Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 44 1 c Solution Created 2026-10-03 Updated 2026-10-07
First, equality of two values forces equality of the corresponding values: neither strict comparison holds for , and the same is then true for in both directions. Thus for a well-defined strictly increasing function on . Since is a convex set and are affine functions, is a real interval andIf contains , let and . The mixture identity gives for . For , express as a convex combination of and and solve the same identity for ; for , express as a mixture of and . Hence the formula holds on all of , not merely between the chosen anchors. We obtain positive affine uniqueness of affine preference representations:If is constant, the same comparisons force to be constant; choose and the appropriate . If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.
Under the expected utility hypothesis, use positive affine uniqueness of affine preference representations to normalize the extreme outcomes to and . The stated indifference relations then giveThe two lotteries have expected utilitiesHence , with normalized expected utility difference . Equivalently, substitution reduces them to lotteries with probabilities and of the better extreme outcome. The inference uses the independence axiom for lottery preferences; the three displayed outcome comparisons alone would not determine preferences over all compound lotteries without the expected utility hypothesis.