Problem set #4 (which is due on Thursday, September 24) consists of two problems: 1) a problem that incorporates both expected utility and stochastic dominance, and 2) an optimal (expected utility-maximizing) portfolio problem. We’ll discuss stochastic dominance in class on Tuesday, but in the meantime, here are some hints for setting up the second problem.
The second problem involves determining how to (optimally) allocate initial wealth W0 = $1,000 to (risky) stock and (safe) bond investments for two investors who are identical in all respects except utility. Let x represent the allocation to stock; then the plan is to invest $1,000x in the stock and $1,000(1-x) in the bond. The key here is to find the value for x which maximizes expected utility. The problem is based on the following facts:
- U(W) = W.5; for Investor A and U(W) = ln W for Investor B;
- W0 = $1,000 for both investors;
- Current bond and stock prices are B0 and S0 respectively;
- End-of-period bond price is B1 = B0(1.02) with probability 1.0; and
- End-of-period stock price is S1 = S0(1.12) with probability .6 and S1 = S0(.88) with probability .4.
In order to compute expected utility of wealth for either investor, you must first determine state-contingent wealth (Ws). Since there is a 60% chance that the stock increases in value by 12%, a 40% chance that the stock decreases in value by 12%, and a 100% chance that the bond increases in value by 2%, this implies the following:
- 60% of the time, Ws = xW0(1.12) + (1-x)W0(1.02) = x1,000(1.12) + (1-x)1,000(1.02) = x1,120 + (1-x)1,020 = 1,020 + 100x.
- 40% of the time, Ws = xW0(.88) + (1-x)W0(1.02) = x1,000(.88) + (1-x)1,000(1.02) = x880 + (1-x)1,020 = 1,020 – 140x.
Therefore, expected utility for Investor A is: E(U(W)) = .6(1,020 + 100x).5 + .4(1,020 – 140x).5, and expected utility for Investor B is E(U(W)) = .6ln (1,020 + 100x) + .4ln(1,020 – 140x). There are two ways to solve for the optimal value of x for each investor – via calculus (applying the power rule and the chain rule) or a spreadsheet model; either approach suffices.