All posts by jgarven

interpreting questions 2 and 3 on Problem Set #3

In question 2 of Problem Set 3, the loss in the FIRE state is $100, and $0 in the NO FIRE state. By “fully insuring” or purchasing “full coverage”, this means that you buy an insurance policy which pays off $100 if the FIRE state occurs and zero otherwise. I would also note that this question is based in large part upon the problem that is shown in Chapter 2 of Doherty (cf. pp. 29–32).

Finally, question 3 addresses how risk aversion changes with respect to increases in the level of initial wealth; as we saw during our last class meeting, diminishing marginal utility implies that the utility consequence of a given risk for a “rich” person is less severe than the utility consequence of such a risk for an otherwise identical “poor” person. We’ll develop this idea further during tomorrow’s class meeting by showing how individuals with larger initial wealth endowments behave in a less risk averse fashion than otherwise identical individuals who have smaller initial wealth endowments.

Interpreting question 1 on Problem Set 3

A student asked me for a clarification of question 1 on problem set #3; this problem set is due at the beginning of class on Thursday, September 14.  Anyway, question 1 reads as follows: “A worker whose utility function U(W) = W.5 has received a job offer which pays $80,000 with a bonus. The bonus is equally likely to be $0, $10,000, $20,000, $30,000, $40,000, $50,000, or $60,000. Assume that initial wealth is $0.”

Here’s how to interpret this problem (in terms of properly delineating state probabilities and state-contingent values for wealth): there are 7 “states of the world”, and since the bonus (over and above the salary of $80,000) is equally likely to be $0, $10,000, $20,000, $30,000, $40,000, $50,000, or $60,000, this implies that ps = 1/7 for each of the 7 states. Furthermore, since initial wealth is $0 and the salary is $80,000, it follows that state contingent wealth for the seven states of the world will be $80,000, $90,000, $100,000, $110,000, $120,000, $130,000, or $140,000.

Risk pooling spreadsheet – solutions for class problem

Here’s a screen shot from the spreadsheet linked below (slightly modified version of the spreadsheet that we built in class). Here, we find that if risks are independent and identically distributed, then by increasing the number of policies in the risk pool, the probability that the average loss exceeds $1,500 declines as we add policies. Without risk pooling, the probability of a “large” loss of $1,500 is 30.85%; with 5 policies, it is 13.18%, and with 10 policies it is 5.69%. However, if risks are positively correlated, then both unique and systematic risks influence this calculation. For example, with 10 policies that have .1 correlation, the probability that the average loss exceeds $1,500 is 12.57% (compared with 5.69% when there is zero correlation):

Risk Pooling – 9-5-2017.xlsx

Extra Credit Opportunity for Finance 4335 (due prior to the beginning of class on Thursday, September 7)

I have decided to offer the following extra credit opportunity for Finance 4335. You can earn extra credit by building an Excel spreadsheet which replicates the Standard Normal Distribution Function table for positive z values ranging from 0.00 to 3.09 (in .01 increments).  Helpful hint – you can obtain cumulative probabilities for all 310 z values (comprising 31 rows and 10 columns) by using the NORMSDIST function that is built into Excel.

This extra credit assignment must be emailed as a file attachment to prior to the beginning of class on Thursday, September 7; I will not accept this assignment in any other way. I will use your grade on this assignment to replace your lowest quiz grade in Finance 4335 (assuming that your grade on the extra credit is higher than your lowest grade).

How government policy exacerbates hurricanes like Harvey

Here’s the (very timely) cover story of the latest issue of The Economist. Quoting from the article, “Underpricing (of flood insurance) encourages the building of new houses and discourages existing owners from renovating or moving out. According to the Federal Emergency Management Agency, houses that repeatedly flood account for 1% of NFIP’s properties but 25-30% of its claims. Five states, Texas among them, have more than 10,000 such households and, nationwide, their number has been going up by around 5,000 each year. Insurance is meant to provide a signal about risk; in this case, it stifles it.”

As if global warming were not enough of a threat, poor planning and unwise subsidies make floods worse.

On the Economics of Price Gouging

This is an oldie (from 2007) but goody – on the economics of price gouging in the wake of a hurricane. The principles discussed are timeless and well worth pondering!

Mike Munger of Duke University recounts the harrowing (and fascinating) experience of being in the path of a hurricane and the economic forces that were set in motion as a result. One of the most important is the import of urgent supplies when thousands of people are without electricity. Should prices be allowed to rise freely or should the government restrict prices? Listen in as Munger and EconTalk host Russ Roberts discuss the human side of economics after a catastrophe.

Harvey’s Test: Businesses Struggle With Flawed Insurance as Floods Multiply

This WSJ article provides a fairly comprehensive look at the financial implications for #Harvey for small business. What’s particularly disconcerting is that NFIP is already for all intents and purposes technically insolvent (current debt to the US Treasury stands at around $25 billion) and Congress is supposed to reauthorize funding for the program’s next five years by September 30. On the lighter side of things, it’s fun to see a couple of academic colleagues’ names in print in this article; specifically, Erwann Michel-Kerjan of the Organization for Economic Cooperation and Development Board on Financial Management of Catastrophes and Ben Collier, who is a faculty member at Temple University’s Fox School of Business.

Hurricane will strain a National Flood Insurance Program out of step with needs of small businesses in era of extreme weather.

Risk and Uncertainty – on the role of Ambiguity

This WSJ article from last spring addresses how to measure uncertainty and also explains the subtle, yet important differences between risk and uncertainty. Risk reflects the “known unknowns,” or the uncertainties about which one can make probabilistic inferences. Ambiguity (AKA “Knightian” uncertainty; see reflects the “unknown unknowns,” where the probabilities themselves are a mystery.

A researcher whose work foreshadowed the VIX now has his eye on an entirely different barometer of market uncertainty—ambiguity.

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

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.[1] 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.[2]

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:

New Picture (1)

[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 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!).

[2]Note that since 33 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’(33) = 1 – (364/365)^[33(32)/2] = 76.5%.