5.6: The two envelope problem; non-baysian approach

In this section I devote myself to Fisher statistics. For a Bayesian approach to the two envelope problem, see Chapter 9. 5.6.1 The two envelope problem The following problem is the famous "two envelope problem".
Problem 5.16 [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{5.22} \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" ).

### 5.6.2 Answer: the two envelope problem

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

where the locally compact space $\Omega$ is arbitrary, that is, it may be $\overline{\mathbb R}_+ =\{ \omega \;|\;\mbox{$\omega \ge 0$} \}$ or the one point set $\{ \omega_0 \}$ or $\Omega =\{ 2^n \;|\; n=0, \pm 1, \pm 2, \ldots \}$. Put $X=\overline{\mathbb R}_+ =\{ x \;|\;\mbox{$x \ge 0$} \}$. Consider two continuous (or generally, measurable ) functions $V_1: \Omega \to \overline{\mathbb R}_+$ and $V_2: \Omega \to \overline{\mathbb R}_+$. such that

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

For each $k=1,2$, define the observable ${\mathsf O}_k=(X(=\overline{\mathbb R}_+), {\mathcal F}(={\mathcal B}_{\overline{\mathbb R}_+}:\mbox{the Borel field}), F_k )$ in $L^\infty (\Omega, \nu )$ such that

\begin{align} & \qquad [F_k(\Xi )](\omega )= \left\{\begin{array}{ll} 1 \qquad & (\mbox{ if } V_k(\omega) \in \Xi) \\ 0 \qquad & (\mbox{ if } V_k(\omega) \notin \Xi) \end{array}\right. \\ & (\forall \omega \in \Omega, \forall \Xi \in {\mathcal F} ={\mathcal B}_{\overline{\mathbb R}_+} \mbox{ i.e., the Bore field in $X(=\overline{\mathbb R}_+)$ } ) \end{align}

Furthermore, define the observable ${\mathsf O}=(X, {\mathcal F}, F )$ in $L^\infty (\Omega ,\nu)$ such that

\begin{align} F(\Xi)=\frac{1}{2} \Big( F_1(\Xi)+F_2(\Xi) \Big) \quad (\forall \Xi \in {\mathcal F}) \tag{5.23} \end{align} That is, \begin{align} & \qquad [F(\Xi )](\omega )= \left\{\begin{array}{ll} 1 \qquad & (\mbox{ if } V_1(\omega) \in \Xi, \;\; V_2(\omega) \in \Xi) \\ 1/2 \qquad & (\mbox{ if } V_1(\omega) \in \Xi, \;\; V_2(\omega) \notin \Xi) \\ 1/2 \qquad & (\mbox{ if } V_1(\omega) \notin \Xi, \;\; V_2(\omega) \in \Xi) \\ 0\qquad & (\mbox{ if } V_1(\omega) \notin \Xi, \;\; V_2 (\omega) \notin \Xi) \end{array}\right. \\ & (\forall \omega \in \Omega, \forall \Xi \in {\mathcal F}={\mathcal B}_X \mbox{ i.e., $\Xi$ is a Borel set in $X(=\overline{\mathbb R}_+)$ } ) \end{align}

Fix a state $\omega (\in \Omega )$, which is assumed to be unknown. Consider the measurement ${\mathsf M}_{L^\infty(\Omega, \nu )} ({\mathsf O}=(X, {\mathcal F} , F ), S_{[\omega]})$. Axiom 1 ($\S$2.7) says that

 $(A_1):$ the probability that a measured value $\left\{\begin{array}{ll} V_1(\omega) \\ V_2(\omega) \end{array}\right\}$ is obtained by the measurement $\left.\begin{array}{l} {\mathsf M}_{L^\infty(\Omega, \nu )} ({\mathsf O}=(X, {\mathcal F} , F ), S_{[\omega]}) \\ {} \end{array}\right.$ is given by $\left\{\begin{array}{ll} 1/2 \\ 1/2 \end{array}\right\}$

If you switch to $\left\{\begin{array}{ll} V_2(\omega) \\ V_1(\omega) \end{array}\right\}$, your gain is $\left\{\begin{array}{ll} V_2(\omega) - V_1(\omega)= \omega \\ V_1(\omega) - V_2(\omega)=- \omega \end{array}\right\}$. Therefore, the expectation of switching is \begin{align} (V_2(\omega) - V_1(\omega))/2 + ( V_1(\omega) - V_2(\omega))/2 = 0 \end{align}

That is, it is wrong "The Other Person's envelope is Always Greener".

Remark 5.17 The condition (a) in Problem 5.16 is not needed. This condition plays a role to confuse the essence of the problem.

### 5.6.3 Another answer: the two envelope problem 5.16Problem

For the preparation of the following section ($\S$ 5.6.4), consider the state space $\Omega$ such that

\begin{align} \Omega=\overline{\mathbb R}_+ \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{\mbox{(identification)}}{\longleftrightarrow} (\omega, 2 \omega ) \in \widehat{\Omega} \tag{5.24} \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}

Fix a state $\omega (\in \Omega )$, which is assumed to be unknown. Consider the measurement ${\mathsf M}_{L^\infty (\Omega, \nu )} ({\mathsf O}=(X, {\mathcal F}, F ), S_{[\omega]})$. Axiom 1 ( measurement: $\S$2.7) says that

