Problem 9.14 [ (= Problem5.16): the two envelope problem]

The host presents you with a choice between two envelopes (i.e., Envelope A and Envelope B). You know one envelope contains twice as much money as the other, but you do not know which contains more. That is, Envelope A [resp. Envelope B] contains $V_1$ dollars [resp. $V_2$ dollars]. You know that
 $(a):$ $\qquad \frac{V_1}{V_2}=1/2$ or, $\frac{V_1}{V_2}=2$
Define the exchanging map $\overline{x}: \{V_1, V_2 \} \to \{V_1, V_2\}$ by \begin{align} \overline{x} = \left\{\begin{array}{ll} V_2,\;\; (\mbox{ if } x=V_1), \\ V_1 \;\; (\mbox{ if } x=V_2) \end{array}\right. \end{align}

You choose randomly (by a fair coin toss) one envelope, and you get $x_1$ dollars (i.e., if you choose Envelope A [resp. Envelope B], you get $V_1$ dollars [resp. $V_2$ dollars] ). And the host gets $\overline{x}_1$ dollars. Thus, you can infer that $\overline{x}_1=2x_1$ or $\overline{x}_1=x_1/2$. Now the host says "You are offered the options of keeping your ${x}_1$ or switching to my $\overline{x}_1$". What should you do?

[(P1):Why is it paradoxical?]. You get $\alpha= x_1$. Then, you reason that, with probability 1/2, $\overline{x}_1$ is equal to either $\alpha/2$ or $2\alpha$ dollars. Thus the expected value (denoted $E_{\mbox{ other}}(\alpha)$ at this moment) of the other envelope is \begin{align} E_{\mbox{ other}} (\alpha)=(1/2)(\alpha/2) + (1/2)(2\alpha)=1.25\alpha \tag{9.18} \end{align} This is greater than the $\alpha$ in your current envelope $A$. Therefore, you should switch to B. But this seems clearly wrong, as your information about A and B is symmetrical. This is the famous two-envelope paradox (i.e., "The Other Person's Envelope is Always Greener" ).

9.5.1:(P1): Bayesian approach to the two envelope problem

Consider the state space $\Omega$ such that

\begin{align} \Omega=\overline{\mathbb R}_+ (=\{ \omega \in {\mathbb R} \;|\; \omega \ge 0 \}) \end{align}

with Lebesgue measure $\nu$. Thus, we start from the classical basic structure

\begin{align} [ C_0(\Omega ) \subseteq L^\infty ( \Omega , \nu ) \subseteq B(L^2 ( \Omega , \nu ) ) ] \end{align}

Also, putting $\widehat{\Omega}=\{ (\omega, 2 \omega ) \;| \; \omega \in \overline{\mathbb R}_+ \}$, we consider the identification:

\begin{align} \Omega \ni \omega \underset{\scriptsize{\mbox{(identification)}}}{\longleftrightarrow} (\omega, 2 \omega ) \in \widehat{\Omega} \tag{9.19} \end{align}

Furthermore, define $V_1:\Omega (\equiv \overline{\mathbb R}_+) \to X(\equiv \overline{\mathbb R}_+)$ and $V_2:\Omega (\equiv \overline{\mathbb R}_+) \to X(\equiv \overline{\mathbb R}_+)$ such that

\begin{align} V_1(\omega ) =\omega , \quad V_2(\omega ) = 2 \omega \qquad (\forall \omega \in \Omega) \end{align}

And define the observable ${\mathsf O}=(X(=\overline{\mathbb R}_+), {\mathcal F}(={\mathcal B}_{\overline{\mathbb R}_+}:\mbox{ the Borel field}), F )$ in $L^\infty (\Omega, \nu )$ such that

\begin{align} & \qquad [F(\Xi )](\omega )= \left\{\begin{array}{ll} 1 \qquad & (\mbox{ if } \omega \in \Xi, \;\; 2 \omega \in \Xi) \\ 1/2 \qquad & (\mbox{ if } \omega \in \Xi, \;\; 2 \omega \notin \Xi) \\ 1/2 \qquad & (\mbox{ if } \omega \notin \Xi, \;\; 2 \omega \in \Xi) \\ 0 \qquad & (\mbox{ if } \omega \notin \Xi, \;\; 2 \omega \notin \Xi) \end{array}\right. \qquad (\forall \omega \in \Omega, \forall \Xi \in {\mathcal F} ) \end{align}

Recalling the identification : $\widehat{\Omega} \ni (\omega, 2\omega ) \longleftrightarrow \omega \in \Omega =\overline{\mathbb R}_+$, assume that

\begin{align} \rho_0(D) =\int_D w_0(\omega ) d \omega \quad (\forall D \in {\mathcal B}_{\Omega }={\mathcal B}_{\overline{\mathbb R}_+ }) \end{align}

where the probability density function $w_0: \Omega ( \approx \overline{\mathbb R}_+ ) \to \overline{\mathbb R}_+$ is assumed to be continuous positive function. That is, the mixed state $\rho_0 (\in {\mathcal M}^m(\Omega(=\overline{\mathbb R}_+ ) ) )$ has the probability density function $w_0$.

