1 The Geometric Distribution and Binomial Distribution Applied to Finance (Preliminary Version) Dec. 2015 Floyd Vest Example 1. Geometric Distribution. Assume you invest in a portfolio which historically has a median yearly total return of 10% (for example the S&P 500 Index of Stocks) . Half of the yearly returns were 10% or greater and half were less than 10%. See the articles in this course for various kinds of averages. For a single year chosen at random, the probability p of 10% or greater is .5, and the probability q of less than 10% is .5 . We know that in general q = 1 – p. Let N be the number of trials of yearly returns until the first return is 10% or greater. It is known that P(N = n) = qpn1 for n-1 consecutive return less than 10% until a return greater or equal 10% on the nth trial occurs. This gives P(N = n) = .5n1 (.5) . We assume that annual returns are independent. See the Side Bar notes and Exercises for independence. Let us examine the investment in the portfolio with successive annual returns. What is the probability P(N =3) of the first return greater than or equal to 10% occurring on the third year? We have P( N = 3) = (.52 )(.5) = .125 . We could build the Table 1: Table 1. We will abbreviate P(N=n) with P(n). P(n = 3) = .125 for the first return greater or equal to 10% in year 3. P(n=2) = .25 for the first return greater or equal to 10% in year 2. P(n=1) = .5 for the first return greater or equal to 10% in year 1. Example 2. For this portfolio, we assume that the proportions of annual returns which lose money with return less than 0% is .31, and the proportion for which returns are greater than or equal to 0% is .69. Similar to the S&P 500 and applying the normal distribution. Here we assume q, the probability of losing money is q = .31 , and p, the probability of not losing money, is p = .69 . We could build the Table 2: Table 2. P(n = 3) = (.31 )2 (.69) = .0067 for the probability of the first nonnegative return in year 3. P(n = 2) = (.31)(.69) = .214 for the probability of the first nonnegative return in year 2. P(n = 1) = .69 for the probability of the first nonnegative return in year 1. Example 3. Compare the probability of the first year with return 10% with the probability of the first year of return 0% for years 1, 2 and 3 in Table 3. Table 3. Year 1 Year 2 Year 3 .5 .25 .125 0% .69 .214 .0067 2 Thus the probability (.0067) of having to wait until the third year for the first is less than the probability (.125) of having to wait until the third year for the first . See the Exercise for discussion of Table 3. Derivation of the mean (expected value) E(N=n) of the random variable N for the Geometric Distribution with probability of success p. N is the number of the trials (periods) with a first success. P(N=n)=q n1 p with the first success at trial n where n = 1, 2, 3, … . It is known that E(N) = n P() n . See the Exercises to verify this . E(N) = 1p 2 qp 3 q23 p 4 q p ... = (1q ) 2 q (1 q ) 3 q23 (1 q ) 4 q (1 q ) ... = 1 – q + 2q - 2q 2 +3q 2 - 3q 3 +4q 3 - 4q 4 +… =1q q23 q ... 11 = 1 qp 1 1 So E(N) = . See the Exercises to show that = . p 1 q TI 83 calculation. We will calculate P(n=3) for p =.69 on the TI 83 function geomtpdf(p,n). Code and commentary: 2nd Dist to geometpdf( Enter. You see geometpdf ( Write in .69,3 to get geometpdf(.69,3. Press Enter, you interpret geometpdf(.69,3) = .066309. This should “agree” with the above value in Table 2. Consider calclating the probability of a success occurring on years n = 1, 2, or 3 by using the cumulative distribution function geometcdf( . Code and commentary 2nd Dist to geometcdf( Enter You see geometcdf( Write in .69,3 Your see geometcdf(.69,3 Enter This gives geomtcdf(.69,3) = .970209. You should be able to express, and calculate this sum with a scientific calculator. Step by Step application of the Geometric Distribution. 1. State the problem. 2. Define a successful trial. Define a failure trial. 3. Give a probability p of a success and a probability q 10%of a failure, where p + q = 1. 4. Define the random variable N which is the number of trials until the first success. Verify or assume that the trials are random or independent. 