## Problem 2 on Problem Set 11 (underinvestment problem)

Problem 2 in Problem Set 11 is a re-scaled version of the class problem discussed last Thursday (see pages 17-19 of the lecture note at http://fin4335.garven.com/fall2023/risk_costly.pdf). In both the lecture note and Problem Set 11, reinvestment has a positive net present value. There’s no underinvestment problem with zero or moderate debt, as in such cases, shareholders capture the full benefit (NPV) from reinvesting. However, excessive debt, leading to default in the loss state, creates the perverse incentive for shareholders to “underinvest” because while they bear the reinvestment cost, the firm’s creditors capture the reinvestment benefit. Coordinating finance and risk management decisions ensures shareholders don’t default on the firm’s promised debt payment while also ensuring that they, and not the creditors, capture the full benefit of the decision to reinvest.  In other words, coordinating finance and risk management decisions solves this moral hazard by mitigating the incentive conflict created by limited liability.

## Problem Set 9 helpful hints – part 2 of 2

Here are some helpful hints for Problem Set 9, problem 2:

• Scenario A requires solving for call and put option prices using the Black-Scholes-Merton option pricing formulas.  See the Part 2 option pricing lecture note, page 21, for a numerical illustration of how to do this.
• Scenario B requires finding the current price of the underlying asset, where the call, put, and exercise prices are all given.  Solve the put-call parity equation ( $C + K{e^{ - rT}} = P + S$) for S.
• Scenario C requires finding the exercise price, where the call, put, and underlying asset prices are all given.  Solve the put-call parity equation for K.
• Scenario D requires finding $\sigma$ for a call option worth $2.38 and a put option worth$3.60.  Feel free to use the Black-Scholes spreadsheet from the course website, or better yet, create your own Excel spreadsheet in which you solve for the call and/or the put by varying $\sigma$ (this can be accomplished either via trial and error or better yet, by using either Solver or Goal Seek).  An important lesson you’ll learn from this part of problem 2 is that call and put option prices are positively related to $\sigma$ .

## Problem Set 9 helpful hints – part 1 of 2

During last Thursday’s class meeting of Finance 4335, we completed our coverage of the Cox-Ross-Rubinstein (CRR) model (as outlined on pages 9-11 of the Part 2 option pricing lecture note and pages 11-13 of the Teaching the Economics and Convergence of the Binomial and Black-Scholes Option Pricing Formulas assigned reading).  This coming Tuesday’s class meeting will be devoted to 1) showing how CRR model probabilities and prices converge to BSM (Black-Scholes-Merton) model probabilities and prices as the number of timesteps becomes arbitrarily large, and 2) showing how the BSM model can be applied to the pricing and management of credit risk.

In the meantime, it’s not too early to begin working on Problem Set 9, Problem 1.  Here are some helpful hints to consider:

• Since the call option described in Problem 1, Part A is initially out-of-the-money (i.e., since S = $18 and K =$20), there will be terminal nodes at which the call option expires in-the-money and others at which it expires out-of-the-money.  By solving the $b = \ln (K/S{d^n})/\ln (u/d)$ equation and rounding to the nearest integer greater than b (referred to in the above-referenced sources as the parameter “a“), this indicates the minimum number of up moves required such that this call option expires inthemoney.  Once you have this information, you can consider only those terminal nodes at which the call option expires in-the-money (which are nodes a through n) and calculate the call option price by applying the CRR call option pricing equation: $C = {e^{ - rT}}\left[ {\sum\limits_{j = a}^n {\frac{{n!}}{{j!\left( {n - j} \right)!}}{q^j}{{\left( {1 - q} \right)}^{n - j}}\left( {{u^j}{d^{n - j}}S - K} \right)} } \right]$
• Part B: Apply the put-call parity equation ( $C+Ke^{-rT} =P+S$) to solve for the put option price.

## Outlines and Study questions

I have prepared a handful of outlines and study questions related to some of the Finance 4335, Part 2 reading assignments:

## Midterm 2 Exam Helpful Hints

Students will likely find the study guide helpful in preparing for the second midterm exam in Finance 4335 (scheduled for Thursday, October 26, in class).  The exam covers:

During the past month, we have also worked on class problems related to the abovementioned topics.  The class problems and their solutions, along with solutions for problem sets 5-7, are available on the Problem Set Solutions page.

