I would also like to call your attention to the new spreadsheet (linked below) entitled “Lecture 4, page 25 standard normal probability example”. As the spreadsheet name indicates, the verbal description of the problem solved in this spreadsheet appears on page 25 of Statistics Tutorial, Part 2. This spreadsheet illustrates, among other things, calculation of z-statistics coupled with the use of Excel’s NORMSDIST function.
I would like to call your attention to the new and improved “Two Asset Portfolio Spreadsheet” that I just posted to the course website (linked as a bullet point under item 3 on the Lecture Notes page). The worksheet entitled “p. 17 spreadsheet” refers to the numerical example on page 17 of the Statistics Tutorial, Part 1 teaching note, and the worksheet entitled “p. 18 spreadsheet” refers to the numerical example on page 18 of the same teaching note. We discussed the page 17 example but skipped the page 18 example during our class meeting.
In our discussion of page 17 numerical example, we discovered that there was no combination of assets 1 and 2 which has lower standard deviation than just holding asset 1 by itself. We determined this numerically by using Solver, and analytically by calculating a ratio involving the variances of the two assets and their covariance. In the PDF file linked below, I provide a brief note which explains the derivation of this ratio. Later in Finance 4335, we’ll see that this variance minimizing ratio plays ab important role in Portfolio Theory.
Occupational licensing has been a “hot topic” in the media related to job creation and entrepreneurship recently. Occupational licensing has extended from its origins in health and safety to occupations like hair braiding (http://www.theatlantic.com/business/archive/2016/08/hair-braider/494084/ ) creating barriers to entry – particularly for the poor. Dr. Kleiner and his coauthor (Alan Krueger – Harvard) are arguably the leading researchers and experts on the topic.
I will offer extra credit if you attend this presentation and submit (via email sent to firstname.lastname@example.org) a 1-2 page executive summary of what you learned from the presentation. The executive summary is due by no later than 5 p.m. on Monday, January 30 (any time after then will be considered “late” and therefore not be eligible for extra credit). The extra credit will replace your lowest quiz grade (assuming the extra grade is higher).
… are available at http://fin4335.garven.com/spring2017/ps1solutions.pdf.
Class will begin tomorrow with two short quizzes; the first quiz will be on readings that were assigned for January 13 (“Optimization and “How long does it take to double (triple/quadruple/n-tuple) your money?), and the second quiz will be on the readings assigned for January 20 (“The New Religion of Risk Management” and “Normal and standard normal distribution“). Furthermore, Problem Set 1 will be due at the beginning of class. (I previously made this announcement 5 days in the blog posting entitled “Plans for the next Finance 4335 class meeting, along with a preview of future topics.”)
Following up on my previous blog posting entitled “Statistical Independence,” consider the so-called “Birthday Paradox”. The Birthday Paradox pertains to the probability that in a set of randomly chosen people, some pair of them will have the same birthday. Counter-intuitively, in a group of 23 randomly chosen people, there is slightly more than a 50% probability that some pair of them will both have been born on the same day.
To compute the probability that two people in a group of n people have the same birthday, we disregard variations in the distribution, such as leap years, twins, seasonal or weekday variations, and assume that the 365 possible birthdays are equally likely. Thus, we assume that birth dates are statistically independent events. Consequently, the probability of two randomly chosen people not sharing the same birthday is 364/365. According to the combinatorial equation, the number of unique pairs in a group of n people is n!/2!(n-2)! = n(n-1)/2. Assuming a uniform distribution (i.e., that all dates are equally probable), this means that the probability that no pair in a group of n people shares the same birthday is equal to p(n) = (364/365)^[n(n-1)/2]. The event of at least two of the n persons having the same birthday is complementary to all n birthdays being different. Therefore, its probability is p’(n) = 1 – (364/365)^[n(n-1)/2].
Given the assumptions listed in the previous paragraph, suppose that we are interested in determining how many randomly chosen people are needed in order for there to be a 50% probability that at least two persons share the same birthday. In other words, we are interested in finding the value of n which causes p(n) to equal 0.50. Therefore, 0.50 = (364/365)^[n(n-1)/2]; taking natural logs of both sides and rearranging, we obtain (ln 0.50)/(ln 364/365) = n(n-1)/2. Solving for n, we obtain 505.304 = n(n -1); therefore, n is approximately equal to 23.
The following graph illustrates how the probability that a pair of people share the same birthday varies as the number of people in the sample increases:
 It is worthwhile noting that real-life birthday distributions are not uniform since not all dates are equally likely. For example, in the northern hemisphere, many children are born in the summer, especially during the months of August and September. In the United States, many children are conceived around the holidays of Christmas and New Year’s Day. Also, because hospitals rarely schedule C-sections and induced labor on the weekend, more Americans are born on Mondays and Tuesdays than on weekends; where many of the people share a birth year (e.g., a class in a school), this creates a tendency toward particular dates. Both of these factors tend to increase the chance of identical birth dates, since a denser subset has more possible pairs (in the extreme case when everyone was born on three days of the week, there would obviously be many identical birthdays!).
