This page is for a short essay on the paper by Falk. 'The Unrelenting Exchange Paradox' from Teaching Statistics Vol 30 No 3 Autumn 2008.
In the paper Ruma Falk discuses the problem described in WP as the Two envelopes problem.
There is some disagreement over whether Falk intends to present an elementary but correct solution to the problem or whether her intention is to show in the simplest possible way where a rather more complex and technical resolution of the paradox lies. My opinion is that her intention is the latter.
My opinion is that Falk intends to show how, through careless thought, the symbol A starts off as an intuitive random variable but slowly morphs into a numerical value which can be used with impunity in an expectation calculation.
To understand this and, in my opinion, to make Falk's point clearer, it is best to consider the simplest version of the problem in which there are two sealed envelopes, one containing £2 and the other £4. A person chooses one uniformly at random and is then given the opportunity, without opening it, to swap it with the other one.
The following argument is put that it is to their advantage to do so and the problem is to find the flaw in this line of reasoning.
1 Denote by A the amount in by selected envelope.
2 The probability that A is the smaller amount is 1/2 and the probability that it is the larger is 1/2.
- This is where Falk says the problem starts. I do not disagree. but the problem is quite subtle (at least to a layman) at this stage.
- A is a random variable, not a fixed number so it could turn out to have one of two different values from the sample space of (£2, £4). If this statement is interpreted to mean, 'The (total) probability that A will turn out to have the value £2 is 1/2 and the probability that A will turn out to have the value £4 is also 1/2', that is fine. The problem is that the statements starts the thought that A might just be a number. It also ignores the fact that it might be a conditional probability that we are interested in.
3 The other envelope may contain either 2A or A/2.
- What is A here?
4 If A is the smaller amount the other envelope contains 2A.
- What is A here?
5 If A is the larger amount the other envelope contains A/2.
- Again, what is A here? The ambiguity now starts to become troublesome. What we are saying is that if we apply the condition that A is £4 then the other envelope contains £2 with certainty.
6 Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7 So the expected value of the money in the other envelope is 1/2 2A + 1/2 A/2 =5/4 A
- As Falk says, this step in illegitimate. It is a misguided attempt to take a short cut in an expectation calculation. The standard way to calculate an expectation is to add up each possible value that A might take multiplied by the probability that A will turn out to have that value. This calculation gives 1/2 £4 + 1/2 £2 = £3. That is exactly the same as the expectation in the envelope you hold, so there is nothing to be gained by swapping.
- Iff the probability that A will turnout to be the larger amount is independent of value that A turns out to be, the above expectation would be valid, or, to put this more bluntly, if the probability that A will be £4 is independent of whether A turns out to be £4 the expectation given in step 7 would be correct.