Solutions Midterm Exam II - Fall 2004 Statistics 1011. A. First compute E(XY )=1(0.1)+2(0.1)+3(0)+4(0.3)+6(0.3)+9(0.2) =5.1. Hence Cov(X, Y )=5.1 − (2.2)(2.2) = 0.26,andsoCor(X, Y )=0.26/0.46 = 0.565. Thus ratings tend to go in the same direction; there isa moderate linear relationship between the two ratings.B. Plainly the expectation will be the same as the original expectation,viz. 2.2. To compute the standard deviation we use the relationshipVar(X/2+Y/2) = (1/4) Var(X)+(1/4) Var(Y )+(1/2) Cov(X, Y ).Thusfrom the first part we have Var(X/2+Y/2) = (1/2).46 + (1/2)0.26 = 0.36.Thus SD=0.6. Since there is a positive correlation, independence wouldreduce the overall variance, but have no effect on the mean.C. First we compute the conditional probabilities P(rating = k|same),and these are 0.1/0.6=1/6, 0.3/0.6=1/2, and 0.2/0.6=1/3 for k =1, 2, 3respectively. Hence E(rating|same)=1(1/6) + 2(1/2) + 3(1/3) = 216.2. A. P (X>9) =R1093x2/1000dx =0.271.B. (i) We assume a binomial model. Thus X ∼ B(100, 0.271) and P (20 ≤X)=P100k=20¡100k¢0.271k0.729n−k.(ii) We use the central limit theorem, X ≈ N(27.1, 27.1(0.729)),andcom-pute P (20 ≤ X)=P ((20 − 27.1)/p(4.4476) ≤ Z)=P (−1.597 ≤ Z)=0.9452.C. (i) OK, integrate xkto get xk+1/(k+1). Evaluating the definite integralwe obtain 10k+1/(k +1) and so the normalizing constant is (k +1)/10k+1.(ii) E(X)=(k +1)/10k+1)R100xk+1dx = 10(k +1)/(k +2).Wesetthisequal to 8 and solve to get k =3.3. A. Of course it depends on one’s stomach for risk. If we consider averagerisk, then since expected profit is 14(0.2)+(-1)0.8=2>0 we would go forit. However, if we consider downside risk, then since the probability oflosing million smackers is 0.8, which is really quite large so we ma y decideagainst this venture.B. Hopefully we have automatic response to this problem: CLT. Use anormal approximation: the E(profit) on one venture is 2 and the standarddeviation is easily worked out to be 6 (lose 1 unit with probability .8 andgain 14 units with probability .2). Th u s, since n=25, using the centrallimit theorem, with X ∼ N(50, 302) we compute P (X ≤ 0) = P (Z ≤−50/30) = 0.0475C. (i) Our friend Bayes theorem once again: we want P (success|makes 2nd round).Displaying the probabilities on a tree diagram or joint table we see thatP (success|mak es 2nd round)=1(0.2)/(1(0.2) + 0.5(0.8)) = 1/3.(ii) We want a conditional expectation: E(profit|wait)=5(1/3)+(−1)2/3=1, less than before so not so smart to wait.14. A.(i) Putting P into first we have E(X)=P (8) + (1 − P )(4) ≥ 5 whenP ≥ 1/4.(ii) Taking P =1/2, for example, we have Var(CREF/2+TIAA/2) =Var(CREF)/4+0=202/4 = 100, and so SD=10.B. (i)Of course we need to consider the volatility drag, viz. µ − σ2/2.With the portfolio CREF/2+TIAA/2,wehaveµ =0.08/2+.04/2=.06and σ2= .22/4=.01,andsoµ − σ2/2=.06 − .01/2=.055.(ii) In general we have µ − σ2/2=0.08P +0.04(1 − P ) − P2(0.04)/2=0.04 + 0.04P − 0.02P2. Recalling that a quadratic ax2+ bx + c has amaxim um at x = −b/(2a) we see that µ − σ2/2 has a maxim um at -0.04/(-0.04)=1. So put all your dough in the risky investment.5. A.Hereweuseourfavouritebinomialmodel: WewantP (X =2)+P (X =3) =¡32¢0.32(0.7) + 0.33+=0.216.B. This one demands a wee bit more concentration, a bit of subtle parti-tioning. (i) Using the partition rule we haveP (impl)=P (impl|CEO)P (CEO)+P (impl|CEOc)P (CEOc)=1(0.3) + 0.5(0.32)0.7=0.3315(This approach has the risk of not counting all the possibilities so here arethe possibilities)CEO CEO COO Probability Implemented ? NumberY Y Y 0.027 Y 3Y Y N 0.063 Y 2Y N Y 0.063 Y 2Y N N 0.147 Y 1N Y Y 0.063 HALF 2N Y N 0.147 N 1N N Y 0.147 N 1N N N 0.343 N 0i) Probability = .027+.063+.063+.147 +.063/2=.3315 as computedabove.ii) x p(x)1 .147/.3315= .44342 (.063+.063+.063/2)/.3315= .47513 .027/.3315=
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