RISK AVERSION To show that(a)implies(d), note that (16) "1(2()=4(21(t) 2(2(t) which is [strictly] decreasing if(and only if) log ui(x)/u2(x)is. The latter follows from(a)and (17) 1(x) =r2(x)-r1(x) u2(x) That(c)is implied by(e)follows immediately upon writing( 9)in the form (18)1-p(x,h)以(x+h)-u(x) P ( x, h) u (x)-ui x-h) To show that(a)implies(e), integrate(a) from w to x, obtaining (19) n1()[>]-1og"(x)frw<x which is equivalent to (20) 1(x) u2(x) for w<x This implies (21) ≤【<]“20=a2( for wsx<y 2(w) as may be seen by applying the Mean value Theorem of differential calculus to the difference of the two sides of ( 1)regarded as a function of y. Condition(e) follows from(21)upon application of the mean value Theorem to the difference of the reciprocals of the two sides of (e)regarded as a function of w. We have now proved that(a)implies(d)implies(b), and(a) implies(e) imp (c). The equivalence of (ae) will follow if we can prove that(b)implies(a), (c)implies(a), or equivalently that not(a)implies not(b)and not(c). But this fol- lows from what has already been proved, for if the weak [strong] form of (a) doe not hold, then the strong [weak] form of (a) holds on some interval with u and u2 interchanged. Then the strong [weak] forms of (b)and(c)also hold on this interval with u, and u2 interchanged, so the weak [strong] forms of(b)and(c)do not hold This completes the proof. We observe that(e)is equivalent to(20),(21), and ()[>2(m) 1(w)-u1(U) forU<w≤x u2(x)