Axiom${}^{{ (m)}}$ 1 ( in $\S$9.1) says that

 $(A_1):$ The probability $P(\Xi)$ $(\Xi \in {\mathcal B}_X ={\mathcal B}_{\overline{\mathbb R}_+ })$ that a measured value obtained by the mixed measurement ${\mathsf M}_{L^\infty (\Omega, d \omega )} ({\mathsf O}=(X, {\mathcal F}, F ), S_{[\ast]}(\rho_0))$ belongs to $\Xi (\in {\mathcal B}_X ={\mathcal B}_{\overline{\mathbb R}_+ })$ is given by \begin{align} P (\Xi ) & = \int_\Omega [F(\Xi )](\omega ) \rho_0 (d \omega ) = \int_\Omega [F(\Xi )](\omega ) w_0 (\omega ) d \omega \nonumber \\ & = \int_{\Xi} \frac{w_0(x/2 )}{4} + \frac{w_0(x )}{2} \;\; d x \quad (\forall \Xi \in {\mathcal B_{\overline{\mathbb R}_+ }}) \tag{9.20} \end{align}

Therefore, the expectation is given by

\begin{align} \int_{\overline{\mathbb R}_+} x P(d x ) = \frac{1}{2} \int_{0}^\infty x \cdot \Big( w_0(x/2 )/2 + w_0(x ) \Big) d x = \frac{3}{2} \int_{\overline{\mathbb R}_+} x w_0(x ) d x \tag{9.21} \end{align}

Furthermore, Theorem 9.11 ( Bayes' theorem ) says that

 $(A_2):$ When a measured value $\alpha$ is obtained by the mixed measurement ${\mathsf M}_{L^\infty (\Omega, d \omega )} ({\mathsf O}=(X, {\mathcal F}, F ),$ $S_{[\ast]}(\rho_0))$, then the post-state $\rho_{\mbox{ post}} (\in {\mathcal M}^m (\Omega ))$ is given by \begin{align} \rho_{\mbox{ post}}^\alpha = \frac{ \frac{w_0(\alpha/2)}{2} } {\frac{h(\alpha/2)}{2} + w_0(\alpha) } \delta_{(\frac{\alpha}{2}, \alpha)} + \frac{ w_0(\alpha) } {\frac{w_0(\alpha/2)}{2} + w_0(\alpha) } \delta_{({\alpha}{},2 \alpha)} \tag{9.22} \end{align}
Hence,
 $(A_3):$ if $[\ast] =$ $\left\{\begin{array}{ll} \delta_{(\frac{\alpha}{2}, \alpha)} \\ \delta_{({\alpha}{}, 2 \alpha)} \end{array}\right\}$, then you change $\left\{\begin{array}{ll} \alpha \longrightarrow \frac{\alpha}{2} \\ \alpha \longrightarrow 2{\alpha} \end{array}\right\}$, and thus you get the switching gain $\left\{\begin{array}{ll} \frac{\alpha}{2} - \alpha (= - \frac{\alpha}{2} ) \\ 2{\alpha} - \alpha (= {\alpha}) \end{array}\right\}$.

Therefore, the expectation of the switching gain is calculated as follows:

\begin{align} & \int_{\overline{\mathbb R}_+} \Big( (-\frac{\alpha}{2}) \frac{ \frac{w_0(\alpha/2)}{2} } {\frac{w_0(\alpha/2)}{2} + w_0(\alpha) } + \alpha \frac{ w_0(\alpha) } {\frac{w_0(\alpha/2)}{2} + w_0(\alpha) } \Big) P(d \alpha ) \nonumber \\ = & \int_{\overline{\mathbb R}_+} (-\frac{\alpha}{2})\frac{w_0(\alpha/2 )}{4} + \alpha \cdot \frac{w_0(\alpha )}{2} \;\; d \alpha =0 \tag{9.23} \end{align}

Therefore, we see that the swapping is even, i.e., no advantage and no disadvantage.

$\square \quad$