Choice Under Uncertainty

States of the World

Definition: States of the World

A state of the world $\omega$ is a complete description of reality.

The set of all possible sets of the world is denoted by $\Omega$.

The definition is too ambitious to be of any practical use.

In practice, we assume $\Omega$ is small (finite).

As any model simplifies reality:

-- Think of a state of the world $\omega$ as describing all relevant information to some decision making problem.

States of the World

Example 1

Decision Problem: In a sunny morning, choose whether to carry an umbrella, a parasol or nothing to work.
Possible States of The World: $\Omega=\{\omega_1,\omega_2,\omega_3\}$.
$\omega_1$: $\longrightarrow\,$ The afternoon is also sunny.
$\omega_2$: $\longrightarrow\,$ The afternoon is cloudy but it does not rain.
$\omega_3$: $\longrightarrow\,$ The afternoon is rainy.

States of the World

Example 2

Decision Problem: Decide whether to lower the target range for the federal funds rate.
Possible States of The World: $\Omega=\{\omega_1, \omega_2, \omega_3, \omega_4\}$.
$\omega_1$: $\longrightarrow\,$ Low inflation and low unemployment.
$\omega_2$: $\longrightarrow\,$ Low inflation and high unemployment.
$\omega_3$: $\longrightarrow\,$ High inflation and low unemployment.
$\omega_4$: $\longrightarrow\,$ High inflation and high unemployment.

Knowledge

"Is there any point to which you would wish to draw my attention?"

"To the curious incident of the dog in the night-time."

"The dog did nothing in the night-time."

"That was the curious incident," remarked Sherlock Holmes.

Source: Doyle, A. C. (1894). The Memoirs of Sherlock Holmes. London, England: George Newnes.

Knowledge

"Reports that say that something hasn't happened are always interesting to me, because as we know, there are known knowns; there are things we know we know. We also know there are known unknowns; that is to say we know there are some things we do not know. But there are also unknown unknowns — the ones we don't know we don't know. And if one looks throughout the history of our country and other free countries, it is the latter category that tend to be the difficult ones."

Donald Rumsfeld, February 12, 2002.

Information Partitions

Individual $i$'s knowledge about the states of the world is represented by a partition $\mathcal P$ of the set $\Omega$.

Definition

A partition $\mathcal P$ of $\Omega$ is a collection of subsets of $\Omega$ such that:
1. If $A\in \mathcal P$ and $B\in \mathcal P$ and $A\neq B$ then $A\cap B=\emptyset$.
2. $\cup_{A\in \mathcal P} A =\Omega$.

If two states are in the same element of our knowledge partition, it means we cannot distinguish between these states. But if two states are in different elements of the partition, we can distinguish them.

Information partition

Weather in the afternoon example

Possible States of The World: $\Omega=\{\omega_1,\omega_2,\omega_3\}$.
$\omega_1$: $\longrightarrow\,$ The afternoon is also sunny.
$\omega_2$: $\longrightarrow\,$ The afternoon is cloudy but it does not rain.
$\omega_3$: $\longrightarrow\,$ The afternoon is rainy.

$\mathcal P_a=\{\{\omega_1\},\{\omega_2,\omega_3\}\}$, we know if the afternoon is sunny or not, but if it is not sunny, we cannot tell if it is rainy or cloudy.
$\mathcal P_b=\{\{\omega_1,\omega_2,\omega_3\}\}$, we do not know anything about the weather in the afternoon.
$\mathcal P_c=\{\{\omega_1,\omega_2\},\{\omega_3\}\}$, we know if the afternoon is rainy or not, but if it is not rainy, we cannot tell if it is sunny or cloudy.

