Probabilities and Beliefs

Extracting $u$ from choices

Under Expected Utility (Standard Model)

The idea is the same as in choice under certainty. There we defined the utility of every basket on the diagonal, $u(x,x)=x$, and gave any other basket the utility of the diagonal basket it is indifferent to. Under expected utility the ruler is no longer the diagonal but the family of lotteries between the best and the worst prize.

Standard Gamble (Probability Equivalent)
Let $b$ be the best prize and $w$ the worst, and set $u(b)=1$ and $u(w)=0$. For any other prize $x$, find the probability $p$ that makes the agent indifferent between $x$ for sure and the lottery that pays $b$ with probability $p$ and $w$ with probability $1-p$: $$\delta_x\sim\left((b,w),(p,1-p)\right).$$ The expected utility of the lottery is $p\cdot u(b)+(1-p)\cdot u(w)=p$, so $u(x)=p$.

This is the von Neumann–Morgenstern construction turned into a question, and it is the standard way of measuring utility in health economics.

Example. Three prizes. Ann's choices between sure prizes reveal the ranking: San Pellegrino $\succ$ KitKat $\succ$ banana.

  1. Fix the scale: $u(\text{banana})=0$ and $u(\text{San Pellegrino})=1$.
  2. Ask Ann: a KitKat for sure, or the lottery $\left((\text{San Pellegrino},\text{banana}),(p,1-p)\right)$?
  3. She takes the KitKat at $p=1/2$, so $u(\text{KitKat})\gt 1/2$: raise $p$. She takes the lottery at $p=3/4$, so $u(\text{KitKat})\lt 3/4$: lower $p$.
  4. Keep going until she is indifferent, say at $p^*=0.7$. Then $u(\text{KitKat})=0.7$.

Extracting $p$ from choices

Under Expected Utility (Standard Model)

Same trick, new unknown. In the standard model beliefs are probabilities. Ann's probability that it rains in Chapel Hill tomorrow is not written anywhere: we extract it from her choices, one indifference at a time.

Matching Probability
Take two prizes, $b\succ w$, and an event $A$. A bet on $A$ pays $b$ if $A$ happens and $w$ if it does not. Find the probability $p$ that makes Ann indifferent between the bet on $A$ and the lottery that pays $b$ with probability $p$ and $w$ with probability $1-p$: $$(b \text{ if } A,\; w \text{ if not } A)\sim\left((b,w),(p,1-p)\right).$$ The expected utility of the bet is $\Pr[A]\cdot u(b)+(1-\Pr[A])\cdot u(w)$ and the expected utility of the lottery is $p\cdot u(b)+(1-p)\cdot u(w)$. They are equal only if $\Pr[A]=p$.

Notice that $u$ cancels: all we used is $u(b)\gt u(w)$. So we do not need to know Ann's utility to extract her beliefs. Any two prizes she can rank will do.

Example. Prizes as before, San Pellegrino $\succ$ banana, and $A$ = rain in Chapel Hill tomorrow.

  1. Ask Ann: a San Pellegrino if it rains tomorrow and a banana if not, or the lottery $\left((\text{San Pellegrino},\text{banana}),(p,1-p)\right)$?
  2. She takes the bet at $p=1/2$, so $\Pr^A[\text{rain}]\gt 1/2$: raise $p$. She takes the lottery at $p=3/4$, so $\Pr^A[\text{rain}]\lt 3/4$: lower $p$.
  3. Keep going until she is indifferent, say at $p^*=0.6$. Then $\Pr^A[\text{rain}]=0.6$.

Beliefs

In the standard model, individuals update their (probabilities) beliefs using:

Bayes' Rule
Given two events $A$ and $B$ such that $B$ is possible, $\Pr(B)\gt 0$, then the probability of $A$ given that $B$ occurred is: $$\Pr(A\mid B)=\dfrac{\Pr(A\cap B)}{\Pr(B)}.$$

Later, we shall see how beliefs (prior probabilities) are “formed” in the standard model…

Bayesian Updating

Let's consider two alternative hypotheses regarding some coin:

(U)
The coin is unbiased, $\Pr[H\mid U]=\Pr[T\mid U]=1/2$.
(B)
The coin is biased towards heads, $\Pr[H\mid B]=3/4$ and $\Pr[T\mid B]=1/4$.

Ann and Bart believe that only these two hypotheses are possible but they ascribe different likelihoods. Ann's prior probabilities are $\Pr^A[U]=\Pr^A[B]=1/2$ and Bart's are $\Pr^B[U]=1/3$ and $\Pr^B[B]=2/3$.

Bayesian Updating

Let $p_t(h^t)$ be Ann's belief that the hypothesis $U$ is true given that the particular history of coin tosses, $h^t$, was observed. We have that: $$p_0(\emptyset)=1/2,$$ $$p_1(T)=\Pr^A[U\mid T]=\dfrac{\Pr^A[U\cap T]}{\Pr^A[T]}=\dfrac{p_0\cdot \tfrac12}{p_0\cdot \tfrac12 +(1-p_0)\cdot \tfrac14},$$ $$p_t(h^{t-1},T)=\dfrac{p_{t-1}(h^{t-1})\cdot \tfrac12}{p_{t-1}(h^{t-1})\cdot \tfrac12 +(1-p_{t-1}(h^{t-1}))\cdot \tfrac14}.$$

In sum: given some coin toss outcome $H$ or $T$ and knowing the prior probability of the hypothesis $U$ being true, we are able to compute the posterior (updated) probability that the hypothesis $U$ is true.

Result: If two individuals assign positive probabilities to the same set of hypotheses, it does not matter if their priors differ. In the limit, as more information becomes available, their posteriors will merge. They will think alike.

Heuristics

Heuristics are rules of thumb that deliver satisfactory outcomes without requiring high computational costs.

Probability Estimation

Common heuristics used to evaluate probabilities:

  1. Availability
  2. Representativeness
  3. Base rate neglect
  4. ‘Law’ of small numbers (gambler's fallacy and ‘hot hand’)

Biases

  1. Overconfidence
    1. Overestimation
    2. Overplacement
    3. Overprecision
  2. Underconfidence
  3. Self-serving bias
    1. Pavlovian withdrawal
    2. Confirmatory bias
    3. Self-attribution bias
    4. Visceral fit
  4. Projection bias
  5. Magical thoughts
    1. Tempting fate
    2. Contagion
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