STOCHASTIC

EFFICIENCY ANALYSIS OF CHOICES INVOLVING RISK 343
the mean-variance criterion becomes a sufficient condition for efficiency. However, by
Theorem 1 and the discussion of Section II, it is evident that even under these assumptions
the (ji, a) is not a necessary condition, since if the two cumulative distributions do not
intersect, one dominates the other for any set of preferences, whether or not the variance
is smaller. But for two such distributions that do intersect, the mean-variance criterion
is equivalent to the condition of Theorem 2, and thus it is both necessary and sufficient
in the face of risk-aversion. This is summarized and formalized in the following theorem.
Theorem 4. Let F(x) and G(y) be two distinct distributions with means ixt and \i2,
and variances o\, o\, respectively, such that F(x) = G(y), for all x and y which satisfy
— = y ^ 2 . Let Hi ^n2, andF(Xi)>G(Xi) for some Xi (i.e., F(x) and G(x) intersect).
Then F dominates G for all concave u(x), if and only ifa\^a\.
That is, F and G belong to the same family of distributions, with two parameters
which are independent functions of n and a 2 , respectively. 1 In that case, Z = is
a
distributed identically with mean 0 and variance 1 (according to some H(Z; 0, 1)), for
both F and G. In these limited cases, when the two distributions F(x) and G(x) intersect,
the mean-variance criterion is both necessary and sufficient for efficiency. The proof is
an immediate application of Theorem 3.
Proof. If at = o2 = a; and iil>fi2; For all x, ^ ± < ^±; thus F(x) g G(x)
a a
for all x, and there is no intersection point. F dominates G by Theorem 1. If ny = ji2,
F and G are identical.
If