] + a^2\right)$$ dont really know where to go from there, perhaps at this point you could expand on your answer. $\endgroup$ Commented Jan 14, 2017 at 4:01
  • $\begingroup$ First of all, do not confuse $Y_s$, the constant, with $y_s$, the variable. $Y_s$ should not appear in your formula! Then, if you can write (you are almost there!) the expression using only $E[y_s]$ and $Var[y_s]$, and sum up for all the $s$ (the agent has many tickets; what happens to $E[y_i y_j]$) for $i\neq j$?), you have answered the question: you have a formula for $E[U(\cdot)]$ that depends solely on $E[y_s]$ and $Var[y_s]$. Almost there! $\endgroup$ Commented Jan 14, 2017 at 4:08

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.