Information Partitions

Cheryl's birthday

Albert and Bernard just met Cheryl.
"When's your birthday?" -- Albert asked Cheryl. Cheryl thought a second and said, "I'm not going to tell you, but I'll give you some clues." She wrote down a list of 10 dates:

May 15, May 16, May 19

June 17, June 18

July 14, July 16

August 14, August 15, August 17

"My birthday is one of these," -- she said.
Then Cheryl whispered in Albert's ear the month and only the month of her birthday. To Bernard, she whispered the day, and only the day.
"Can you figure it out now?" -- she asked Albert.
Albert: I don't know when your birthday is, but I know Bernard doesn't know, either.
Bernard: I didn't know originally, but now I do.
Albert: Well, now I know, too!
When is Cheryl's birthday?

Information Partitions

Cheryl's birthday — Solution

May 15 May 16 May 19 Jun 17 Jun 18 Jul 14 Jul 16 Aug 14 Aug 15 Aug 17

"whispered in Albert's ear the month"

"To Bernard, she whispered the day"

Albert: I don't know when your birthday is,
but I know Bernard doesn't know, either.

Bernard: I didn't know originally, but now I do.

Albert: Well, now I know, too!

Uncertainty and Time

Let's consider an economy with three periods $t=0,1$ and $2$; and two sources of uncertainty: inflation (low or high) and unemployment (low or high). Therefore there are four states of the world:

$\omega_1$ (low inflation and low unemployment),
$\omega_2$ (low inflation and high unemployment),
$\omega_3$ (high inflation and low unemployment),
$\omega_4$ (high inflation and high unemployment).

The set of the states of the world is $$\Omega = \{ \omega_1, \omega_2, \omega_3, \omega_4 \}.$$ Usually, in an economy uncertainty is gradually resolved over time.

Uncertainty and Time

Continuation

In an economy, uncertainty is resolved over time as new information is revealed over time. For example:

In period $t=0$, agents do not learn any information.
In period $t=1$, agents learn the inflation level.
In period $t=2$, agents learn the unemployment level.

The knowledge of the agents is described by the following partitions:

$\mathcal P_0=\{\{ \omega_1, \omega_2, \omega_3, \omega_4 \}\}$.
$\mathcal P_1=\{\{ \omega_1, \omega_2 \}, \{ \omega_3, \omega_4 \}\}$.
$\mathcal P_2=\{\{ \omega_1 \}, \{ \omega_2 \}, \{ \omega_3 \}, \{ \omega_4 \}\}$.

The partitions reflect the gradual information revelation:

$\mathcal P_0$: At $t=0$, all states are indistinguishable (one element containing all four states).
$\mathcal P_1$: At $t=1$, agents know inflation level, so low inflation states $\{\omega_1, \omega_2\}$ are distinguishable from high inflation states $\{\omega_3, \omega_4\}$.
$\mathcal P_2$: At $t=2$, agents know both inflation and unemployment, so all states are distinguishable.

Filtrations

A General Definition

A filtration is a sequence of partitions $\mathcal{F} = (\mathcal{P}_0, \mathcal{P}_1, \ldots, \mathcal{P}_T)$ where each partition represents the information available at time $t$, and the partitions become progressively finer over time.

Formally, a filtration satisfies:

Each $\mathcal{P}_t$ is a partition of $\Omega$.
For any $t < s$, the partition $\mathcal{P}_s$ is finer than $\mathcal{P}_t$.

What does "finer" mean?

Partition $\mathcal{P}_s$ is finer than $\mathcal{P}_t$ if every element of $\mathcal{P}_s$ is a subset of some element of $\mathcal{P}_t$.

Intuition: As time progresses, agents can distinguish between more states. They never "forget" information—they only gain more precision about which state occurred.

