Wanting Too Much

Alright, time to go into a bit of detail about the two envelope paradox. Much of this analysis I actually originally read on a paper by David Chalmers. Anyway, the first thing to realize is that this paradox can show up even if the amount of money in the first envelope is taken from a well defined probability distribution. That is, suppose the smaller cheque has value between x and x+dx with probability P(x)dx, and the larger cheque is exactly twice the smaller one. Now, we have to ask the following thing: given that we have opened an envelope with value between y and y+dy, what are the odds that it is the smaller cheque? Well, if it is the smaller cheque, then there was a P(y)dy chance of it being written down, but if it is the larger cheque, then there was a P(y/2)d(y/2)=1/2 P(y/2)dy chance of it happening. Thus the ratio of probabilities is 2P(y):P(y/2), though one might naively have expected P(y):P(y/2). In particular, if the first cheque is randomly chosen from 0 to 10 with uniform distribution, and you see a cheque of value 2, then it is twice as likely that the other cheque has 4 than it has 1. This may seem nonintuitive, but this must occur in order to balance out the case that if you see a cheque with value over 10 and then the other cheque must be smaller. Note that this actually will make the initial paradox somewhat worse.
Now we can see that if the distribution function for the smaller cheque is P(x), then the distribution function for a random cheque will be Q(x)=(P(x)+P(x/2)/2)/2. The relative ratio between the two terms was derived last paragraph, and the overall factor of 1/2 comes from normalizing Q(x) assuming P(x) already was properly normalized. Let the first cheque be given a value A and the unknown cheque a value B (B=2A or B=A/2 always). Let E() be the expectation value operator, so that E(B) denotes the expectation value of B, and E(B|A=x) will denote the expectation value of B given that A is the specific value x. The paradox will be at its strongest if E(B|A=x) is greater than x for all values of x (and the paradox will still exist if that had been true on average). Now lets calculate how much money we expect to get by switching blindly without even looking at the value of the first cheque by integrating over the distribution function for the first cheque Q(A). The specific value of E(B|A=x) is (2xP(x)+x/2P(x/2)/2)/(P(x)+P(x/2)/2) being (value if B is larger)(chance B is larger)+(value if B is smaller)(chance B is smaller) divided by normalization, so E(B|A=x) = x/2(8P(x)+P(x/2))/(2P(x)+P(x/2)). Now, the expected gain for switching cheques blindly is:
∫ (E(B|A=x)-x) Q(x)dx
=∫ (x/2(8P(x)+P(x/2))/(2P(x)+P(x/2))-x)(2P(x)+P(x/2))/4 dx
=∫ (x(4P(x)+P(x/2)/2-2P(x)-P(x/2))/4) dx
=1/2 ∫ (xP(x)-x/4P(x/2)) dx

Next is the trick that I really appreciate about this entire thing, assuming that the integral ∫ xP(x) dx converges, we may break this up into two integrals to find
=1/2 ∫ xP(x)dx-1/2 ∫ x/2P(x/2)d(x/2)

Which is simply zero. Note that if P(x) was something like 1/x^2 from 1 to infinity, it will be a properly normalized distribution function, but the paradox will still arise. In order to avoid the paradox, the first cheque needs to have a finite expectation value, otherwise the paradox that E(B) is greater than E(A) isn't really a problem.
More simply put, given infinite expectations, any finite value is disappointing.
Anyway, back to the puzzle I posted last time, you may further assume that the distribution function that the cheques are chosen from have finite expectation value, there is still a strategy that works.

No comments: