Dr. V.R. Bencivenga Economics 329 Spring 2012 MIDTERM #2: RIGHT ANSWERS 1. a. Define B = rate of return on bond fund. P(B > 3) = .15 + .1 + .05 = .3 Bond mutual fund 1 3 5 1 .05 .1 .15 .3 Stock mutual fund 4 .1 .2 .1 .4 7 .15 .1 .05 .3 .3 .4 .3 Define S = rate of return on stock fund. P(S > 4) = .15 + .1 + .05 = .3 Bond mutual fund 1 3 5 1 .05 .1 .15 .3 Stock mutual fund 4 .1 .2 .1 .4 7 .15 .1 .05 .3 .3 .4 .3 b. P(B > 3  S > 4) = .05 Bond mutual fund 1 3 5 1 .05 .1 .15 .3 Stock mutual fund 4 .1 .2 .1 .4 7 .15 .1 .05 .3 .3 .4 .3 P(B > 3  S > 4) = .15 + .1 + .05 + .15 + .1 = .55 Bond mutual fund 1 3 5 1 .05 .1 .15 .3 Stock mutual fund 4 .1 .2 .1 .4 7 .15 .1 .05 .3 .3 .4 .3 c. No, B and S are not independent. Independence requires P(Bi  Sj) = P(Bi)P(Sj) for all i, j pairs. However (as one example), P(B = 1  S = 1) = .05 ≠ P(B = 1)P(S = 1) = (.3)(.3) = .09. Probability distributions are not required for full credit.d. Bond mutual fund 1 3 5 1 .05 .1 .15 .3 Stock mutual fund 4 .1 .2 .1 .4 7 .15 .1 .05 .3 .3 .4 .3 marginal distribution of B marginal distribution of S E[B] = 1(.3) + 3(.4) + 5(.3) = 3 Var(B) = (1 – 3)2(.3) + (3 – 3)2(.4) + (5 – 3)2(.3) = 2.4 E(S) = 1(.3) + 4(.4) + 7(.3) = 4 Var(S) = (1 – 4)2(.3) + (4 – 4)2(.4) + (7 – 4)2(.3) = 5.4 e. Conditional probability distribution of S given B = 5: S 1 4 7 P(S|B = 5) .15/.3 = .5 .1/.3 = 1/3 .05/.3 = 1/6 E(S|B = 5) = 1(.5) + 4(1/3) + 7(1/6) = 3 Var(S|B = 5) = (1 – 3)2(.5) + (4 – 3)2(1/3) + (7 – 3)2(1/6) = 5 f. Note that only four terms out of nine are non-zero—in the remaining five terms, at least one of the two random variables B and S has a value equal to its mean, and so that deviation equals zero. Cov(B,S) = (1-3)(1-4)(.05) + (5 – 3)(1 – 4)(.15) + (1 – 3)(7 – 4)(.15) + (5 – 3)(7 – 4)(.05) = – 1.2 g. Define X = rate of return on the portfolio (assuming equal investment in bond fund and stock fund) )SB(21X  therefore 5.3)43(21])S[E]B[E(21]X[E  Bond mutual fund 1 3 5 1 X = 1 X = 2 X = 3 .05 .1 .15 Stock mutual fund 4 X = 2.5 X = 3.5 X = 4.5 .1 .2 .1 7 X = 4 X = 5 X = 6 .15 .1 .05 P(X > 3.5) = .1 + .15 + .1 + .05 = .4 h. Define Y = dollar value of your portfolio one year from now (assuming $1000 in bond fund, $2000 in stock fund) S20B103000100S12000100B11000Y       31101103000)4(20)3(10SE20BE103000YE  1920)2.1(400)4.5(400)4.2(100)S,B(Cov)20)(10(2)S(Var20)B(Var10)Y(Var222. The probability that at least one defective item is selected is one minus the probability that none of the items selected is defective. Let X = number of defective items selected. X is a hypergeometric random variable. Then 63062604CCC1)0X(P1)1X(P- Let Y = number of items of extremely high quality selected. Y is a hypergeometric random variable. Then 630612018CCC)6Y(P- 3. a. X = number of teams to succeed in developing the new product X ~ b(n= 4, p = .07) The new product will be developed if at least one team develops the new product 252.)93)(.07(.C1)0X(P1)1X(P4004 b. Find n such that 3.)93)(.07(.C1)0X(P1)1X(Pn00n From part a, we know n = 4 is too small. Try n = 5: 3043.)93)(.07(.C1)1X(P5005 Therefore, the answer is n = 5. c. If the firm sets up n teams, expected profits minus costs of the teams equals: expected profits minus costs = – $1n + $20 P(X  1) + $0 P(X = 0) = – n + 20(1 – P(X = 0)) = – n + 20(1 – nC0(.070)(.93n) = – n + 20(1 – .93n) Evaluating we have: n expected profits minus costs 4 1.039 5 1.086 answer 6 1.060 NOT REQUIRED (but part of a full answer, if you ever encounter a similar problem in the real world!): The net profit function is concave in n since (i) expected benefits are positive for n = 1, and (ii) 20(1 - .93n) starts at zero and asymptotes to 20, while – n is negative and unbounded in absolute value. Therefore, n = 5 is the number of teams that maximizes net profits.4. a. X = number of defaults is a binomial random variable with n = 10 and p = .08 P(X = 0) = 10C0 .080 .9210 = .434 P(X = 1) = 10C1 .081 .929 = .378 P(X > 1) = 1 – .434 – .378 = .188 b. P(X = 0) = 15C0 .080 .9215 = .286 (smaller) c. Find the smallest n such that P(X = 0) .2 nn00n92.92.08.C)0X(P     20n0834.609.1n2.ln}92.{lnn2.92.n Note the direction of the inequality changes when we divide through by ln(.92), which is negative. Also, note we must round the solution for n up to the nearest integer. It is also acceptable to answer this question with a “trial and error” approach. 5. a. Since we want to calculate the probability of a certain number of occurrences per unit time, and the assumptions of the Poisson random variable are met, X = number of schools of fish found is a Poisson random variable. The Poisson parameter is the expected number of schools of fish located in one day of searching. The value of this parameter is: (area searched per day)/(average area per school) = 10,000/100,000 = .1 school per day b. 0952.9048.1!0)1(.e1)0X(P1)1X(P01. c. The value of the Poisson parameter must be adjusted for the second calculation. The expected number of schools of fish located in three days of searching is .3. 2592.7408.1!0)3(.e1)0X(P1)1X(P03. 6. a. State and graph the probability density function of X. The number of days ranges from 1 to 7 (the length of the interval of possible number of days is 6). otherwise07x1if6/1X f(x)