Filtrations

Interpreting Our Example

In our inflation-unemployment example: $$\mathcal{F} = (\mathcal{P}_0, \mathcal{P}_1, \mathcal{P}_2)$$ where: $$\begin{aligned} \mathcal{P}_0 &= \{\{\omega_1, \omega_2, \omega_3, \omega_4\}\} \\ \mathcal{P}_1 &= \{\{\omega_1, \omega_2\}, \{\omega_3, \omega_4\}\} \\ \mathcal{P}_2 &= \{\{\omega_1\}, \{\omega_2\}, \{\omega_3\}, \{\omega_4\}\} \end{aligned}$$

Notice that:

Each element of $\mathcal{P}_1$ is contained in an element of $\mathcal{P}_0$ (both are contained in the single element $\{\omega_1, \omega_2, \omega_3, \omega_4\}$).
Each element of $\mathcal{P}_2$ is contained in an element of $\mathcal{P}_1$ (e.g., $\{\omega_1\} \subset \{\omega_1, \omega_2\}$).

This confirms $\mathcal{F}$ is indeed a filtration.

Filtration

Information Tree

$\Omega$ $\{\omega_1,\omega_2\}$ $\{\omega_3,\omega_4\}$ $\{\omega_1\}$ $\{\omega_2\}$ $\{\omega_3\}$ $\{\omega_4\}$ $t = 0$ $t = 1$ $t = 2$

Lotteries

Assume we have $n$ states of the world and a probability distribution over the states. A lottery is a real-random variable. Or put simply,

Definition: A lottery is a list of prizes and probabilities,

$$\ell=\left((x_1,x_2,\ldots,x_n),(p_1,p_2,\ldots,p_n)\right),$$

where $x_k \in \mathbb{R}^m$ is the prize the lottery gives when state $k$ occurs, and $p_k\ge 0$ is the probability that state $k$ occurs.

Lotteries

In practice, it is very useful to depict lotteries as a graph.

$\ell$ $x_1$ $x_2$ $\ldots$ $x_n$ $p_1$ $p_2$ $p_n$

The Certain Lottery

The lottery that gives prize $x$ with probability one (with certainty) is denoted by: $$\delta_x=\left((x),(1)\right).$$

$\delta_x$ $x$ $1$

Expectation and Variance

The expected value of a lottery $\ell_1=\left((x_1,x_2,\ldots,x_n),(p_1,p_2,\ldots,p_n)\right)$ is: $$\mathbb{E}[\ell_1]=p_1\cdot x_1+p_2\cdot x_2 +\ldots p_n\cdot x_n = \sum_{i=1}^n p_i\cdot x_i;$$ and the variance of a lottery $\ell$ is: $$\begin{aligned}\mathrm{Var}[\ell_1]=&p_1\cdot (x_1-\mathbb{E}[\ell_1])^2+p_2\cdot (x_2-\mathbb{E}[\ell_1])^2 +\ldots p_n\cdot (x_n-\mathbb{E}[\ell_1])^2 = \\ =& \sum_{i=1}^n p_i\cdot (x_i-\mathbb{E}[\ell_1])^2.\end{aligned}$$

Composition of Lotteries

Given two lotteries, $\ell_1=\left((x_1,x_2,\ldots,x_n),(p_1,p_2,\ldots,p_n)\right)$ and $\ell_2=\left((y_1,y_2,\ldots,y_m),(q_1,q_2,\ldots,q_n)\right)$ and a number $\alpha$ in the interval $(0,1)$, we can create a compound lottery $\ell$ that plays $\ell_1$ with prob. $\alpha$ and $\ell_2$ with prob. $1-\alpha$.

The compound lottery $\ell$ plays $\ell_1$ with prob. $\alpha$ and $\ell_2$ with prob. $1-\alpha$:

$\ell$ $\ell_1$ $\ell_2$ $x_1$ $x_2$ $y_1$ $y_2$ $\alpha$ $1-\alpha$ $p$ $1-p$ $q$ $1-q$

$$\ell = \alpha\ell_1 \oplus (1-\alpha)\ell_2 = \left((x_1,x_2,y_1,y_2),(\alpha p,\,\alpha(1-p),\,(1-\alpha)q,\,(1-\alpha)(1-q))\right)$$

The reduced form — $\ell$ connects directly to prizes:

$\ell$ $x_1$ $x_2$ $y_1$ $y_2$ $\alpha \cdot p$ $\alpha \cdot (1-p)$ $(1-\alpha) \cdot q$ $(1-\alpha) \cdot (1-q)$

$$\ell = \alpha\ell_1 \oplus (1-\alpha)\ell_2 = \left((x_1,x_2,y_1,y_2),(\alpha p,\,\alpha(1-p),\,(1-\alpha)q,\,(1-\alpha)(1-q))\right)$$

Preferences Over Lotteries

Given two lotteries $\ell_a$ and $\ell_b$ such that a decision maker (DM) chooses $\ell_a$ over $\ell_b$,
the following statements are equivalent:

The DM judges $\ell_a$ no worst than $\ell_b$ (everyday language);
The DM prefers $\ell_a$ to $\ell_b$ (economics language);
$\ell_a\succeq_{\text{DM}} \ell_b$ (mathematics language).

For simplicity we write:

$\ell_a\succ \ell_b$ when $\ell_a\succeq\ell_b$ but $\ell_b\not\succeq\ell_a$ (strict preference)
$\ell_a\sim \ell_b$ when $\ell_a\succeq\ell_b$ and $\ell_b\succeq\ell_a$ (indifference).

Preferences Over Lotteries

A preference of the DM, $\succeq_{\text{DM}}$, over the set of lotteries is just the DM's ranking of lotteries.
We wish (for convenience) a numerical score that reflects the DM's ranking.

von Neumann & Morgenstern's Assumptions

Completeness: For any two lotteries $\ell_1$ and $\ell_2$,
$\ell_1\succeq \ell_2$ and/or $\ell_2\succeq \ell_1$.
Transitivity: For any lotteries $\ell_1$, $\ell_2$ and $\ell_3$,
if $\ell_1\succeq \ell_2$ and $\ell_2\succeq \ell_3$ then $\ell_1\succeq \ell_3$.
Continuity: If $\ell_1\succeq\ell_2\succeq \ell_3$ then exists $p\in[0,1]$ such that $\ell_2\sim p\ell_1\oplus(1-p)\ell_3$.
Independence: If $\ell_1\succ\ell_2$ then for any $\ell_3$ and any $0<p<1$, $p\ell_1\oplus (1-p)\ell_3\succ p\ell_2\oplus(1-p)\ell_3$.

If $\succeq$ satisfy all of the above, there exists $u:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\left((x_1,\ldots,x_n),(p_1,\ldots,p_n)\right)\succ \left((y_1,\ldots,y_m),(q_1,\ldots,q_m)\right)$$ if and only if $$\sum_{k=1}^n u(x_k)\cdot p_k> \sum_{k=1}^m u(y_k)\cdot q_k.$$

Expected Utility

We write: $$ U(\ell_1)=u(x_1)\cdot p_1+\ldots + u(x_n)\cdot p_n, $$

and refer to $U$ as the expected utility

and to $u$ as the:

utility for money or
Bernoulli utility or
von Neumann-Morgenstern utility (vNM)

Risk Aversion

$$u(x)=2\frac{x^{\alpha}-100^{\alpha}}{102^{\alpha}-100^{\alpha}}+100$$

$$X=\begin{cases}100 & \text{with probability } 1/2\\ 102 & \text{with probability } 1/2\end{cases}$$

$u(100)$ $u(102)$ $U(X)\!=\!{\scriptsize\tfrac{u(100)}{2}+\tfrac{u(102)}{2}}$ $u(101)$

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Risk Aversion

Consider two lotteries $A=(x_0,1)$ and $B=\left((x_0+M,x_0-M),\left(\frac{1}{2},\frac{1}{2}\right)\right)$ with $M>0$. That is, $A$ is a sure lottery (no risk) and $B$ is a risky lottery in which you win or lose $M$ with equal probabilities.

Notice both lotteries have the identical expected value, $\E[A]=\E[B]=x_0$.