 $(A_2):$ the probability that a measured value $\left\{\begin{array}{ll} x=V_1(\omega)=\omega \\ x=V_2(\omega ) =2 \omega \end{array}\right\}$ is obtained by ${\mathsf M}_{L^\infty (\Omega, \nu )} ({\mathsf O}=(X, {\mathcal F}, F ), S_{[\omega]})$ is given by $\left\{\begin{array}{ll} 1/2 \\ 1/2 \end{array}\right\}$

If you switch to $\left\{\begin{array}{ll} V_2(\omega) \\ V_1(\omega) \end{array}\right\}$, your gain is $\left\{\begin{array}{ll} V_2(\omega) - V_1(\omega) \\ V_1(\omega) - V_2(\omega) \end{array}\right\}$. Therefore, the expectation of switching is

\begin{align} (V_2(\omega) - V_1(\omega))/2 + ( V_1(\omega) - V_2(\omega))/2 = 0 \end{align}

That is, it is wrong "The Other Person's envelope is Always Greener".

Remark 5.18 The readers should note that Fisher's maximum likelihood method is not used in the two answers ( in $\S$5.6.2 and $\S$5.6.3). If we try to apply Fisher's maximum likelihood method to Problem 5.16 ( Two envelope problem), we get into a dead end. This is shown below.

### 5.6.4 Where do we mistake in (P1) of Problem 5.16?

Now we can answer to the question:
Where do we mistake in (P1) of Problem 5.16?
Let us explain it in what follows. Assume that
 $(a):$ a measured value $\alpha$ is obtained by the measurement ${\mathsf M}_{L^\infty(\Omega, \nu )} ({\mathsf O}=(X, {\mathcal F}, F ), S_{[\ast]})$
Then, we get the likelihood function $f(\alpha, \omega )$ such that \begin{align} f(\alpha, \omega) \equiv \inf_{\omega_1 \in \Omega } \Big[\lim_{\Xi \to \{ x \}, [{F}(\Xi )](\omega_1) \not= 0} \frac{[{F}(\Xi )](\omega)}{ [{F}(\Xi )](\omega_1) } \Big] = \left\{\begin{array}{ll} 1 \quad & (\omega = \alpha/2 \mbox{ or }\alpha ) \\ 0 & \mbox{( elsewhere )} \end{array}\right. \end{align}
Figure 5.10 Two envelope problem

Therefore,Fisher's maximum likelihood method says that

 $(B_1):$ unknown state $[\ast]$ is equal to $\alpha/2$ or $\alpha$

$\Big($ If $[\ast]=\alpha/2$ [resp. $[\ast]=\alpha$ ], then the switching gain is $(\alpha/2-\alpha)$ [resp. $(2\alpha-\alpha)$] $\Big)$. However, Fisher's maximum likelihood method does not say

 $(B_2):$ $\left\{\begin{array}{ll} \mbox{"the probability that$[\ast]=\alpha/2$"=1/2 } \\ \mbox{"the probability that$[\ast]=\alpha$"=1/2} \\ \mbox{"the probability that$[\ast]$is otherwise"=0} \end{array}\right.$

Therefore, we can not calculate ( such as (5.22)):

\begin{align} (\alpha/2-\alpha) \times \frac{1}{2} + (2\alpha-\alpha)\times \frac{1}{2} =1.25 \alpha \end{align}
 $(C_1):$ Thus, the sentence "with probability $1/2$" in [(P1):Why is it paradoxical?] is wrong.

Hence, we can conclude that

 $(C_2):$ If "state space" is specified, there will be no method of a mistake.
since the state space is not declared in [(P1):Why is it paradoxical?].
Remark 5.19 For the Bayesian solution of the two envelope problem 5.16Problem}, see Ch.10.
$\fbox{Note 5.5}$ The readers may think that
 $(\sharp_1):$ the answer of Problem 5.16Problem} is a direct consequence of the fact that the information about A and B is symmetrical (as mentioned in [(P1): Why is it paradoxical?] in Problem 5.16). That is, it suffices to point out the symmetry.
This answer ($\sharp_1$) may not be wrong. But we think that the ($\sharp_1$) is not sufficient. That is because
 $(\sharp_2):$ in the above answer ($\sharp_1$), the problem "What kind of theory (or, language, world view) is used?" is not clear. On the other hand, the answer presented in Section \ref{S5.6.2} is based on {quantum lanuage}.
This is quite important. For example, someone may paradoxically assert that it is impossible to decide "Geocentric model vs. Heliocentrism", since motion is relative. However, we can say, at least, that
 $(\sharp_3):$ Heliocentrism is more handy (than Geocentric model) nunder Newtonian mechanics.
That is, I think that
 $(\sharp_4):$ Geocentric model may not be wrong under Aristotle's world view.
Therefore, I think that the true meaning of the Copernican revolution is \begin{align} { \fbox{Aristotle's world view} \xrightarrow[\mbox{ (the Copernican revolution)}]{} \fbox{Newtonian mechanical world view} \tag{5.25} } \end{align} and not \begin{align} \fbox{Geocentric model} \xrightarrow[\mbox{ (the Copernican revolution)}]{} \fbox{Heliocentrism} \tag{5.26} \end{align} Thus, this (5.26) is merely one of the symbolic events in the Copernican revolution (5.25). The readers should recall my only one assertion in this note, i.e., Figure 1.1 (The history of the world views) in $\S$1.1. Therefore, we assert that the answer in Sec.5.6.2 is recommended The readers will see that the qeustion ($\sharp_2$) is essential to the solution of Zeno's paradox (in Chapter 14).