Note that since 21 students are enrolled in Finance 4335 this semester, this implies that the probability that two Finance 4335 students share the same birthday is roughly p’(21) = 1 – (364/365)^[21(20)/2] = 43.8%.
Tomorrow’s Finance 4335 class meeting will be devoted to a tutorial on the topics of probability and statistics. Much of our class discussion will focus on the implications of statistical independence for probability distributions such as the binomial and normal distributions which we will rely upon throughout the semester.
Whenever risks are statistically independent of each other, this implies that they are uncorrelated; i.e., random variations in one variable are not meaningfully related to random variations in another. For example, auto accident risks are largely uncorrelated random variables; just because I happen to get into a car accident, this does not make it any more likely that you will suffer a similar fate (that is unless we happen to run into each other!). Another example of statistical independence is a sequence of coin tosses. Just because a coin toss comes up “heads,” this does not make it any more likely that subsequent coin tosses will also come up “heads.”
Computationally, the joint probability that we both get into car accidents or heads comes up on two consecutive tosses of a coin is equal to the product of the two event probabilities. Suppose your probability of getting into an auto accident during 2017 is 1%, whereas my probability is 2%. Then the likelihood that we both get into auto accidents during 2017 is .01 x .02 = .0002, or .02% (1/50th of 1 percent). Similarly, when tossing a “fair” coin, the probability of observing two “heads” in a row is .5 x .5 = 25%. The probability rule which emerges from these examples can be generalized as follows:
Suppose Xi and Xj are uncorrelated random variables with probabilities pi and pj respectively. Then the joint probability that both Xi and Xj occur is equal to pipj.
As SAP CEO Bill McDermott explains in the video below, taking time to read the Wall Street Journal (WSJ) is all about intellectual curiosity. Furthermore, I will frequently post links to WSJ articles throughout the course of the semester which help to bridge the gap between Finance 4335 specifically (as well as your business studies generally) and the “real” world. WSJ subscription instructions are at http://risk.garven.com/2017/01/02/how-to-obtain-a-spring-2017-wall-street-journal-subscription/.
Problem Set 1 is due at the beginning of class on Friday, January 20. Here is a hint for solving the 4th question on problem set 1.
The objective is to determine how big a hospital must be so that the cost per patient-day is minimized. We are not interested in minimizing total cost; if this were the case, there would be no hospital because marginal costs are positive, which implies that total cost is positively related to the number of patient-days.
The cost equation C = 4,700,000 + 0.00013X2 tells you the total cost as a function of the number of patient-days. This is why you are asked in part “a” to derive a formula for the relationship between cost per patient-day and the number of patient days. Once you have that equation, then that is what you minimize, and you’ll be able to answer the question concerning optimal hospital size.
The next Finance 4335 class meeting on Friday, January 20 will be devoted to covering various topics in probability and statistics that are important for Finance 4335. Class will begin with two short quizzes; the first quiz will be on readings that were assigned for January 13 (“Optimization and “How long does it take to double (triple/quadruple/n-tuple) your money?), and the second quiz will be on the readings assigned for January 20 (“The New Religion of Risk Management” and “Normal and standard normal distribution“). Furthermore, Problem Set 1 will be due at the beginning of class that day.
While I have your attention, let me briefly explain what the main “theme” will initially be in Finance 4335 (up to the first midterm exam, which is scheduled for Friday, February 17). Specifically, we will delve into decision theory. Decision theory addresses decision making under risk and uncertainty, and not surprisingly, risk management lies at the very heart of decision theory. Initially, we’ll focus our attention upon variance as our risk measure. Most basic finance and economics models implicitly or explicitly assume that risk = variance. We’ll learn that while this is a reasonable assumption to make under some circumstances, other circumstances exist where it is not an appropriate assumption. Furthermore, since individuals and firms are typically exposed to multiple sources of risk, we need to take into consideration the portfolio effects of risk. To the extent to which risks are not perfectly positively correlated, this implies that risks often “manage” themselves by canceling each other out. Thus the risk of a portfolio is typically less than the sum of the individual risks which comprise the portfolio.
The decision theory provides us with a very useful framework for thinking about concepts such as risk aversion and risk tolerance. The calculus comes in handy by providing an analytic framework for determining “optimal” exposure to risk. More specifically, the calculus helps us determine how much risk we hold onto and how much risk we transfer to others. Such decisions occur regularly in our daily lives, encompassing all sorts of practical problems such as deciding 1) how to allocate assets in a 401-K or IRA, 2) whether to insure our health, life, and property, 3) whether to work for a startup or an established business, and so forth. There’s also quite a bit of ambiguity when we make decisions without complete information, but this course will at least help you think critically about costs, benefits, and trade-offs related to decision-making whenever you encounter risk and uncertainty.
After the first midterm, we’ll move on to other topics including demand for insurance, asymmetric information, portfolio theory, capital market theory, option pricing theory, and corporate risk management.