The difference in expected utility of these lotteries is approximately,

$$U(B)-U(A)=\frac{M}{2}\cdot\left(Mu(x_0+M)-Mu(x_0-M)\right)$$

Expected Utility Theory

Attitudes Towards Risk

Diminishing marginal utility, $u$ is concave, $u^{\prime\prime}<0\Rightarrow$, the consumer is risk-averse. $$U(X)<u(\mathrm{E}[X]) \text{ for all } X$$
Increasing marginal utility, $u$ is convex, $u^{\prime\prime}>0\Rightarrow$, the consumer is risk-loving. $$U(X)>u(\mathrm{E}[X]) \text{ for all } X$$
Constant marginal utility, $u$ is affine (linear plus a constant), $u^{\prime\prime}=0\Rightarrow$, the consumer is risk-neutral, $$U(X)=u(\mathrm{E}[X]) \text{ for all } X$$

Measuring the Degree of Risk-Aversion

The Arrow-Pratt or Absolute Measure of Risk Aversion

Definition

The Arrow-Pratt absolute measure of risk-aversion of an agent with VN-M utility $u$ at wealth level $w$ is: $$\rho_{u}(w)=\dfrac{-u^{\prime\prime}(w)}{u^{\prime}(w)}.$$

If for two individuals with VN-M utilities $u$ and $\widetilde{u}$ we have that $\rho_{u}(w)>\rho_{\widetilde{u}}(w)$ for all wealth levels $w$ then we say that the agent with utility $u$ is more risk-averse than the agent with utility $\widetilde{u}$.

Measuring the Degree of Risk-Aversion

The Relative Measure of Risk-Aversion

Definition

The relative measure of risk-aversion of an agent with utility $u$ at wealth level $w$ is: $$r_{u}(w)=\dfrac{-u^{\prime\prime}(w)\,w}{u^{\prime}(w)}.$$

How to use expected utility

Step 1: Set up the decision problem
- States of the world: $\Omega = \{\omega_1, \ldots, \omega_n\}$
- Prior probabilities: $\Pr(\omega_1), \ldots, \Pr(\omega_n)$
- Information partition: $\mathcal{P} = \{I_1, \ldots, I_k\}$
Step 2: Define strategies
- A strategy $s$ specifies one action for each information set: $s: \mathcal{P} \to A$
- You choose your action based only on what you know (your information set)
Step 3: Evaluate each strategy
- For strategy $s$, expected utility is: $$\mathbb{E}[u|s] = \sum_{j} \Pr(I_j) \cdot \mathbb{E}[u \,|\, I_j, s]$$ where $\mathbb{E}[u \,|\, I_j, s] = \sum_{\omega \in I_j} \Pr(\omega|I_j) \cdot u(s(I_j), \omega)$
Step 4: Choose optimal strategy
- Select the strategy $s^*$ that maximizes expected utility.

How to use expected utility

Example: Umbrella or Parasol?

Decision Problem: Before leaving home, choose whether to carry an umbrella, a parasol, or nothing.

Set-up:

States (afternoon weather): $\Omega = \{\omega_1, \omega_2, \omega_3\}$ where $\omega_1$ = sunny, $\omega_2$ = cloudy, $\omega_3$ = rainy
Probabilities: $\Pr(\omega_1) = 1/2$, $\Pr(\omega_2) = 1/4$, $\Pr(\omega_3) = 1/4$
Actions: $A = \{\text{umbrella}, \text{parasol}, \text{nothing}\}$
Information partition: $\mathcal{P} = \{\{\omega_1\}, \{\omega_2, \omega_3\}\}$ (you can tell if it will be sunny or not)

Strategies: A strategy assigns an action to each information set.

$I_1 = \{\omega_1\}$: You know it will be sunny $\Rightarrow$ choose parasol
$I_2 = \{\omega_2, \omega_3\}$: You know it won't be sunny $\Rightarrow$ choose umbrella

Strategy $s^*$: parasol if sunny, umbrella if not sunny.

