User:IssaRice/Two envelopes problem

https://en.wikipedia.org/wiki/Two_envelopes_problem

Step	Wikipedia	Formal interpretation
1	I denote by A the amount in my selected 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.
If A is the smaller amount, then the other envelope contains 2A.
If A is the larger amount, then the other envelope contains A/2.
Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
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 greater than A, so I gain on average by swapping.
After the switch, I can denote that content by B and reason in exactly the same manner as above.
I will conclude that the most rational thing to do is to swap back again.
To be rational, I will thus end up swapping envelopes indefinitely.
As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.