User:IssaRice/Two envelopes problem

This page looks at the two envelopes problem, as stated on Wikipedia. The "Wikipedia" column contains the steps given in the fallacious argument, and the "Formal interpretation" column gives the equivalent step translated into the language of probability theory.

Step	Wikipedia	Formal interpretation
1	I denote by A the amount in my selected envelope.	First, let $x>0$ be the amount of money in the lesser envelope, so that $2x$ is the amount in the greater envelope. We need to define a sample space. Let $\Omega =\{(e_{1},c_{1}),(e_{1},c_{2}),(e_{2},c_{1}),(e_{2},c_{2})\}$ be a set. The intended interpretation here is that $(e_{i},c_{j})$ is the world in which envelope $i$ contains the larger amount ($2x$) and we chose envelope $j$. Since all choices are made randomly, we can assume a uniform distribution, i.e. $\Pr((e_{i},c_{j}))=1/2$ for all $i,j$. We need to decide what sort of object $A$ is. Since later on, we start talking about hypotheticals (e.g. "If A is the smaller amount") it makes sense to let $A:\Omega \to \mathbf {R}$ be a random variable. We define $A$ as $A((e_{i},c_{j}))={\begin{cases}2x&{\text{if }}i=j\\x&{\text{if }}i\neq j\end{cases}}$ For instance, if the larger amount was in envelope 1 and we chose envelope 2, then we are in the world $\omega =(e_{1},c_{2})$ so $A(\omega )=x$ as expected.
2	The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.	We can interpret this as saying that $\Pr(A=x)=1/2$ and $\Pr(A=2x)=1/2$. Both of these are true, e.g. $\Pr(A=x)=\Pr(\{\omega :A(\omega )=x\})=\Pr(\{(e_{1},c_{2}),(e_{2},c_{1})\})=1/2$.
3	The other envelope may contain either 2A or A/2.	Here it helps to introduce a random variable tracking the amount in the unchosen envelope. Call it $B:\Omega \to \mathbf {R}$. We have $B((e_{i},c_{j}))={\begin{cases}x&{\text{if }}i=j\\2x&{\text{if }}i\neq j\end{cases}}$ Now the claim from Wikipedia is saying that either $B=2A$ or $B=A/2$. Since this is an equality between random variables, things are starting to get a little tricky. Formally, $B=2A$ is an event; in fact, it is the set $\{\omega :B(\omega )=2A(\omega )\}=\{(e_{1},c_{2}),(e_{2},c_{1})\}$ (for both of those pairs, $B(\omega )=2x$ and $A(\omega )=x$ so $2x=2x$ indeed). Similarly, $B=A/2$ is the set $\{(e_{1},c_{1}),(e_{2},c_{2})\}$. Now, "either $B=2A$ or $B=A/2$" means that $B=2A$ and $B=A/2$ (interpreted as events) include all the possible worlds, i.e. that $(B=2A)\cup (B=A/2)=\Omega$. This is indeed the case.
4	If A is the smaller amount, then the other envelope contains 2A.	This is saying that "if $A=x$ then $B=2A$". What could that possibly mean? Like the previous step, we have to interpret this one in terms of events. This time, the "if ... then ..." means we are talking about set inclusion. In other words, the claim is that $(A=x)\subseteq (B=2A)$. Well, $A=x$ is the event $\{(e_{1},c_{2}),(e_{2},c_{1})\}$ and $B=2A$ is also the event $\{(e_{1},c_{2}),(e_{2},c_{1})\}$. So in fact $(A=x)=(B=2A)$.
5	If A is the larger amount, then the other envelope contains A/2.	Similar to the previous step. This is saying that $(A=2x)\subseteq (B=A/2)$. This is true, and in fact equality holds.
6	Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.	This is saying that $\Pr(B=2A)=1/2$ and $\Pr(B=A/2)=1/2$. Both statements are true.
7	So the expected value of the money in the other envelope is: ${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$	This is where things start to go downhill. On the surface, the statement is claiming that ${\textstyle \mathbf {E} [B]={\frac {5}{4}}A}$. But this is nonsense, since the left-hand side is a real number and the right-hand side is a random variable! In other words, not even our types match. Is there some way to rescue this statement by interpreting it a different way? The left-hand side looks like $\Pr(B=2A)\cdot B+\Pr(B=A/2)\cdot B$. In fact, a valid way to calculate $\mathbf {E} [B]$ is to partition $\Omega$ into subsets on which $B$ takes a constant value (this is a very special case of the law of total expectation). So we have $\mathbf {E} [B]=\Pr(B=2A)\cdot B_{B=2A}+\Pr(B=A/2)\cdot B_{B=A/2}$ where $B_{B=2A}\in \mathbf {R}$ is the value that $B$ takes on the set $B=2A$ (i.e. the value $2x$) and $B_{B=A/2}\in \mathbf {R}$ is the value that $B$ takes on the set $B=A/2$ (i.e. the value $x$). So we have ${\textstyle \mathbf {E} [B]={\frac {1}{2}}\cdot 2x+{\frac {1}{2}}\cdot x={\frac {3}{2}}x}$. But now, the Wikipedia column tries to replace $B_{B=2A}$ and $B_{B=A/2}$ by the equivalent values for $A$, i.e. by $2A_{B=2A}$ and $A_{B=A/2}/2$. This is fine, since $A_{B=2A}=x$ (hence $2A_{B=2A}=2x=B_{B=2A}$) and $A_{B=A/2}=2x$ (hence $A_{B=A/2}/2=2x/2=x=B_{B=A/2}$). So finally, what the Wikipedia column wants to do is: ${\begin{aligned}\mathbf {E} [B]&=\Pr(B=2A)\cdot B_{B=2A}+\Pr(B=A/2)\cdot B_{B=A/2}\\&={\frac {1}{2}}\cdot 2A_{B=2A}+{\frac {1}{2}}\cdot A_{B=A/2}/2\end{aligned}}$ Now the problem is apparent! We have $A_{B=2A}=2x\neq x=A_{B=A/2}$. So we are not allowed to combine these into a single "$A$".
8	This is greater than A, so I gain on average by swapping.
9	After the switch, I can denote that content by B and reason in exactly the same manner as above.
10	I will conclude that the most rational thing to do is to swap back again.
11	To be rational, I will thus end up swapping envelopes indefinitely.
12	As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.