3
$\begingroup$

I am trying to prove the expected utility theorem with three outcomes. The expected utility with $n$ outcomes is rather cumbersome and long in the economics textbook Mas-Colell. But I was hoping that the proof with three outcomes is shorter, however, I am having some difficulty proving it.

Suppose I have three lotteries $x \succeq y \succeq z$. We can think of $x$ as the "best" lottery and $z$ as the "worst". The can set $u(x)=1$ and $u(z)=0$ and then $u(y)=p$ where $p$ is the probability at which a gamble with a $p$ chance of $x$ and a $1-p$ chance of $z$ is indifferent to $y$.

How do I proceed from here to prove the Expected utility theorem?

For $n$ outcomes, the EU states that given $\succeq $ satisfies independence and continuity axioms, there is a utility function $ u:Z\rightarrow \mathbb{R}$ such that if $$p\succeq q\Leftrightarrow \sum_{i=1}^{n}p_{i}u(z_{i})\geq \sum_{i=1}^{n}q_{i}u(z_{i}).$$

Edit: Letting $u(x)= 1$ and $u(z) = 0$ linearly scales the utility function. So we want to show $$u(y) = u\left ( px + (1-p)z \right ) = pu(x) + (1-p)u(z) = p \text{ by the independence axiom}.$$

Since we have assumed $x \succeq y \succeq z$ then it follows that $u(x) \geq u(y) \geq u(z)$, where $p \in [0,1]$. Does this make sense?

I am not sure about the part: $u\left ( px + (1-p)z \right ) = pu(x) + (1-p)u(z)$. Is this a legitimate way of writing the proof for the case of three outcomes?

$\endgroup$
3
$\begingroup$

In order to do that, you need to define $u(·)$ as a utility function on "sure things" rather than on lotteries. In your example, you need to think in terms of the set of possible prizes to the lotteries. Say the set of possible prizes is given by $R$ and asume that is finite. For any $r\in R$, define $w_r$ as a lottery that pays $r$ in every state of nature. Then, it should be clear that

$$x \sim \sum_i p_i w_i,$$

(and similarly for $y$ and $z$) where $p_i$ denotes the probability of state $i.$ Given that, as you mention, it must be the case that two equally desirable lotteries receive the same expected utility, the utility assigned to $x$ needs to be

$$H(x)=H\left(\sum_i p_i w_i\right).$$

The first chapter of Huang, C. and Litzenberger, R.H. Foundations for Financial Economics, North Holland, 1988 explains in detail why the function representing the preferences $(H(·))$ has to be linear, and, with linearity stablished, you get the expected utility representation because

$$H\left(\sum_i p_i w_i\right)=\sum_i p_i H(w_i).$$

The only thing left is to define $u(r) \equiv H(w_r),$ i.e, $u(r)$ is defined on outcomes, while $H(w_r)$ is defined on lotteries. This is why we say that $u(r)$ is a utility function on sure things, because $r$ receives the same utility level that a lottery paying precisely $r$ in every state of nature.

Of course, expected utility is most useful in settings with large outcomes spaces and/or large lotteries spaces. In a restricted setting such as yours, you're probably best served by assigning values ("utilities") to lotteries themselves (as you did in your question). Again, a more rigorous proof can be found in Huang and Litzenberger, op.cit.

$\endgroup$
  • 1
    $\begingroup$ Why do you say OP's example has 3 lotteries but just 2possible outcomes? Also what's $z_3$? $\endgroup$ – Herr K. Nov 26 '18 at 18:34
  • $\begingroup$ @Patricio I have made an edit. $\endgroup$ – OGC Nov 26 '18 at 19:48
  • $\begingroup$ @OGC, As I mention in my answer, in order to obtain an expected utility representation, you need to define $u(·)$ on "sure things." I'm edditing my answer to hopefully make myself clearer. $\endgroup$ – Patricio Nov 27 '18 at 10:13
  • $\begingroup$ @HerrK., my mistake, I'll edit the answer $\endgroup$ – Patricio Nov 27 '18 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.