Name SOLUTION Business Statistics Midterm Exam Fall 2013 Russell Do not turn over this page until you are told to do so. You will have 2 hours to complete the exam. There are a total of 100 points divided into three parts. The true and false questions are worth 10 points, the multiple choice are 3 points each for a total of 30 points and the long answer questions are worth 60 points. You can use one side of an 8.5x11 sheet of notes during the exam. No other notes are permitted. You may use a calculator. Please write clearly and provide answers in the space provided. If you need additional space use the back of the exam pages and clearly organize your work. Students in my class are required to adhere to the standards of conduct in the Booth Honor Code and the Booth Standards of Scholarship. The Booth Honor Code also requires students to sign the following Booth Honor pledge, "I pledge my honor that I have not violated the Honor Code during this examination. I also understand that discussing the contents of this exam before all students have completed the exam would be a violation of the Honor Code". Please sign here to acknowledge
I. True or False Clearly indicate the best answer by circling T or F indicating that the statement is true or false respectively. If neither T nor F is clearly indicated the problem will be marked as incorrect. Each problem is worth 1 point. T F For a given set of returns data, converting the data from returns, to returns measured in percent will result in a sample mean and sample standard deviation that is 100 times larger. T F If a histogram has a long right tail and short left tail, the median will be above the mean. T F If the sample covariance is near 1 then there must be a very strong positive relationship between the two variables. T F For any continuous random variable X, X Pr.68. T F If two random variables X and Y are independent then the correlation must be zero. T F If you play 100 slot machines and whether or not you win is an iid Bernoulli with probability.01, the probability that you don t get any winners is.366. T F T F T F T F Standardized values from a data set obtained by subtracting the mean and dividing by the standard deviation that are larger than three are always very rare events. If X and Y are identically distributed random variables, they have the same mean and variance. Daily temperatures in Chicago could be modeled well as an IID process. For a sample average ( x ) constructed from 50 iid draws from the random E X x. variable X, 2
II. Multiple choice: Clearly circle the answer that is best. Each problem is worth 3 points for a total of 30. No partial credit will be given in this section. If no answer is clearly circled the problem will be marked as incorrect. For problems 1 and 2 consider the data set 1 3 2 3 5. 1. What is the sample average? a. 3 b. 2 c. 2.5 d. 2.8 2. What is the sample variance? a. 2.1 b. 2.2 c. 2.3 d. 4 Refer to the following table for questions 3 and 4. Let X denote the number of service visits required to fix a problem with a product purchased from a company. x 1 2 3 4 P(x).6.2.15.05 3. The probability that it will take more than 2 visits? a..8 b..2 c..55 d. 1.0 4. Given that it takes more than one visit, what is the probability that it will take 4 visits? a..5625 b..3352 c..3751 d..125 3
Suppose annual returns for a mutual fund are N(.1792,.12 2 ). Use the following table to answer questions 5-8. Normal with mean = 0.179200 and standard deviation = 0.120000 x P( X <= x) -0.1000 0.0100 0.0000 0.0677 0.1000 0.2546 0.2000 0.5688 0.3000 0.8430 0.4000 0.9671 5. What is the probability that the return on the fund is negative? a. 23.34% b. 6.77% c. 1% d. 93.33% 6. What is the probability that the return on the fund is between 0 and.2? a. 49.11% b. 35.43% c. 56.88% d. 50.11% 7. What is the probability that the return will be larger than.2? a. 0.7652 b. 0.2134 c. 0.5375 d. 0.4312 8. Find the value such that there is a 1% chance that the return is less than that value. a. 0 b..1 c..3 d. -.1 f. None of the above. 4
