Information about tomorrow’s midterm exam in Finance 4335

Tomorrow afternoon’s midterm exam in Finance 4335 consists of four problems. Since your exam grade will be based upon the three
highest scoring problems of these four, feel free to either work all four problems or just three of the four problems. Each problem is worth 32 points; thus three problems times 32 points each totals 96
points. I will add an additional 4 points on your exam if you also legibly write your name on the cover page in the space provided. Thus, the maximum number of points possible on this exam is 100 points.  Furthermore, I have also posted the formula sheet for Midterm Exam 1, which will be included as part of the exam booklet. I recommend that y’all familiarize yourselves with this document sometime prior to tomorrow’s exam.

I’d like to make an important point about the formulas provided on the formula sheet.  What you’ll find there is not a complete census of all formulas used to date in Finance 4335.  For example, I don’t include any insurance pricing formulas.  By now, I assume that everyone knows that an actuarially fair price for an insurance policy is simply the expected value of the claim/indemnity under said policy.  On the other hand, an actuarially “unfair” policy has a “premium loading” which represents a “markup” from the actuarially fair price.  Also, I recommend when performing expected utility calculations, I recommend that you go out to no less than the 3rd digit to the right of the decimal point.

Unfortunately, I cannot be at the exam tomorrow due to an important medical issue affecting a close member of my family.  Professor Paul Anderson has graciously agreed to proctor tomorrow’s exam in my place.

See y’all next week!

Dr. Garven

Solutions for Fall 2018 Finance 4335 Midterm 1 exam and for Problem Set 5

I have posted the solutions for the Fall 2018 Finance 4335 Midterm 1 exam; the direct link is http://fin4335.garven.com/spring2019/midterm1_fall2018_solutions; also linked from the sample exams webpage located at http://fin4335.garven.com/sample-exams/.

I have also posted the solutions for Problem Set 5 at the following location: http://fin4335.garven.com/spring2019/ps5solutions.

I’ll post the formula page for Thursday’s midterm exam sometime tomorrow; I’ll notify y’all about this via the blog the moment it gets posted.

Finance 4335 course synopsis for Midterm Exam 1

Next week (2/19 and 2/21), we’ll focus our attention on the first midterm exam in Finance 4335. Class on 2/19 will be devoted to a review session for the exam, and the exam will be administered during class on Thursday, February 21.

Even though I list it as an “optional” reading, I highly recommend that y’all read my teaching note due on 2/19 entitled “Finance 4335 course synopsis for Midterm Exam 1”. In order to properly prepare for next Tuesday’s review session, I also recommend reviewing problem sets 2 – 5 (actually, problem set 5 is due on Tuesday!) and midterm 1 from Fall 2018; solutions for problem sets in FIN 4335 are available at http://risk.garven.com/category/problem-set-solutions. After class next Tuesday, I will post the solutions for problem set 5 and for midterm 1 from Fall 2018.

New reading entitled “Modeling Risk Preferences Using Taylor Series Expansions of Utility Functions”

I decided to write a 4 page teaching note entitled “Modeling Risk Preferences Using Taylor Series Expansions of Utility Functions” which more clearly documents the relationships between expected utility, the mean-variance model, and the mean-variance-skewness-kurtosis model. I plan to begin class tomorrow covering this reading and then moving on to the mean-variance model and stochastic dominance.

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 risk@garven.com sometime prior to the beginning of next Thursday’s class meeting, at which time I’ll also collect your completed problem sets.

Hints for solving problem set #4

Problem set #4 (which is due on Thursday, February 14) consists of two problems: 1) an optimal (expected utility maximizing) portfolio problem, and 2) a stochastic dominance problem.  We’ll discuss stochastic dominance next Tuesday, but in the meantime allow me to provide you with some hints for setting up the first problem.

The first problem involves determining how to (optimally) allocate initial wealth W0 = $100 to (risky) stock and (safe) bond investments for two investors who are identical in all respects except utility. Let \alpha represent the allocation to stock; then the plan is to invest $100\alpha in the stock and $100(1-\alpha) in the bond. The key here is to find the value for \alpha 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 = $100 for both investors;
  • Current bond and stock prices are B0 and S0 respectively;
  • End-of-period bond price is B1 = B0(1.05) with probability 1.0; and
  • End-of-period stock price is S1 = S0(1.3) with probability .6 and S1 = S0(.7) 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 30%, a 40% chance that the stock decreases 30%, and a 100% chance that the bond increases in value by 5%, this implies the following:

  • 60% of the time, Ws = \alphaW0(1.3) + (1-\alpha)W0(1.05) = \alpha100(1.30) + (1-\alpha)100(1.05) = \alpha130 + (1-\alpha)105 = 105 + 25\alpha.
  • 40% of the time, Ws = \alphaW0(.7) + (1-\alpha)W0(1.05) = \alpha100(.7) + (1-\alpha)100(1.05) = \alpha70 + (1-\alpha)105 = 105 – 35\alpha.

Therefore, expected utility for Investor A is: E(U(W)) = .6(105 + 25\alpha).5 + .4(105 – 35\alpha).5, and expected utility for Investor B is E(U(W)) = .6ln (105 + 25\alpha) + .4ln(105 – 35\alpha). It is up to you to solve for the optimal value of \alpha for each investor.  There are two ways to do this – via calculus or a spreadsheet model.  Actually, I would encourage y’all to work this problem both ways if time permits because doing so will help you develop an even better grasp of the underlying principles and concepts in Finance 4335.  However, at your option, you may rely solely on building your own spreadsheet model.  If you do this, in order to receive full credit, you need to email your spreadsheet model to risk@garven.com along with turning in the completed problem set.

Finance 4335