A formula sheet will be included on the last page of the exam booklet; this same formula sheet can be downloaded from http://fin4335.garven.com/fall2023/formulas_part2.pdf.  On this exam, you must only complete three of four problems. If you complete all four problems on the exam, only the three highest-scoring problems will count toward your Midterm Exam 2 grade.  Each problem will be worth 32 points, and you will receive 4 points for including your name on the exam booklet. Thus, the maximum number of points possible on Midterm Exam 1 will be 100.

Whenever you take an exam in Finance 4335, it is important to not only show your work but also provide complete answers for each question; i.e., besides producing appropriate numerical results, also clearly explain your results using plain English.

## Important Notice: Re-download Problem Set 7 from the Finance 4335 course website!

Earlier this afternoon, I found a typo in Problem 2 of Problem Set 7.  I fixed that typo and also lightly edited Problem 2 for greater clarity.   I didn’t make any changes to Problem 1.

If you downloaded Problem Set 7 before 4:30 p.m. today, please replace it with this updated version. If you’ve already worked on Problem Set 7, don’t worry about Problem 1, but you might want to check your answers for Problem 2 in light of the edits I made.

## An important clarification of the logical principles behind the stochastic dominance model

On problem set 4, part D, most of you had no apparent difficulty in correctly establishing that the sum of the differences between the cumulative distribution (CDF) for risk 2 and the CDF for risk 1 is positive.   However, many of you drew the wrong conclusion, claiming that since the sum of differences between $F({W_{2,s}})$ and $F({W_{1,s}})$ came out to a positive number, it followed that risk 2 second order stochastically dominates risk 1.  Actually, this result implies the opposite; i.e., that risk 1 second order stochastically dominates 2. The purpose of this blog posting is to clarify everyone’s understanding of the logic behind the stochastic dominance model.

The one-page exam formula sheet includes section 4 which explains that risk i dominates risk j, in both the first and second cases, when 1) the cumulative distribution function (CDF) of the ith risk is either less than or equal to the CDF of the jth risk for all states (first order dominance), or 2) the sum of the differences between the jth risk CDF and the ith risk CDF for all states is positive (second order dominance): While the math behind first and second order stochastic dominance is summarized in my optional reading entitled “Technical Note on Stochastic Dominance and Expected Utility”, the intuition for first and second order stochastic dominance can be seen in the figures featured on pages 9 and 12 of my Decision-Making under Risk and Uncertainty, part 4 lecture note. In the above figure from page 9 of my Decision-Making under Risk and Uncertainty, part 4 lecture note, the G risk has 50% of a $0 payoff, and 25% each of a$10 payoff and a $100 payoff. The F risk involves removing 25 percentage points off the$0 payoff and adding 25 percentage points extra to the $100 payoff, and both F and G have a 25% probability of$10 payoffs.  Graphically, this ensures that F first order stochastically dominates G; i.e., G(Ws) is greater than or equal to F(Ws) for all s, which also implies that EF[U(W)] > EG[U(W)]. Intuitively, the picture which gets rendered by this analysis shows that most of the probability mass of the stochastically dominant risk (in this case, F) lies below the probability mass of the stochastically dominated risk (in this case, G). Furthermore, since risk F first order stochastically dominates risk G, risk F also second order stochastically dominates risk G because G(Ws) – F(Ws) > 0 for $0 and$10 payoffs, and G(Ws) – F(Ws) = 0 for the \$100 payoff.

Next consider the figure from page 12 of my Decision-Making under Risk and Uncertainty, part 4 lecture note: Here, G(Ws) – F(Ws) > 0 for payoffs ranging from 1-5, G(Ws) – F(Ws) < 0 for payoffs ranging from 5-8, and G(Ws) – F(Ws) = 0 payoffs ranging from 8-12.  Thus, there is no first order dominance.  However, since the positive difference between G(Ws) – F(Ws) for payoffs ranging from 1-5 exceeds the negative difference between G(Ws) – F(Ws) for payoffs ranging from 5-8, the sum of G(Ws) – F(Ws) over the entire range of payoffs comes out positive.  Thus, risk F second order stochastically dominates risk G, which also implies that EF[U(W)] > EG[U(W)].