Capítulo 7: Problemas e Exercícios Estatísticos

104 Chapter 7 Problems 1 Let X 1 if the coin toss lands heads and let it equal 0 otherwise Also let Y denote the value that shows up on the die Then with pi j PX i Y j Ereturn 6 6 1 1 2 1 0 2 j j j jp j p j 1 42 105 12 52512 2 a 6 6 9 324 b X 6 S6 W9 R c EX 666PS 0 W 0 R 3 639PS 0 W 3 R 0 369PS 3 W 0 R 0 657PS 0 W 1 R 2 567PS 1 W 0 R 2 648PS 0 W 2 R 1 468PS 2 W 0 R 1 549PS 1 W 2 R 0 459PS 2 W 1 R 0 558PS 1 W 1 R 1 9 6 9 6 6 1 216 324 420 6 384 9 360 6 200669 21 3 3 2 2 2 3 1988 3 If the first win is on trial N then the winnings is W 1 N 1 2 N Thus a PW 0 PN 1 12 b PW 0 PN 2 14 c EW 2 EN 0 4 1 1 2 0 0 0 1 1 0 0 0 1 1 0 0 0 1 2 16 1 2 1 4 1 1 2 y y y E XY xy dxdy y dy y E X x dxdy y dy y E Y y dxdy ydy y Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 105 5 The joint density of the point X Y at which the accident occurs is fx y 1 9 32 x y 32 fx fy where fa 13 32 a 32 Hence we may conclude that X and Y are independent and uniformly distributed on 32 32 Therefore EX Y 3 2 3 2 3 2 0 1 4 2 3 2 3 3 x dx xdx 6 10 10 1 1 i i i i E X E X 1072 35 8 Enumber of occupied tables 1 1 N N i i i i E X E X Now EXi Pith arrival is not friends with any of first i 1 1 pi1 and so Enumber of occupied tables 1 1 1 N i i p 7 Let Xi equal 1 if both choose item i and let it be 0 otherwise let Yi equal 1 if neither A nor B chooses item i and let it be 0 otherwise Also let Wi equal 1 if exactly one of A and B choose item i and let it be 0 otherwise Let X 10 1 i i X Y 10 1 i i Y W 10 1 i i W a EX 10 1 i i E X 103102 9 b EY 10 1 i i E Y 107102 49 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 106 Chapter 7 c Since X Y W 10 we obtain from parts a and b that EW 10 9 49 42 Of course we could have obtained EW from EW 10 1 i i E W 102310710 42 9 Let Xj equal 1 if urn j is empty and 0 otherwise Then EXj Pball i is not in urn j i j 1 1 n i j i Hence a Enumber of empty urns 1 1 1 n n j i j i b Pnone are empty Pball j is in urn j for all j 1 1 n j j 10 Let Xi equal 1 if trial i is a success and 0 otherwise a 6 This occurs when PX1 X2 X3 1 It is the largest possible since 18 1 3 1 i i P X P X Hence PXi 1 6 and so PX 3 PX1 X2 X3 1 PXi 1 6 b 0 Letting X1 1 if 6 0 otherwise U X2 1 if 4 0 otherwise U X3 1 if 3 0 otherwise U Hence it is not possible for all Xi to equal 1 11 Let Xi equal 1 if a changeover occurs on the ith flip and 0 otherwise Then EXi Pi 1 is H i is T Pi 1 is T i is H 21 pp i 2 Enumber of changeovers 1 n i i i E X E X 2n 11 p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 107 12 a Let Xi equal 1 if the person in position i is a man who has a woman next to him and let it equal 0 otherwise Then EXi 1 if 1 2n 2 2 1 1 1 2 1 otherwise 2 2 12 2 n i n n n n n Therefore 1 n i i E X 2 1 n i i E X 1 2 3 2 2 2 2 1 4 2 n n n n n 3 2 4 2 n n n b In the case of a round table there are no end positions and so the same argument as in part a gives the result 2 1 2 3 1 2 12 2 4 2 n n n n n n n where the right side equality assumes that n 1 13 Let Xi be the indicator for the event that person i is given a card whose number matches his age Because only one of the cards matches the age of the person i 1000 1000 1 1 i i i i E X E X 1 14 The number of stages is a negative binomial random variable with parameters m and 1 p