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

In a previous “helpful” hint pertaining to the first problem in Problem Set 4, I noted (among other things) that expected utility for Investor A is $E(U(W)) = .6{(105 + 25\alpha )^{.5}} + {\rm{ }}.4{(105 - 35\alpha )^{.5}}$.  Since thee square root utility function is itself a function of $\alpha$; i.e., $U(W) = W{(\alpha )^{.5}},$ this means that we must apply the chain rule in order to compute the first order condition and solve for $\alpha$: $\displaystyle\frac{{dU}}{{d\alpha }} = \displaystyle\frac{{dU}}{{dW}}\displaystyle\frac{{dW}}{{d\alpha }} = .5W{(\alpha )^{ - .5}}\displaystyle\frac{{dW}}{{d\alpha }}$.  Applying the chain rule, the first order condition for Investor A is: $\displaystyle\frac{{\partial E(U(W))}}{{\partial \alpha }}{\rm{ }} = {\rm{ }}.3{(105 + 25\alpha )^{ - .5}}(25) - .2{(105 - 35\alpha )^{ - .5}}(35) = 0.$

In order to determine Investor A’s optimal risk exposure, all that remains to be done is to solve this equation for $\alpha$.

Rinse and repeat to determine Investor B’s optimal exposure to risk.  Since Investor B’s utility function is $U(W) = \ln W$, B’s expected utility is $E(U(W)) = .6\ln (105 + 25\alpha ) + .4\ln (105 - 35\alpha)$.  Applying the chain rule to $U(W) = \ln W(\alpha )$, it follows that $\displaystyle\frac{{dU}}{{d\alpha }} = \displaystyle\frac{{dU}}{{dW}}\displaystyle\frac{{dW}}{{d\alpha }} = \displaystyle\frac{1}{{W(\alpha )}}\displaystyle\frac{{dW}}{{d\alpha }}$.  Given these clues, you should be able to determine optimal $\alpha$ for Investor B. However, before you begin, keep in mind that since Investor A’s Arrow-Pratt coefficient is ${R_A}(W) = .5/W$ and Investor B’s Arrow-Pratt coefficient is ${R_A}(W) = 1/W$ (see pages 8, 10, and 12 of http://fin4335.garven.com/fall2019/lecture6.pdf to see how these measures were determined), we know that Investor B is more risk averse than Investor A.  Therefore, Investor B will select a lower level of exposure to risk then Investor B.