Rabin's calibration theorem

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Consider an expected-utility decision-maker with wealth level ${\displaystyle w}$ and differentiable, concave Bernoulli utility function ${\displaystyle u}$. Imagine that she rejects the following lottery:

${\displaystyle L={\begin{cases}{\text{win }}\$125{\text{ with }}50\%{\text{ chance,}}\\{\text{lose }}\$100{\text{ with }}50\%{\text{ chance.}}\end{cases}}}$

This implies that

${\displaystyle 0.5u(w+125)+0.5u(w-100)<u(w)}$

${\displaystyle \iff u(w+125)-u(w)<u(w)-u(w-100)}$

${\displaystyle \iff {\frac {u(w+125)-u(w)}{125}}<{\frac {100}{125}}{\frac {u(w)-u(w-100)}{100}}.}$

Since ${\displaystyle u}$ is concave, we have

${\displaystyle u'(w+125)\leq {\frac {u(w+125)-u(w)}{125}}<{\frac {100}{125}}{\frac {u(w)-u(w-100)}{100}}\leq {\frac {100}{125}}u'(w-100).}$

This means, therefore, that on each interval ${\displaystyle [w-100,w+125]}$ of length ${\displaystyle 225}$ for which the above holds (i.e., the lottery is rejected), increasing wealth by ${\displaystyle \$225}$ reduces marginal utility ${\displaystyle u'}$ to at most ${\displaystyle {\frac {100}{125}}=0.8}$ times its previous value. If the lottery continues to be rejected throughout the relevant range of wealth levels, receiving ${\displaystyle \$1000}$ would reduce marginal utility to at most ${\displaystyle 0.8^{4}\approx 41\%}$ of its current value, and receiving ${\displaystyle \$10000}$ would reduce it to at most ${\displaystyle 0.8^{44}\approx 0.0054\%}$.

For example, suppose that the decision maker rejects the above lottery ${\displaystyle L}$ for all wealth levels ${\displaystyle w\leq \$300,000}$. This will imply that, at wealth level ${\displaystyle w=\$290,000}$, she rejects the following lottery:^[4]

${\displaystyle L'={\begin{cases}{\text{win }}\$160{\text{ billion with }}50\%{\text{ chance,}}\\{\text{lose }}\$1000{\text{ with }}50\%{\text{ chance.}}\end{cases}}}$

Such rejection is intuitively absurd and empirically counterfactual.^[7][8]

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Below is a tightened version of Rabin's original theorem given by Balter, Chau, and Schweizer.^[9]

Theorem: Suppose an expected-utility maximizer has an increasing and concave utility function ${\displaystyle u}$. Let ${\displaystyle W\subseteq \mathbb {R} }$ be an interval, and consider the binary lottery ${\displaystyle L:=(-l,g;{\tfrac {1}{2}},{\tfrac {1}{2}})}$, where ${\displaystyle g>l>0}$. If the decision maker rejects ${\displaystyle L}$ at every wealth level ${\displaystyle w\in W}$, then given any ${\displaystyle m,n\in \mathbb {N} ^{+}}$, for any ${\displaystyle w\in W}$ such that ${\displaystyle [w-ng,w+mg]\subseteq W}$,

${\displaystyle \underbrace {u(w+mg)-u(w)} _{{\text{utility gain of }}m{\text{ steps up}}}\leq r(w)\sum _{i=0}^{m-1}q^{i}\quad {\text{and}}\quad r(w)\sum _{i=1}^{n}q^{-i}\leq \underbrace {u(w)-u(w-ng)} _{{\text{utility loss of }}n{\text{ steps down}}},}$

where ${\displaystyle q:=l/g<1}$ and ${\displaystyle r(w):=u(w+g)-u(w)}$.

Since ${\displaystyle 0<q<1}$, the first geometric sum converges as ${\displaystyle m\to \infty }$, while the second diverges to ${\displaystyle \infty }$ as ${\displaystyle n\to \infty }$. Thus, the utility gain from arbitrarily many upward steps remains bounded, whereas the utility loss from sufficiently many downward steps becomes arbitrarily large.

Corollary: Let ${\displaystyle L':=(-ng,mg;{\tfrac {1}{2}},{\tfrac {1}{2}})}$. Under the assumptions of the theorem, if ${\displaystyle q^{m}+q^{-n}\geq 2}$, then the decision maker rejects ${\displaystyle L'}$ at every wealth level ${\displaystyle w\in W}$ such that ${\displaystyle [w-ng,w+mg]\subseteq W}$.

In particular, if ${\displaystyle W=\mathbb {R} }$ and ${\displaystyle q^{-n}\geq 2}$, then the decision maker rejects ${\displaystyle L'}$ for every ${\displaystyle m\in \mathbb {N} ^{+}}$ and at every wealth level. Thus, for a reasonably large fixed loss ${\displaystyle ng}$, the decision maker will reject the lottery no matter how large the possible gain ${\displaystyle mg}$ becomes.

For example, suppose the decision maker rejects the 50–50 lottery ${\displaystyle (-10,11;{\tfrac {1}{2}},{\tfrac {1}{2}})}$ at every wealth level. Then the theorem implies the striking conclusion that the decision maker must also reject every 50–50 lottery of the form ${\displaystyle (-100,G;{\tfrac {1}{2}},{\tfrac {1}{2}})}$, no matter how large the finite gain ${\displaystyle G}$ is.

To see this, here ${\displaystyle q=10/11}$, and ${\displaystyle q^{-8}=(11/10)^{8}>2}$. The corollary therefore implies rejection of ${\displaystyle (-88,11m;{\tfrac {1}{2}},{\tfrac {1}{2}})}$ for every positive integer ${\displaystyle m}$. For any finite ${\displaystyle G}$, choose ${\displaystyle m}$ such that ${\displaystyle 11m\geq G}$. Since ${\displaystyle (-88,11m;{\tfrac {1}{2}},{\tfrac {1}{2}})}$ is better in both outcomes than ${\displaystyle (-100,G;{\tfrac {1}{2}},{\tfrac {1}{2}})}$, monotonicity implies rejection of the latter as well.

Zvi Safra and Uzi Segal extended Rabin's calibration results to any model which has a Gâteaux differentiable utility ${\displaystyle V}$ from lotteries over final wealth. This includes many non-expected-utility models, such as rank-dependent expected utility or Gul's disappointment aversion.^[5]

James C. Cox and others have also shown that any utility representation which is nonlinear in either probabilities or payoffs suffers from similar calibrational problems.^[6]