Hence its expected value is m1 p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 108 Chapter 7 15 Let Xij i j equal 1 if i and j form a matched pair and let it be 0 otherwise Then EXij Pi j is a matched pair 1 n n 1 Hence the expected number of matched pairs is 1 1 2 1 2 i j i j i j i j n E X E X n n 16 EX 2 2 2 2 1 2 2 x y y x e y e dy π π 17 Let Ii equal 1 if guess i is correct and 0 otherwise a Since any guess will be correct with probability 1n it follows that EN 1 1 n i i E I n n b The best strategy in this case is to always guess a card which has not yet appeared For this strategy the ith guess will be correct with probability 1n i 1 and so EN 1 1 1 n i n i c Suppose you will guess in the order 1 2 n That is you will continually guess card 1 until it appears and then card 2 until it appears and so on Let Ji denote the indicator variable for the event that you will eventually be correct when guessing card i and note that this event will occur if among cards 1 thru i card 1 is first card 2 is second and card i is the last among these i cards Since all i orderings among these cards are equally likely it follows that EJi 1i and thus EN 1 1 1 n n i i i E J i 18 Enumber of matches 52 1 1 match on card 0 i i i E I I 52 1 13 4 since EIi 113 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 109 19 a Etime of first type 1 catch 1 1 1 1 p using the formula for the mean of a geometric random variable b Let Xj 1 a type is caught before a type 1 0 otherwise j Then 1 1 j j j j E X E X 1 type before type 1 j P j 1 1 j j j P P P where the last equality follows upon conditioning on the first time either a type 1 or type j is caught to give Ptype j before type 1 Pjj or 1 1 j j P P P 20 Similar to b of 19 Let Xj 1 ball removed before ball 1 0 j 1 1 1 ball before ball 1 j j j j j E X E X P j 1 or 1 j P j j 1 1 j W j W W j Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 110 Chapter 7 21 a 3 97 100 1 364 365 3 365 365 b Let Xj 1 if day is someones birthday 0 j 100 365 365 1 1 364 365 1 365 j j E X E X 22 From Example 3g 1 6 6 6 6 6 5 4 3 2 23 5 8 5 8 1 1 1 1 i i i i E X Y E X E Y 2 3 3 147 5 8 11 20 120 110 24 Number the small pills and let Xi equal 1 if small pill i is still in the bottle after the last large pill has been chosen and let it be 0 otherwise i 1 n Also let Yi i 1 m equal 1 if the ith small pill created is still in the bottle after the last large pill has been chosen and its smaller half returned Note that X 1 1 n m i i i i X Y Now EXi Psmall pill i is chosen after all m large pills 1m 1 EYi Pith created small pill is chosen after m i existing large pills 1m i 1 Thus a EX nm 1 1 1 1 m i m i b Y n 2m X and thus EY n 2m EX 25 PN n PX1 X2 Xn 1 n EN 1 1 1 n n P N n e n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 111 26 a Emax 1 0 max P t dt 1 0 1 max P t dt 1 0 1 1 n n t dt n b Emin 1 0 min 4 p t t 1 0 1 1 1 t n dt n 27 Let X denote the number of items in a randomly chosen box Then with Xi equal to 1 if item i is in the randomly chosen box EX 101 101 1 1 101 10 10 i i i i E X E X Hence X can exceed 10 showing that at least one of the boxes must contain more than 10 items 28 We must show that for any ordering of the 47 components there is a block of 12 consecutive components that contain at least 3 failures So consider any ordering and randomly choose a component in such a manner that each of the 47 components is equally likely to be chosen Now consider that component along with the next 11 when moving in a clockwise manner and let X denote the number of failures in that group of 12 To determine EX arbitrarily number the 8 failed components and let for i 1 8 Xi 1 if failed component is among the group of 12 components 0 otherwise i Then X 8 1 i i X and so EX 8 1 i i E X Because Xi will equal 1 if the randomly selected component is either failed component number i or any of its 11 neighboring components in the counterclockwise direction it follows that EXi 1247 Hence EX 81247 9647 Because EX 2 it follows that there is at least one possible set of 12 consecutive components that contain at least 3 failures Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 112 Chapter 7 29 Let Xii be the number of coupons one needs to collect to obtain a type i Then 1 2 3 4 1 2 3 4 1 2 3 4 8 12 83 34 min 4 min 2 12 34 min 43 min 85 34 min 87 12 min 1 i i i j j i E X i E X i E X X E X X i j E X X E X X X j E X X X i E X X X X a Emax Xi 2 8 2 83 4 4 2 43 2 85 2 87 1 437 35 b EmaxX1 X2 8 8 4 12 c EmaxX3 X4 83 83 43 4 d Let Y1 maxX1 X2 Y2 maxX3 X4 Then EmaxY1 Y2 EY1 EY2 EminY1 Y2 giving that EminY1 Y2 12 4 437 123 35 35 30 EX Y2 VarX Y VarX VarY 2σ2 31 10 1 1 Var 10 Var i i X X Now VarX1 2 2 1 7 2 E X 1 4 9 16 25 366 494 3512 and so 10 1 Var i i X 35012 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 113 32 Use the notation in Problem 9 X 1 n j j X where Xj is 1 if box j is empty and 0 otherwise Now with EXj PXj 1 1 1 n i j i we have that VarXj EXj1 EXj Also for j k EXjXk 1 1 1 1 2 k n i j i k i i Hence for j k CovXj Xk 1 1 1 1 2 1 1 1 1 k n n n i j i k i j i k i i i i VarX 1 1 2Cov n j j j k j E X E X X X 33 a EX2 4X 4 EX2 4EX 4 VarX E2X 4EX 4 14 b Var4 3X Var3X 9VarX 45 34 Let Xj 1 if couple are seated next to each other 0 otherwise j a 10 1 2 20 1019 19 j E X PXj 1 2 19 since there are 2 people seated next to wife j and so the probability that one of them is her husband is 2 19 b For i j EXiXj PXi 1 Xj 1 PXi 1PXj 1Xi 1 2 2 19 18 since given Xi 1 we can regard couple i as a single entity Var 2 10 1 2 2 2 2 2 10 1 10 9 19 19 19 18 19 j j X Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 114 Chapter 7 35 a Let X1 denote the number of nonspades preceding the first ace and X2 the number of nonspades between the first 2 aces It is easy to see that PX1 i X2 j PX1 j X2 i and so X1 and X2 have the same distribution Now EX1 48 5 by the results of Example 3j and so E2 X1 X2 106 5 b Same method as used in a yields the answer 39 265 5 1 14 14 c Starting from the end of the deck the expected position of the first from the end heart is from Example 3j 53 14 Hence to obtain all 13 hearts we would expect to turn over 52 53 14 1 13 14 53 36 Let Xi 1 roll lands on 1 0 otherwise i Yi 1 roll lands on 2 0 otherwise i CovXi Yj EXi Yj EXiEYj i 1 since X 0 when 36 1 1 0 36 36 j i j Y i j i j Cov Cov i j i j i j i j X Y X Y 36 n 37 Let Wi i 1 2 denote the ith outcome CovX Y CovW1 W2 W1 W2 CovW1 W1 CovW2 W2 VarW1 VarW2 0 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 115 38 EXY 2 0 0 2 x y e x dydx 2 2 2 0 0 1 3 1 8 8 4 x y x e dx y e dy Γ EX 2 0 0 2 x x x x e xf x dx f x dy x 2e2x 1 2 EY 