5. Specify that P(N = n) = q n1 p as the probatility that the first success occurs on trial n. This could be named Geometpdf(p,n). Write and describe the numerical inputs. Input the numbers. 0% 6. Write the numerical formula for 3 geometpdf(p,n) = q n1 p = P(N=n) which gives the probability of the first success on the nth trial, or gemoetcdf(p,n) = P(n=1) + P(n=2) + … + P(N = n) which is the probability of a first success on or before the n th trial. Solve. 7. Write an interpretation of the solution in context, and check the conclusion. Example 4. The Binomial Distribution: What is the probability of two success of out of 5 2 trials, for p, p = = .2 and q = .8 ? We need to consider the number of combinations in which 5 5! 2 out of 5 can happen. This is 52 C = = 10. We will use (5/2) for 52C . 52 (2!)(3!) So there are 10 ways to get two successes out of 5 trials. Since there are only 10 of these, you could count them with a diagram. See the Exercises. Since these possibilities are disjoint, we can simply add the probabilities getting P(2 successes out of 5 trials) = (5/2)(.2 2)(.8 5 2 ) = .2084 . See the Exercises for calculations with a scientific calculator. In general for the binomial distribution P(x in n) = (n/x) pqx n x where p is the probability of a success and q = 1- p is probability of a failure in a single trial, and x successes out of n trials. Calculating P(x in n on the TI83), for P(x in n) = (n/x) we use three parameters, n trials, p = probability of a success, and x = the number of successes. binompdf(n, p, x) == (n/x) . For calculating the probability of x or fewer successes out of n trials, we could use the cumulative distribution function binomcdf(n, p, x). For the probability of 2 or fewer successes in 5 trials from above we use n = 5, p = .2, and x = 2. binomcdf(5, .2, 2) = .94208 . You will find binomcdf by the code 2nd Distr … . See the Exercises for calculating with a scientific calculator. Derivation of the mean = E(X) of a binomial distribution. We can consider 1 trial with the probability of success of p and probability of failure q. We will denote a success by 1 and a failure by 0. Mean = [P(0)][0] + [P(1)][1] = q[0] + p[1] = p. Consider a succession of one trial events XXX1, 2, ... n We know 4 E(X) = EXEXEX(12 ) ( ) ... (n ) = p + p + … +p = np . See the Exercises for examining this claim. Example 5. We return to the p = probability of a return less than 0% = .31 in the above Example 2. The probability of a return 0% = .69 . We ask for the probability of a 0% return in 2 out of 3 trials with p = .69 and q = .31. By the binomial distribution, 3! P(2 successes out of 3 trials) = (3/2)p 2q (3 2) (.69)2 (.31) .44277 We check: (2!)(1!) binompdf(n, p, q) = binompdf(3,.69,2) = .44277 . Resorting to the Normal Distribution. Consider with p = .06, P(2000 out of 35000) = (35000/2000)(.06 2000)(.94 33,000 ) . When calculating binompdf(35000, .06, 2000) we get ERR:DOMAIN. The numbers are too large for the calculator. Assume we wanted P(n 2000 ), this would require even more calculations. We will use a Normal Distribution with mean = np = 2100 and sd = npq 44.43 . Resorting to the 2000 2100 Normal Distribution, P(n<2000) = P[ z ] = P(z < -2.25) = .012. The Normal 44.43 Approximation gives a good approximation if np and nq are large enough. (See Bock in References.) Bock says if np = 10 and nq = 10, or greater, you get a good approximation. See the Exercises. Side Bar Notes: Lee reports some Spearman Rank Correlation Tests of randomness of annual returns on a stock in which the hypothesis of randomness could not be rejected (pp.772 to 779). See References for Lee. He reports a Chi Square Test of Independence for Cross Classified Attributes (p. 513.) Two events are said to be independent when the probability of one event is not affected by the occurrence of the other. Lee page 168, Bock p. 388. See the References. Studies have been conducted showing that in some cases, stocks that do well, continued to do well for a period of time. People forget to help themselves. One cost of helping yourself is for financial responsibility and retirement, and knowledge of financial mathematics.