Category Archives: Math and Statistics

Risk pooling spreadsheet – solutions for yesterday’s class problem

Here’s a screenshot from a spreadsheet that I coded for the risk pooling class problem (linked below) that we discussed during class yesterday. We found in class yesterday (and now find in this spreadsheet today), 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 Spreadsheet.xlsx

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

Following up on the 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 77 students are enrolled in two sections of Finance 4335 this semester, this implies that the probability that two Finance 4335 students share the same birthday is roughly p’(77) = 1 – (364/365)^[77(76)/2] = 99.97%,

Statistical Independence

During today’s Finance 4335 class meeting, I introduced the concept of statistical independence. On Thursday, 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 the coming year is 1%, whereas my probability is 2%. Then the likelihood that we both get into auto accidents during the coming year 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.

A much more “rigorous” way to calculate 1+1 = 2

One of my Baylor faculty colleagues pointed out an entertaining and somewhat whimsical parody on the use of math in applied economics and finance which first appeared in the Nov.-Dec. 1970 issue of The Journal of Political Economy, entitled “A First Lesson in Econometrics” (at least I found it entertaining :-)).  Anyway, check it out!

File Attachment: JPEMathParody.pdf (30 KB)

Lagrangian Multipliers

There is a section in the assigned “Optimization” reading tomorrow on pp. 74-76 entitled “Lagrangian Multipliers” which (as noted in footnote 9) may be skipped without loss of continuity.  The primary purpose of this chapter is to re-acquaint students with basic calculus and how to use the calculus to solve so-called optimization problems.  Since the course only requires solving unconstrained optimization problems, there’s no need for Lagrangian multipliers.

Besides reading the articles entitled “Optimization” and “How long does it take to double (triple/quadruple/n-tuple) your money?” in preparation for tomorrow’s meeting of Finance 4335, make sure that you fill out and email the student information form as a file attachment to risk@garven.com prior to the beginning of tomorrow’s class.  As I explained during yesterday’s class meeting, this assignment counts as a problem set, and your grade is 100 if you turn this assignment in on time (i.e., sometime prior to tomorrow’s class meeting) and 0 otherwise.

Calculus and Probability & Statistics recommendations…

Since many of the topics covered in Finance 4335 require a basic knowledge and comfort level with differential calculus and probability & statistics, the second class meeting (August 23) will include a mathematics tutorial, and the third and fourth class meetings (August 28-30) will cover probability & statistics. I know of no better online resource for brushing up on (or learning for the first time) these topics than the Khan Academy.

So here are my suggestions for Khan Academy videos which cover these topics (unless otherwise noted, all sections included in the links which follow are recommended):

Finally, if your algebra is a bit rusty, I would also recommend checking out the Khan Academy’s review of algebra.