# Hints for solving problem set #4 (calculus hint)

In a previous “helpful” hint pertaining to the second problem in Problem Set 4, I noted (among other things) that expected utility for Investor A is E(U(W)) = .6(1,020 + 100x).5 + .4(1,020 – 140x).5, where x corresponds to the proportion of the portfolio that is to be allocated to the risky asset. “Optimal” exposure to risk is determined by maximizing E(U(W)); this is accomplished by differentiating E(U(W)) with respect to x, setting that result equal to 0, and solving for x. The mathematical logic applied here is the same as the approach shown on pp. 14-15 in the Mathematics Tutorial for determining the profit-maximizing production decision of a firm. Here, since the square root utility function is itself a function of x; i.e., U(W(x)), this means that we must apply the chain rule in order to differentiate E(U(W)) with respect to x:

$\displaystyle\frac{{dE(U(W))}}{{dx}} = \frac{{dE(U(W))}}{{dW}}\frac{{dW}}{{dx}} = .5W{(x)^{ - .5}}\frac{{dW}}{{dx}}.$

Since there are two states, this means that the are two state-contingent values for W(x); specifically, 60% of the time, W(x) = 1,020 + 100x , and 40% of the time, W(x) = 1,020 -140x. Once the chain rule has been applied to differentiating both terms on the right-hand side of the E(U(W)) equation for Investor A, set that result (also known as the “first-order condition”) equal to 0 and solve for Investor A’s optimal exposure to the risky asset. Once that proportion has been determined, the allocation to the safe asset is equal to 1-x.

Rinse and repeat to determine Investor B’s optimal exposure to risk. Since Investor B’s utility function is E(U(W)) = .6ln (1,020 + 100x) + .4ln(1,020 – 140x), it follows that the first-order condition is $\displaystyle \frac{{dE(U(W))}}{{dx}} = \frac{{dE(U(W))}}{{dW}}\frac{{dW}}{{dx}} = \frac{1}{{W(x)}}\frac{{dW}}{{dx}}.$ Set Investor B’s first-order condition equal to 0, solve for x and 1-x, and you’re good to go!