Hints for solving problem set #4 (Hint #2)

A student asked me whether it is possible for the value of \alpha in problem set 4 to be negative. Here, \alpha represents the optimal level of exposure to the risky asset; 1-\alpha represents the optimal level of exposure to the safe bond. While it is certainly theoretically possible for \alpha to have a negative value, for this particular problem it turns out that \alpha > 0. The reason \alpha is positive in this case is that the 60/40 probabilities imply a positive expected return on the risky asset which exceeds the expected return on the bond. Thus, even though the stock is risky, a square root utility investor is willing to invest some of her money in the stock because, in an expected utility sense, some positive exposure to risk is worthwhile.

If you obtain a negative value for \alpha, this means that you must have made a math error somewhere. In the case of investor A, E(U(W)) = .6(105 + 25\alpha ).5 + .4(105 – 35\alpha ).5, then one can find the optimal value for investor A’s \alpha by applying the chain rule individually to both the .6(105 + 25\alpha ).5 and the .4(105 – 35\alpha ).5 terms, setting the resulting equation equal to zero (this is the so-called “first order condition”) and solving for \alpha .  Obviously the same logic applies to solving investor B’s problem; the only difference is that E(U(W)) = .6ln(105 + 25\alpha ) + .4ln(105 – 35\alpha ) for investor B.

This raises an interesting question; specifically, what would have to be different about this problem in order to obtain a negative value for \alpha?  If this problem had been originally parameterized such that the expected return on the stock was less than the expected return on the bond, this would guarantee a negative value for \alpha. For example, suppose that everything stayed the same, but that the state probabilities for the stock were 55/45 rather than 60/40. Note that with 55/45 state probabilities, the stock has an expected return of .55(.3) + .45 (-.3) = 3%, which is less than the guaranteed 5% return on the bond.  Under this scenario,  investor A’s optimal \alpha is -48.33%, which implies that she would optimally sell short $48.33 of stock and invest her initial wealth of $100 plus the $48.33 in proceeds from the short sale in the bond. From date 0 to date 1, she would earn 5% or $7.42 on her $148.33 bond investment. At date 1, she would close out her short position by buying the stock back at either $48.33 x (1.30) = $62.83 (in which case she would lose $7.08 on her $100 net investment) or at $48.33 x (.7) = $33.83 (in which case she would gain $21.92 on her $100 net investment). Thus the expected return on her portfolio is .55(-7.08%) + .45(21.92%) = 5.97%, and the standard deviation is 14.43% (short selling is risky because you might get stuck having to close out the short position at a high price; this is why the standard deviation is so high).  On the other hand, investor B’s optimal \alpha is -24% when the state probabilities for the stock are 55/45 rather than 60/40; I will leave it as an exercise for the reader to determine the expected return and risk for investor B’s optimal portfolio under this alternative scenario.

As I noted in my previous Problem Set #4 hint, if you get stuck on the math at all, you might consider inputting the data into an Excel spreadsheet and use Solver to find the optimal value for \alpha.  If you do this, be sure to email your spreadsheet model to fin4335@gmail.com sometime prior to the beginning of next Thursday’s class meeting, at which time I’ll also collect your completed problem sets.

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