Dutch Book Argument

This is an leadin example (a.k.a. de Finetti’s betting model) of STAT 775 Bayesian Analysis course.

For more reference, click here (Stanford Encyclopedia of Philosophy).

Suppose we have two actor: Client and Broker, and a series of events \(E_1,E_2,\cdots\) which are uncertain for both of them. These two actors will do transactions based on the events. The rules are as follows:

The broker specifies a collection of ‘previsions’: \(q_1,q_2,\cdots\), and then the client (knowing \(q_i\)) enter the transactions with a set of stakes \(s_1,s_2,\cdots\). For each transaction, the client gives \(q_is_i\) to the broker and the broker gives \(s_i\mathbb{I}(E_i)\) to the client. Our goal is to prevent the “dutch book” meaning that win or lose is not certain.

First, we consider the range of \(q_i\). For any \(q_i<0\) or \(q_i>1\), it will lead to Dutch Book in this transaction. Thus, we first have \(0\leqslant q_i\leqslant 1\).

For each transaction, the client will gain \[ s_i\mathbb{I}(E_i)-q_is_i=s_i(\mathbb{I}(E_i)-q_i). \]

For two disjoint events \(E_1,E_2\), we assume \(E_3=E_1\cup E_2\). Then, after the first three transaction, we know the clinet will gain \[\begin{equation} s_1\mathbb{I}(E_1)+s_2\mathbb{I}(E_2)+s_3\mathbb{I}(E_3)-(q_1s_1+q_2s_2+q_3s_3)\\ = \left\{ \begin{aligned} &s_1+s_3-(q_1s_1+q_2s_2+q_3s_3)=:g_1, \mathrm{if\ } E_1\\ &s_2+s_3-(q_1s_1+q_2s_2+q_3s_3)=:g_2, \mathrm{if\ } E_2\\ &s_1+s_2-(q_1s_1+q_2s_2+q_3s_3)=:g_3, \mathrm{if\ } (E_1\cup E_2)^c \end{aligned} \right. \end{equation}\]

Thus, we have \[ \left( \begin{matrix} 1-q_1&-q_2&1-q_3\\ -q_1&1-q_2&1-q_3\\ 1-q_1&1-q_2&-q_3 \end{matrix} \right) \left( \begin{matrix} s_1\\s_2\\s_3 \end{matrix} \right) = \left( \begin{matrix} g_1\\g_2\\g_3 \end{matrix} \right). \] We name it \(RS=g\). If \(R\) is invertable, we have \(S=R^{-1}g\), which will lead to Dutch Book. So, we need \(R\) singular. We compute the determinant of \(R\), and that is \(\det (R)=q_1+q_2-q_3=0\). It meas that if we need the “coherence” we need \(q_3=q_1+q_2\), which is the axiom of probability.

Related problems (Supplies to the Axioms):

Problem 1

Recall the simple betting model from the class. For an uncertain event \(E\) consider the whole space \(\Omega = E \cup E^c\).

1. Show that if the broker’s prevision \(q(E)\) for \(E\) exceeds unity, then the client can set stakes to guarantee loss to the broker.

2. Show also that if \(q(\Omega)\) is less than unity, then again there is a Dutch book.

Solution:

1. For the transaction based on event \(E\), the client will gain \(s_i(\mathbb{I}(E)-q(E)).\) Since \(q(E)>1\), the client can set \(s_i<0\) so that he is certain to win.

2. For the transaction based on event \(\Omega\), the client will gain \(s_i(\mathbb{I}(\Omega)-q(\Omega))=s_i(1-q(\Omega)).\) Since \(q(\Omega)<1\), the client can set \(s_i>0\) so that he is certain to win.

Problem 2

Again in the simple betting model, suppose that the broker’s previsions \(q(E_i)\) over various events \(E_i\) do satisfy the axioms of probability. Confirm that there is no Dutch book for these previsions. Hint: Let \(G\) be the client’s total gain over a finite set of transactions, and work out distributional properties (e.g. moments) of \(G\) relative to the probability distribution defined by the previsions.

Solution:

Suppose \(E_1,\cdots,E_m\) are disjoint events, and let \(E_{0}:=(\cup_{i=1}^m E_i)^c\). After the \(m\) transaction, the the client will gain \[G_k=s_k-\sum_{i=1}^ms_iq_i,\] if the event \(E_k\) happens, where \(k \in \{1,2,\cdots,m\}\). \[G_0=-\sum_{i=1}^ms_iq_i,\] if \(E_0\) happens.

Thus, the expectation of the money the client gains will be \[ G = \sum_{k=0}^m G_kP(E_k). \] Based on the probability of distribution defined by prevision, we have \[ \begin{aligned} E(G)&=\sum_{k=0}^mG_kq_k\\ &=\sum_{k=1}^m(s_k-\sum_{i=1}^ms_iq_i)q_k+(1-\sum_{k=1}^m q_k)(-\sum_{i=1}^ms_iq_i)\\ &=\sum_{k=1}^ms_kq_k-(\sum_{k=1}^mq_k)(\sum_{i=1}^ms_iq_i)+(1-\sum_{k=1}^m q_k)(-\sum_{i=1}^ms_iq_i)\\ &=0 \end{aligned}. \] Hence, there is no Dutch Book for these previsions.