F 0 if z

Then F(l/2) = p > 0 = Gil/2) and JxdF(x) « 2(1 - p) > I = JxdG(x). Hence F( ♦) does not first-order stochastically dominate G(0, but the mean of F( •) is larger than that of G(0.

6.D.3 Any elementary increase in risk from a distribution F( -) is a mean-preserving spread of F(0. in Example 6.D.2, we saw that any mean-preserving spread of F( -) is second-order stochastically dominated by F(0. Hence the assertion follows.

(a) By a direct calculation, the means of L and L' are 2 - p^ + p^ and 2 ~ pj + p^- Thus the two lotteries have an equal mean if and only if p^ - p0 = p^ -p^. Hence they have an equal mean if and only if they are both on a segment that is parallel to the segment connecting the $2-vertex and the middle point of the ($l,$3)-face, as depicted below:

(b) If the decision-maker exhibits risk aversion, then he prefers getting $2 with probability one to the lottery yielding $1 with probability 1/2 and S3 with probability 1/2. Hence the indifference lines are steeper than the segment connecting the $2-vertex and the middle point of the (Si,S3)-face. Hence, when L and L1 have an equal mean, L is preferred to L' if and only if L" is located on the right of L'. Therefore, L second-order stochasticall-y dominates Ly if and only if L is located on the right of L\ as depicted in

the figure below:

Was this article helpful?

0 0

Post a comment