Use the following histogram for problems 9 and 10. 12 Histogram of x 10 Frequency 8 6 4 2 0-4 -2 0 2 x 4 6 8 9. The distribution looks roughly normal with mean and standard deviation given by: a. =2 and =4 b. =3 and =4 c. =2 and =3 d. =3 and =3 10. By viewing the histogram above, what can you say about the data: a. iid b. Independent but not identically distributed c. Identically distributed but not independent d. Dependent e. Not enough information to tell. 5
III. Long answer questions. Try to do work in the space provided under each question and be show all work in order to facilitate partial credit. Be sure to clearly indicate your final answer and complete all computations. This section is worth 60 points. Problem 1. (12 points) Here are the summary statistics for the returns on two assets Descriptive Statistics: usa, france Variable N N* Mean StDev usa 107 0 0.01346 0.03328 france 107 0 0.01383 0.05494 Covariances: france, usa france usa france 0.00301820 usa 0.00091587 0.00110774 Consider a portfolio that invests 20% into a risk free asset that will obtain a.05 return over the next year, 50% in the USA asset and 30% in the French asset. Hence, P.2(.05).5 R.3R. USA France a. Find the mean of P E(P) = 0.2*0.05 + 0.5*E(R USA ) + 0.3* E(R France ) = 0.2*0.05 + 0.5*0.01346 + 0.3*0.01383 = 0.0209 b. Find the variance of P. Riskfree rate is a constant no variance or covariance with the risky assets. Var(P) = (0.5 2 )*Var(R USA ) + (0.3 2 )* Var(R France ) + 2*0.5*0.3*Covar(R USA, R France ) = (0.5 2 )*(0.03328 2 ) + (0.3 2 )*(0.05494 2 ) + 2*0.5*0.3*0.00091587 = 0.000823 6
Suppose instead that you borrow at the risk free rate and use the money to invest in the USA and France assets. Suppose that your weights are P.2(.05).75 R.45R. c. What is the mean of P now? E(P) = -0.2*0.05 + 0.75*E(R USA ) + 0.45* E(R France ) = (-0.2)*0.05 + 0.75*0.01346 + 0.45*0.01383 = 0.00632 USA France d. What is the variance of P now? Var(P) = (0.75 2 )*Var(R USA ) + (0.45 2 )* Var(R France ) + 2*0.75*0.45*Covar(R USA, R France ) = (0.75 2 )*(0.03328 2 ) + (0.45 2 )*(0.05494 2 ) + 2*0.75*0.45*0.00091587 = 0.00185 7
Problem 2. (26 points) Consider the following joint probability distribution for an indicator variable for whether the economy is in an expansion or a contraction in quarter i and quarter i+1. X i =1 denotes a expansion and X i =0 denotes a contraction in quarter i. The series is a first order Markov model so that Pr X i 1 Xi, Xi 1,... X0 Pr Xi 1 Xi. Consider the following joint model for X i and X i+1 : X i 0 1 X i+1 0.35.15 1.15.35 a. Are consecutive expansion/contraction indicators dependent or independent? Explain. Pr(X i = 0) = 0.35 + 0.15 = 0.5; Pr(X i+1 = 0) = 0.35 + 0.15 = 0.5 Pr(X i = 0, X i+1 = 0) = 0.35 is not equal to the product of the marginals, hence dependent b. Explain why X i and X i+1 are identically distributed and state the distribution. Pr(X i = 0) = Pr(X i+1 = 0) = 0.5 Pr(X i = 1) = Pr(X i+1 = 1) = 0.5 They have the same (unconditional) model c. What is the probability of an expansion? Pr(X i = 1) = 0.5 d. What is the mean of X i? E(X i ) = Pr(X i = 0)*0 + Pr(X i = 1)*1 = 0.5*0 + 0.5*1 = 0.5 e. What is the variance of X i? Var(X i ) = Pr(X i = 0)*(0 E(X i )) 2 + Pr(X i = 1)* (1 E(X i )) 2 = 0.5*0.5 2 + 0.5*0.5 2 = 0.25 8
f. Without doing any calculations, what is the sign of the correlation between X i and X i+1? State you logic. Positive, since 70% (sum of diagonals) of the time they are the same value. There is a good chance they are the same value. g. What is the conditional distribution of X i+1 given quarter i is a expansion quarter? Pr(X i+1 = 0 X i = 1) = 0.15/0.5 = 0.3 Pr(X i+1 = 1 X i = 1) = 0.35/0.5 = 0.7 h. What is the conditional distribution of X i+1 given that quarter i is a contraction? Pr(X i+1 = 0 X i = 0) = 0.35/0.5 = 0.7 Pr(X i+1 = 1 X i = 0) = 0.15/0.5 = 0.3 i. What is the expected value for X i+1 given that quarter i is a contraction? E(X i+1 X i = 0) = Pr(X i+1 = 0 X i = 0)*0 + Pr(X i+1 = 1 X i = 0)*1 = 0.7*0 + 0.3*1 = 0.3 j. What is the probability of getting the sequence 0, 0, 0, 1, 1, 1, 1? Markov, so only depend on the immediately past observation. Pr(X 1 = 0)*Pr(X 2 = 0 X 1 = 0)*Pr(X 3 = 0 X 2 = 0)*Pr(X 4 = 1 X 3 = 0)*Pr(X 5 = 1 X 4 = 1)*Pr(X 6 = 1 X 5 = 1)*Pr(X 7 = 1 X 6 = 1) = (0.5)*(0.7)*(0.7)*(0.3)*(0.7)*(0.7)*(0.7) = 0.0252 9
Problem 3. (10 points) Whether or not any customer decides to purchase a phone after entering the store is iid Bernoulli with probability.2. Suppose on Friday 76 potential customers enter the store and on Saturday 122 customers enter the store. a. Let Y Fri denote the total phone purchases on Friday. What is the model fory Fri? Binomial(76, 0.2) b. Let Y Sat denote the total phone purchases on Friday. What is the model fory Sat? Binomial(122, 0.2) c. Let T denote the total sales on Friday and Saturday. What is the model for T? Binomial(198, 0.2) d. What are the mean and variance of T? E(T) = np = 198*0.2 = 39.6 Var(T) = np(1 p) = 31.68 10
Problem 4. (12 points) Let s consider an extension of the random walk model yt yt 1 xt that is a little more realistic. Let s allow for the variance of the change in the asset price to vary over time. Typically we find that declining prices tend to be associated with more volatile stock returns. We capture this by saying that x t has a distribution that depends on the past values of x t in the following way: x t Pr(x t x t-1 <0) And 1.4 0.2-1.4 x t P x x 0 t t 1 1.3 0.40-1.3 a. What are the mean and variance of the change x t following declining prices? E(x t x t-1 < 0) = 0.4*1 + 0.2*0 + 0.4*(-1) = 0 Var(x t x t-1 < 0) = 0.4*(1 0) 2 + 0.2*(0 0) 2 = 0.4*(-1 0) 2 = 0.8 b. What are the mean and variance of the change in x t following rising (or no change in) prices? E(x t x t-1 >= 0) = 0.3*1 + 0.4*0 + 0.3*(-1) = 0 Var(x t x t-1 >= 0) = 0.3*(1 0) 2 + 0.4*(0 0) 2 = 0.3*(-1 0) 2 = 0.6 11
c. Now, given that y t-1 =100 and y t-2 =99 find the conditional model for y t. Since y t-1 = y t-2 + x t-1, this means x t-1 is 1 and is greater than zero. We need to use Pr(x t x t-1 >=0). y t = y t-1 + x t = 100 + x t y_t Pr(y_t y_t 1, y_t 2) 101 0.3 100 0.4 99 0.3 d. Again, given that y t-1 =100 and y t-2 =99 find the conditional model for y t+1. Using part c, we have to try different cases for x t is greater than or equal to zero, versus if x t is less than zero. If x t <0, we have to use the conditional probabilities in the first T- chart. If x t >=0, we have to use the conditional probabilities in the second T-chart. We end up with an asymmetrical conditional distribution. y_t+1 Pr(y_t+1 y_t 1, y_t 2) 102 0.09 101 0.24 100 0.37 99 0.18 98 0.12 e. Suppose that you are holding a call option at 101 and a put option at 99 (this is called a straddle). This position will deliver a positive payoff when the price rises above 101 or falls below 99. What is the chance that this straddle position delivers a positive payoff in period t+1 if y t-1 =100 and y t-2 =99? This is just asking for the probability that y t+1 is above 101 or below 99, i.e. Pr(y t+1 > 101) + Pr(y t+1 <99) = 0.09 + 0.12 = 0.21 12