This problem requires the von Neumann-Morgenstern theorem to solve.
Let L = {A, B, C, D} be a set of lotteries. You are indifferent between receiving A for sure and a lottery that gives you B with probability 0.9 and C with probability 0.1. You are also indifferent between receiving A for sure and a lottery that gives you B with probability 0.6 and D with probability 0.4. You preferences satisfy the vonn Neumann-Morgenstern axioms.
(i) What do you prefer most, C or D?
(ii) Calculate the (relative) difference in utility between B and C, and between B and D.
(iii) If we stipulate that your utility of B is 1 and your utility of C is 0, what are your utilities of A and D?
I did the problem like this:
For (i), it was obvious that D is preferred to D, because in the lottery with D I was willing to accept a lower chance of receiving B than I was in the lottery with C.
u(A) = 0.9u(B) + 0.1u(C)
u(A) = 0.6u(B) + 0.4u(D)
Therefore,
0.9u(B) + 0.1u(C) = 0.6u(B) + 0.4u(D)
Subtract both sides by 0.6u(B) and get
0.3u(B) + 0.1u(C) = 0.4u(D)
Let u(B) = 1, and u(C) = 0. Therefore, by the above equation:
0.3 = 0.4u(D)
Divide both sides by 0.4 and get
0.75 = u(D)
Now for the answer to (ii), if u(B) = 1, u(C) = 0, and u(D) = 0.75, then
u(B) - u(c) = 1
and
u(B) - u(d) = 0.25
Therefore, the difference between B and C is 4 times the difference between B and D.
Finally, for (iii),
Since u(A) = 0.9u(B) + 0.1u(C)
u(A) = 0.9
We can confirm this by looking at the other equation too:
u(A) = 0.6u(B) + 0.4u(D) = 0.6 + 0.4*0.75 = 0.6 + 0.3 = 0.9
So, u(A) = 0.9, and u(D) = 0.75.
I looked at the answer in the book, and this is exactly what the author put. But then I looked at the errata list for the book on his website (http://www.martinpeterson.org/errata.pdf)
and it says the following:
p. 115,... in the solution to 5.7 all that can be concluded is that 0.9B + 0.1C = 0.6B + 0.4D.
Did I make the same mistake as the author? If so, what is the mistake?


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