∫(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

=1/2∫xP(x)(S(x)-S(2x))dx

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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