The exchange paradox is a paradox within the probability theory and the decision theory which demonstrates problems that can arise when working with expected values. It can appear in different forms and hence under different names. Often it is presented in a form where two envelopes are used and it is then called the two-envelopes paradox.

The setup

The setup can appear in different forms using e.g. envelopes, boxes, and wallets. Using envelopes, it can be described as follows: In a game a player is presented two envelopes containing money. He is told that one envelope contains twice as much money as the other envelope, but he does not know which one contains the larger amount. The player then may pick one envelope at will, and after he has made a decision, he is offered to exchange his envelope with the other envelope.

The problem

Assuming that the player wants to maximise his gain, the crucial question is:
* Should the player swap, that is, exchange the envelopes?

Usually the answer to this question is made dependent of two circumstances, namely:
# Is the player allowed to see what's inside the envelope he has selected at first?
# How is the money in the envelopes selected, i.e., what is the probability distribution?

Discussion

The setup can be separated in two random events:
# Selection of the amounts of money in the envelopes.
# Selection of one of the two envelopes.

In the following it is assumed that the selection of the amounts of money in the envelopes is realised as follows: A random number "X" is chosen with some for now unspecified probability distribution. This amount of some currency units is put in one envelope in a symbolic way by writing it on a sheet of paper. Then the amount 2"X" is put in the other evelope in the same way. When the player selects one envelope, it is assumed that he selects the amounts "X" or 2"X" with a probability of 50 % each. The content of this selected envelope is noted as "A", the content of the other envelope as "B". If he selects the envelope with the amount "X" at first, this event is noted as ("A", "B") = ("X", 2"X"), if he selects the envelope with the amount 2"X" at first, this event is noted as ("A", "B") = (2"X", "X").

Case 1: Envelope not opened

In this case the expected values for "A" and "B" can be calculated as follows: For each value of "X" both envelopes contain either "X" or 2"X" with a probability of 0.5, hence:$operatorname E\left(A\right) = operatorname E\left(B\right) = 0.5 cdot X + 0.5 cdot 2X = 1.5,X ,$so swapping makes no difference Eric Schwitzgebel and Josh Dever. [http://www.faculty.ucr.edu/~eschwitz/SchwitzPapers/TwoEnvelope060707.pdf "The Two Envelope Paradox and Using Variables Within the Expectation Formula"] . 2006-07-07.] [Priest and Restall. [http://consequently.org/papers/envelopes.pdf "Envelopes and Indifference"] . 2003-02-15.] .

Case 2: Envelope opened

If the player opens his envelope, he knows the value of "A" = "a". Hence, the amount in the other envelope must be either "B" = 2"a" or "B" = "a"/2. Note here that "a" should not be zero otherwise swapping is easily the best choice. Now swapping would be reasonable if the expected value of "B" was greater than "a". It can be calculated using conditional probabilities as follows:

:$operatorname E\left(B|A=a\right) = P\left(B=2a|A=a\right)cdot 2a + P\left(B=a/2|A=a\right)cdot a/2$

The probabilities in this formula depend on the distribution of "X" as follows:

:$P\left(B=2a|A=a\right) = P\left(X=a|A=a\right)=frac\left\{P\left(A=a|X=a\right)P\left(X=a\right)\right\}\left\{P\left(A=a\right)\right\} = frac\left\{frac\left\{1\right\}\left\{2\right\}P\left(X=a\right)\right\}\left\{P\left(A=a\right)\right\}.$

The event "A" = "a" can occur if "X" = "a" and the player has selected the envelope with the smaller amount or if "X" = "a/2" and the player has selected the envelope with the larger amount, each with a probability of 50 %, hence:

:$P\left(A=a\right) = P\left(A=a|X=a\right)P\left(X=a\right) + P\left(A=a|X=a/2\right)P\left(X=a/2\right) = frac\left\{1\right\}\left\{2\right\} left\left(P\left(X=a\right) + P\left(X=a/2\right) ight\right).$

Inserting this result in the previous equation yields:

:$P\left(B=2a|A=a\right) = frac\left\{P\left(X=a\right)\right\}\left\{P\left(X=a\right)+P\left(X=a/2\right)\right\}.$

The conditional probability $P\left(B=a/2|A=a\right)$ can be calculated similarly, or alternatively by utilising the fact that this probability and the previous must sum up to 1:

:$P\left(B=a/2|A=a\right) = 1 - P\left(B=2a|A=a\right) = frac\left\{P\left(X=a/2\right)\right\}\left\{P\left(X=a\right)+P\left(X=a/2\right)\right\}.$

With this probabilities it follows that

:$operatorname E\left(B|A=a\right) = P\left(B=2a|A=a\right)cdot 2a ,+, \left(1-P\left(B=2a|A=a\right)\right)cdot a/2 = frac\left\{1+3P\left(B=2a|A=a\right)\right\}\left\{2\right\} cdot a ,.$

Now it can be seen that for the case "A" = "a" the expected value of the other envelope is larger than "a" if $1 + 3P\left(B=2a|A=a\right) > 2$ or equivalently

:$P\left(B=2a|A=a\right) > 1/3,.$

This implies for the distribution of "X" that the player should swap an envelope containing the amount "a" if

:$P\left(X=a\right) > frac\left\{1\right\}\left\{2\right\} P\left(X=a/2\right).$

For a finite discrete distribution of "X" this constraint is not sufficient for the maximal value of "a", as in this case the other envelope must contain "a"/2. For the minimal value of "a", the other envelope must contain 2"a". However, there are infinite distributions of "X" for which the above constraint is always fulfilled, e.g. the discrete distribution [John Norton. [http://www.pitt.edu/~jdnorton/papers/Exchange_paradox.pdf "When the sum of our expectations fails us: The exchange paradox"] . Pacific Philosophical Quarterly 79, 1998, p. 38, eq. 5.]

:$p_n = P\left(X=2^n\right) = k p_\left\{n-1\right\} = \left(1-k\right)k^n , quad 0.5 < k < 1 , n ge 0 ,.$

In this case, it seems that the player should swap for any amount "a" he finds in his envelope, i.e. always, so he actually doesn't have to look inside his envelope, which is a contradiction to the above discussion of that case, so this problem is considered a paradox.

References

Bibliography

* Ronald Christensen and Jessica Utts, "Bayesian solution of the "Exchange Paradox". "American Statistician", Vol. 46 (1992) pp 274-276.

*Two envelopes problem

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