Keep The Right One

Alright, time to finish off the stuff I've been writing about the two envelope paradox. The analysis last time showed that we had best assume that the smaller cheque is coming from a normalized distribution function P(x) with a finite expectation value (∫ xP(x)dx converges). Now, we must come up with a strategy on wether to keep the cheque or not, and this strategy can only be based on the value of the cheque that we see (known as A). Let us choose to accept the switch when cheque A has a value x with probability S(x). S(x) must be a function from the positive reals to [0,1], but is otherwise arbitrary. Now, we know that the expected gain from switching is E(B|A=x)-x (from last time), times Q(x)=(P(x)+P(x/2)/2)/2 (the distribution function on the cheque A), times S(x) (the chance we actually take the switch). So, the expected gain is
∫(E(B|A=x)-x) Q(x) S(x) dx
=1/2 ∫(xP(x)-x/4P(x/2))S(x) dx
=1/2 ∫xP(x)S(x)dx-1/2∫x/2P(x/2)S(x)d(x/2)

The split into two integrals is valid this time as ∫ xP(x)dx converges and the bounded function S(x) is not going to make that any worse. Now, change variables x/2 -> x in the second integral and recombine to get

So we see that as long as S(x)-S(2x) is positive, we will have a positive expected gain for any distribution function P(x). Specifically you may choose any decreasing function to get this effect, like S(x)=1-tanh(x). In effect, this means that you are more likely to keep a larger valued cheque than a smaller cheque. This sort of solution also solves a similar game where I write down any two random numbers (from some probability distribution) on cheques and offer you a "keep or switch" scenario.

Anyway, one can also find that the optimal choice of S(x) is to set it to 1 whenever P(x)-1/4P(x/2)>0 and 0 whenever P(x)-1/4P(x/2)<0 (when P(x)-1/4P(x/2)=0, choose S(x) arbitrarily, it doesn't matter), however this assumes that you know the function P.

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