2 0 0 2 x Y Y e yf y dy f y dx x 2 0 2 x y y e dxdy x 2 0 0 2 x y e x dydx x 2 2 0 1 2 1 4 4 4 xe x dx ye dy Γ CovX Y 1 1 1 1 4 2 4 8 39 CovYn Yn VarYn 3σ2 CovYn Yn1 CovXn Xn1 Xn2 Xn1 Xn2 Xn3 CovXn1 Xn2 Xn1 Xn2 VarXn1 Xn2 2σ2 CovYn Yn2 CovXn2 Xn2 σ2 CovYn Ynj 0 when j 3 40 fYy ey 1 x y y e dx e y In addition the conditional distribution of X given that Y y is exponential with mean y Hence EY 1 EX EEXY EY 1 Since EXY EEXYY EYEXY EY2 2 since Y is exponential with mean 1 it follows that EY2 2 Hence CovX Y 2 1 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 116 Chapter 7 41 The number of carp is a hypergeometric random variable EX 60 10 6 VarX 2080 3 7 336 99 10 10 99 from Example 5c 42 a Let Xi 1 pair consists of a man and a woman 0 otherwise i EXi PXi 1 10 19 EXiXj PXi 1 Xj 1 PXi 1PXj 1X2 1 10 9 19 17 i j 10 1 100 19 i E X Var 2 10 2 1 10 10 10 9 10 900 18 10 1 10 9 19 19 19 17 19 17 19 Xi b Xi 1 pair consists of a married couple 0 otherwise i EXi 1 19 EXiXj PXi 1PXj 1Xi 1 1 1 19 17 i j 10 1 10 19 i E X Var 2 10 2 1 1 15 1 1 1 180 18 10 10 9 19 19 19 17 19 17 19 Xi 43 ER nn m 12 VarR 2 2 1 1 1 2 n m i i nm n m n m n m The above follows from Example 3d since when F G all orderings are equally likely and the problem reduces to randomly sampling n of the n m values 1 2 n m Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 117 44 From Example 8l n nm n m n m Using the representation of Example 2l the variance can be computed by using EI1Ilj 0 1 1 2 n m n n m n m n m 1 1 1 j n j EIiIij 0 1 1 1 2 3 mn m n n m n m n m n m 1 1 1 j n j 45 a 1 2 2 3 1 2 2 3 Cov 1 2 Var Var X X X X X X X X b 0 46 EI1I2 12 1 2 2 bank rolls Pbank rolls i E I I i i roll is greater than 2 bank rolls i P i P i 2 1 E I EI12 EI1 EI2 47 a It is binomial with parameters n 1 and p b Let xij equal 1 if there is an edge between vertices i and j and let it be 0 otherwise Then Di k i Xi k and so for i j CovDi Dj Cov i k r j k i r j X X i k r j k i r j Cov X X CovXi j Xi j VarXi j p1 p where the third equality uses the fact that except when k j and r i Xi k and Xr j are independent and thus have covariance equal to 0 Hence from part a and the preceding we obtain that for i j ρDi Dj 1 1 1 1 1 p p n p p n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 118 Chapter 7 48 a EX 6 b EXY 1 1 6 7 c 2 3 4 1 4 1 4 1 4 1 4 1 2 3 4 5 6 5 5 5 5 5 5 5 5 49 Let Ci be the event that coin i is being flipped where coin 1 is the one having head probability 4 and let T be the event that 2 of the first 3 flips land on heads Then PC1T 1 1 1 1 2 2 P T C P C P T C P C P T C P C 2 2 2 34 6 34 6 37 3 395 Now with Nj equal to the number of heads in the final j flips we have EN10T 2 EN7T Conditioning on which coin is being used gives EN7T EN7TC1PC1T EN7TC2PC2T 28395 49605 40705 Thus EN10T 60705 50 fXYxy 0 1 x y y x y x y y e e y y e e e y dx 0 x Hence given Y y X is exponential with mean y and so EX 2Y y 2y2 51 fXYxy 0 1 y y y e y y e y dx 0 x y EX 3Y y 3 3 0 1 4 y x dx y y 52 The average weight call it EW of a randomly chosen person is equal to average weight of all the members of the population Conditioning on the subgroup of that person gives EW 1 1 member of subgroup r r i i i i i E W i p w p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 119 53 Let X denote the number of days until the