### Friendly reminder about assigned readings and quiz due Thursday, February 9

Here’s a friendly reminder about assigned readings and quiz due Thursday, February 9. Be sure cover the following two readings prior to tomorrow’s class; the Quiz is based on these readings.

1. Expected Utility, Mean-Variance, and Stochastic Dominance, by James R. Garven
2. Modeling Risk Preferences Using Taylor Series Expansions of Utility Functions, by James R. Garven

### Some important intuitions from today’s class meeting of Finance 4335

1. The most important concept covered in class today a is that people vary in terms of their preferences for bearing risk. Although we focused most of our attention on modeling risk-averse behavior, we also considered examples of risk neutrality (where you only care about expected wealth and are indifferent about the riskiness of wealth) and risk loving (where you actually prefer to bear risk and are willing to pay money for the opportunity to do so).
2. Related to point 1: irrespective of whether you are risk averse, risk neutral, or risk loving, the foundation for decision-making under conditions of risk and uncertainty is expected utility. Given a choice among various risky alternatives, one selects the choice which has the highest utility ranking.
3. If you are risk averse, then $E(W) > {W_{CE}}$ and the difference between $E(W)$ and ${W_{CE}}$ is equal to the risk premium $\lambda$. Some practical implications — if you are risk averse, then you are okay with buying “expensive” insurance at a price that exceeds the expected value of payment provided by the insurer, since (other things equal) you’d prefer to transfer risk to someone else if it’s not too expensive to do so. On the other hand, you are not willing to pay more than the certainty equivalent for a bet on a sporting event or a game of chance.
4. If you are risk neutral, then $E(W) = {W_{CE}}$ and $\lambda = 0$; risk is inconsequential and all you care about is maximizing the expected value of wealth.
5. If you are risk loving, then $E(W) < {W_{CE}}$ and $\lambda < 0$; you are quite willing to pay for the opportunity to (on average) lose money.

### Important notes about tomorrow’s Risk Pooling class Problem

In tomorrow’s recorded lecture, there is some discussion of the Risk Pooling Class Problem (located at http://fin4335.garven.com/spring2023/rpclassproblem.pdf) based on the assigned reading entitled Supply of Insurance. When you watch the video, I highly recommend that you pause at the 34:40 mark and try to complete this class problem on your own before moving on. Right after the 34:40 mark, I spend some time solving the class problem using an Excel spreadsheet. You’ll come away with a much better understanding of the risk pooling concept if you struggle a bit with it first, before viewing my explanation of this problem.

### Finance 4335 Contingency planning for tomorrow’s #Icepocalypse

Most of the information you need about remote instruction tomorrow for Finance 4335 is listed in the "Virtual Instruction on Tuesday, January 31" email I sent via Canvas earlier this evening.

In the meantime, here are a couple of other housekeeping notes:

1. I changed the due date for Quiz 3 from tomorrow at 11 am to Thursday at 11 am.
2. Also, since I am cautiously optimistic that we’ll be back to in-person instruction by Thursday, plan to turn in Problem Set 2 at the beginning of class on Thursday.

### Problem Set 2 (2nd of 2) helpful hints)

The first set of problem set 2 hints appears at http://risk.garven.com/2023/01/25/problem-set-2-helpful-hints

Here’s the second problem set 2 hint, based on the following questions:

Finance 4335 Student Question: “I am in the process of completing Problem Set 2, but I am getting stumped, are there any class notes that can help me with 1E and question 2? I am having trouble finding the correct formulas. Thank you.”

Dr. Garven Answer (for question 2): Yesterday, I provided some hints about problem set #2 @ http://risk.garven.com/2023/01/25/problem-set-2-helpful-hints.  Regarding question #2, here is what I wrote there: “The second problem involves using the standard normal probability distribution to calculate the probabilities of earning various levels of return by investing in risky securities and portfolios; see pp. 15-21 of the http://fin4335.garven.com/spring2023/lecture4.pdf lecture note for coverage of that topic.”

The inputs for this problem are all specified in parts A and B; in part A, expected return is 10% and standard deviation is 20%, whereas in part B,  expected return is 6.5% and standard deviation is 10%.  The probability of losing money in parts A and B requires calculating the z statistics for both cases.  In part A, the z stat is z = (0-mu)/sigma= (0-10)/20, and in part B it is z = (0-mu)/sigma= (0-6.5)/10.  Once you have the z stats, you can obtain probabilities of losing money in Parts A and B by using the z table from the course website (or better yet, your own z table).

Part C asks a different probability question – what the probability of earning more than 6% is, given the investment alternatives described in parts A and B.  To solve this, calculate 1-N(z) using the z table, where z = (6-mu)/sigma.

Dr. Garven Answer (for question 1E): Regarding 1E – since returns on C and D are uncorrelated, this means that they are statistically independent of each other.  Thus, the variance of an equally weighted portfolio consisting of C and D is simply the weighted average of these securities’ variances.  See the http://fin4335.garven.com/spring2023/lecture3.pdf lecture note, page 9, the final two bullet points on that page.  Also see the portfolio variance equation on page 13, which features two variance terms and one covariance term – Since C and D are statistically independent, the third term there (2w(1)w(2)sigma(12)) equals 0.

### The Birthday Paradox: an interesting probability problem involving “statistically independent” events

During this week’s statistics tutorials, we discussed (among other things) the concept of statistical independence, and focused attention on some important implications of statistical independence for probability distributions such as the binomial and normal distributions.

In this blog posting, I’d like to call everyone’s attention to an interesting (non-finance) probability problem related to statistical independence. Specifically, 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.[1] Thus, we assume that birthdates 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].[2]

Given these assumptions, 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:[1] 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 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 birthdates 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!).

[2] Note that since 29 students are enrolled in Finance 4335 this semester, the probability that at least two Finance 4335 students share the same birthday is p’(29) = 1 – (364/365)^[29(28)/2] = 67.2%; this probability can also be inferred from the above figure.

### Z Table Extra Credit Assignment (due at the start of class on Tuesday, January 31)

Here’s an extra credit opportunity for Finance 4335. Working on your own (i.e., this is not a group project; I will only give credit for spreadsheets that are uniquely your own), build your own “z” table in Excel (patterned after the table at http://fin4335.garven.com/stdnormal.pdf); the top row should have values ranging from 0.00 to 0.09, and the first column should have z values ranging from -3.0 to +3.0, in increments of 0.1).

Conveniently, Excel has the standard normal distribution function built right in; e.g., if you type “=normsdist(z)”, Excel returns the probability associated with whatever z value you provide. If you type “=normsdist(0)”, .5 is returned, since half of the area under the curve lies to the left of the expected value E(z) = 0. Similarly, if you type “=normsdist(1)”, then .8413 is returned because 84.13% of the area under the curve lies to the left of z = 1. Perhaps you recall from your QBA course that 68.26% of the area under the curve lies between z = -1; this “confidence interval” of +/- 1 standard deviation away from the mean (E(z)=0) is calculated in Excel with the following code: “=normsdist(1)-normsdist(-1)”, and so forth.