MMP.19:5.3 - Bound benefit without inventing a joint law

Suppose binary success probabilities under actions 1 and 0 are p1=7/10 and p0=2/5. Let b=P(Y_0=0,Y_1=1), the probability of succeeding only under action 1. The four type masses must be

P(0,1)=b                 P(1,1)=p1-b
P(1,0)=p0-p1+b           P(0,0)=1-p0-b.

Nonnegativity gives max(0,p1-p0) <= b <= min(p1,1-p0), hence 3/10 <= b <= 3/5. At b=3/10 the masses in order (00,01,10,11) are (3/10,3/10,0,2/5); at b=3/5 they are (0,3/5,3/10,1/10). Both attain the supplied marginals, establishing the bounds for this unrestricted response-type class.

Now suppose assignment A is independent of the response pair (Y_0,Y_1), P(A=1)>0, and a case is observed with A=1,Y=1. The probability that this case would fail under action 0 is P(Y_0=0 given A=1,Y=1)=b/p1, hence between 3/7 and 6/7. Conditioning retains the types with Y_1=1; the independent assignment probability cancels from numerator and denominator. The same endpoint distributions attain these conditional bounds, since their denominator is the fixed positive p1=7/10.

The mean effect is p1-p0=3/10 in every compatible model. An expected-success criterion with an action-1 cost of 1/10 therefore has net gain 1/5 without identifying b. A criterion that explicitly requires b to exceed 2/5 remains unsettled by these bounds.

If subject knowledge warrants that action 1 never changes a success into failure, P(1,0)=0 fixes b=3/10. That is an additional monotonicity assumption; it was not learned from the two marginal probabilities.