9.3: St. Petersburg two envelope problem

This section is extracted from the following:

 $(\sharp):$ S. Ishikawa; The two envelopes paradox in non-Bayesian and Bayesian statistics $\quad$ ( arXiv:1408.4916v4 [stat.OT] 2014 )

Now, we will review the St. Petersburg two envelope problem (cf. D.J. Chalmers, "The St. Petersburg Two-Envelope Paradox," Analysis, Vol.62, 155-157, (2002)).

Problem 9.8 [The St. Petersburg two envelope problem] The host presents you with a choice between two envelopes (i.e., Envelope A and Envelope B). You are told that each of them contains an amount determined by the following procedure, performed separately for each envelope:
 $(\sharp):$ a coin was flipped until it came up heads, and if it came up heads on the $k$-th trial, $2^k$ is put into the envelope. This procedure is performed separately for each envelope.
You choose randomly (by a fair coin toss) one envelope. For example, assume that the envelope is Envelope A. And therefore, the host get Envelope $B$. You find $2^m$ dollars in the envelope $A$. Now you are offered the options of keeping A (=your envelope) or switching to B (= host's envelope ). What should you do?

You reason that, before opening the envelopes A and B, the expected values $E(x)$ and $E(y)$ in A and B is infinite respectively. That is because \begin{align} 1 \times \frac{1}{2}+ 2 \times \frac{1}{2^2}+ 2^2 \times \frac{1}{2^3}+ \cdots =\infty \end{align} For any $2^m$, if you knew that A contained $x=2^m$ dollars, then the expected value $E(y)$ in B would still be infinite. Therefore, you should switch to B. But this seems clearly wrong, as your information about A and B is symmetrical. This is the famous St. Petersburg two-envelope paradox (i.e., "The Other Person's Envelope is Always Greener" ).

(P2): St. Petersburg two envelope problem: classical mixed measurement

Define the state space $\Omega$ such that $\Omega=\{\omega= 2^k \; |\; k=1,2, \cdots \}$, with the discrete metric and the counting measure $\nu$. And define the exact observable ${\mathsf O}=(X, {\mathcal F}, F )$ in $L^\infty(\Omega, \nu )$ such that

\begin{align*} & X= \Omega, \quad {\mathcal F}=2^X \equiv \{ \Xi \;|\; \Xi \subseteq X \} \\ & [F(\Xi)](\omega) = \chi_{{}_\Xi} (\omega ) \equiv \begin{cases} 1 \quad & ( \omega \in \Xi ) \\ 0 \quad & (\omega \notin \Xi ) \end{cases} \qquad (\forall \Xi \in {\mathcal F}, \forall \omega \in \Omega ) \end{align*}

Define the mixed state $w$ ($\in L^1_{+1}(\Omega, \nu )$, i.e., the probability density function on $\Omega$) such that

\begin{align*} w_0(\omega) = 2^{-k} \quad (\forall \omega = {2^k} \in \Omega ). \end{align*}

Consider the mixed measurement ${\mathsf M}_{L^\infty (\Omega, \nu )} ({\mathsf O}=(X, {\mathcal F}, F ), {\overline S}_{[\ast]}(w_0) )$. Axiom${}^{{\rm (m)}}$ 1(C$_1$) says that

 $(A):$ the probability that a measured value $2^k$ is obtained by ${\mathsf M}_{L^\infty (\Omega )} ({\mathsf O}=(X, {\mathcal F}, F ),$ ${\overline S}_{[\ast]}(w_0))$ is given by $2^{-k}$.
Therefore, the expectation of the measured value is calculated as follows. \begin{align*} E= \sum_{k=1}^\infty 2^k \cdot 2^{-k}= \infty \end{align*}

Note that you knew that A contained $x=2^m$ dollars (and thus, $E =\infty > 2^m$). There is a reason to consider that the switching to $B$ is an advantage.

Remark 9.9 After you get a measured value $2^m$ from the envelope $A$, you can guess (also see Bayes theorem later) that the probability density function $w_0$ changes to the new $w_1$ such that $w_1 ( 2^m) = 1, w_1 ( 2^k) = 0 ( k \not= m )$. Thus, now your information about $A:w_1$ and $B:w_0$ is not symmetrical (under Bayesian statistics (cf. $\S$9.4)). Hence, in this case, it is true: "The Other Person's envelope is Always Greener".

$\fbox{Note 9.2}$ There are various criterions except the expectaion. For example, consider the criterion such that
 $(\sharp):$ the probability that the switching is disadvantageous" $< \frac{1}{2}$
Under this criterion, it is reasonable to judge that $$\begin{cases} m = 1 &\Longrightarrow \mbox{switching to B} \\ m=2,3,... & \Longrightarrow \mbox{keeping A} \end{cases}$$