How to use expected utility

Example: Computing Expected Utility

Utility matrix $u(\text{action}, \omega)$:

	$\omega_1$ (sunny)	$\omega_2$ (cloudy)	$\omega_3$ (rainy)
umbrella	6	7	10
parasol	12	6	2
nothing	10	8	0

Expected utility of strategy $s^*$ (parasol if sunny, umbrella if not):

$$\begin{aligned} \mathbb{E}[u|s^*] &= \Pr(I_1) \cdot u(s^*(I_1), \omega_1) + \Pr(I_2) \cdot \mathbb{E}[u|I_2, s^*] \\ &= \frac{1}{2} \cdot u(\text{parasol}, \omega_1) + \frac{1}{2} \cdot \left[\Pr(\omega_2|I_2) \cdot u(\text{umb}, \omega_2) + \Pr(\omega_3|I_2) \cdot u(\text{umb}, \omega_3)\right] \\ &= \frac{1}{2} \cdot 12 + \frac{1}{2} \cdot \left[\frac{1}{2} \cdot 7 + \frac{1}{2} \cdot 10\right] = 6 + \frac{17}{4} = 10.25 \end{aligned}$$

How to use expected utility

Example: Federal Reserve

Decision Problem: The Federal Reserve must decide whether to lower interest rates or keep them unchanged.

Set-up:

States: $\Omega = \{\omega_1, \omega_2, \omega_3, \omega_4\}$
- $\omega_1$ = Low inflation, low unemployment
- $\omega_2$ = Low inflation, high unemployment
- $\omega_3$ = High inflation, low unemployment
- $\omega_4$ = High inflation, high unemployment
Probabilities: $\Pr(\omega_1) = 0.4$, $\Pr(\omega_2) = 0.1$, $\Pr(\omega_3) = 0.3$, $\Pr(\omega_4) = 0.2$
Actions: $A = \{\text{lower}, \text{keep}\}$
Information partition: $\mathcal{P} = \{\{\omega_1, \omega_2\}, \{\omega_3, \omega_4\}\}$ (FED knows if inflation is high or not)

Strategies: A strategy assigns an action to each information set.

$I_1 = \{\omega_1, \omega_2\}$: FED knows inflation is low $\Rightarrow$ lower rates
$I_2 = \{\omega_3, \omega_4\}$: FED knows inflation is high $\Rightarrow$ keep rates

Strategy $s^*$: lower if low inflation, keep if high inflation.

How to use expected utility

Example: FED Expected Utility

Utility matrix $u(\text{action}, \omega)$:

	$\omega_1$ (LI, LU)	$\omega_2$ (LI, HU)	$\omega_3$ (HI, LU)	$\omega_4$ (HI, HU)
lower	6	10	2	4
keep	8	5	9	7

Expected utility of strategy $s^*$ (lower if low inflation, keep if high):

$$\begin{aligned} \mathbb{E}[u|s^*] &= \Pr(I_1) \cdot \mathbb{E}[u|I_1, s^*] + \Pr(I_2) \cdot \mathbb{E}[u|I_2, s^*] \\[4pt] \mathbb{E}[u|I_1, s^*] &= \Pr(\omega_1|I_1) \cdot u(\text{lower}, \omega_1) + \Pr(\omega_2|I_1) \cdot u(\text{lower}, \omega_2) \\[4pt] &= 0.8 \cdot 6 + 0.2 \cdot 10 = 6.8 \\[4pt] \mathbb{E}[u|I_2, s^*] &= \Pr(\omega_3|I_2) \cdot u(\text{keep}, \omega_3) + \Pr(\omega_4|I_2) \cdot u(\text{keep}, \omega_4) \\[4pt] &= 0.6 \cdot 9 + 0.4 \cdot 7 = 8.2 \\[4pt] \mathbb{E}[u|s^*] &= 0.5 \cdot 6.8 + 0.5 \cdot 8.2 = 7.5 \end{aligned}$$