prisoner is free and let I denote the initial door chosen Then EX EXI 15 EXI 23 EXI 32 2 EX5 4 EX3 2 Therefore EX 12 54 Let Ri denote the return from the policy that stops the first time a value at least as large as i appears Also let X be the first sum and let pi PX i Conditioning on X yields ER5 12 5 2 i i E R X i p ER5p2 p3 p4 12 5 i i ip 7p7 5 6 36 E R 5436 6536 8536 9436 10336 11236 12136 5 6 36 E R 19036 Hence ER5 193 633 In the same fashion we obtain that ER6 6 10 1 36 36 E R 30 40 36 30 22 12 implying that ER6 17026 654 Also ER8 8 15 1 140 36 36 E R or ER8 14021 667 In addition ER9 9 20 1 100 26 36 E R or ER9 10016 625 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 120 Chapter 7 And ER10 10 24 1 64 36 36 E R or ER10 6412 533 The maximum expected return is ER8 55 Let N denote the number of ducks Given N n let I1 In be such that Ii 1 if duck is hit 0 otherwise i ENumber hitN n 1 n i i E I 10 1 6 1 1 n i i E I n n since given N n each hunter will independently hit duck i with probability 6n ENumber hit 10 6 0 6 1 1 6 n n n e n n 56 Let Ii 1 elevator stops at floor 0 otherwise i Let X be the number that enter on the ground floor 1 1 1 1 k N N i i i i N E I X k E I X k N N 10 1 0 1 10 k N k i i k N E I N N e N k N Ne10N N1 e10N 57 1 N i i E X E N E X 125 58 Let X denote the number of flips required Condition on the outcome of the first flip to obtain EX EXheadsp Extails1 p 1 11 pp 1 1p1 p 1 p1 p 1 pp Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 121 59 a Etotal prize shared Psomeone wins 1 1 pn1 b Let Xi be the prize to player i By part a 1 1 1 1 1 n n i i E X p But by symmetry all EXi are equal and so EX 1 1 pn1n 1 c EX p E11 B where B which is binomial with parameters n and p represents the number of other winners 60 a Since the sum of their number of correct predictions is n one for each coin it follows that one of them will have more than n2 correct predictions Now if N is the number of correct predictions of a specified member of the syndicate then the probability mass function of the number of correct predictions of the member of the syndicate having more than n2 correct predictions is Pi correct PN i PN n i i n2 2PN i PN iN n2 b X is binomial with parameters m 12 c Since all of the X 1 players including one from the syndicate that have more than n2 correct predictions have the same expected return we see that X 1 Payoff to syndicate m 2 implying that EPayoff to syndicate m 2 EX 11 d This follows from part b above and c of Problem 56 61 a PM x 1 1 1 1 1 1 n n n n pF x P M x N n P N n F x p p p F x b PM xN 1 Fx c PM xN 1 FxPM x d PM x PM xN 1PN 1 PM xN 1PN 1 Fxp FxPM x1 p again giving the result PM x 1 1 pF x p F x Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 122 Chapter 7 62 The result is true when n 0 so assume that PNx n xnn 1 Now PNx n 1 1 1 0 1 P N x n U y dy 0 x P N x y n dy 0 x P N u n du 1 0 1 x un n du by the induction hypothesis xnn which completes the proof b ENx 0 0 0 1 n x n n n P N x n P N x n x n e 63 a Number the red balls and the blue balls and let Xi equal 1 if the ith red ball is selected and let it by 0 otherwise Similarly let Yj equal 1 if the jth blue ball is selected and let it be 0 otherwise Cov Cov i j i j i j i j X Y X Y Now EXi EYj 1230 EXiYj Pred ball i and blue ball j are selected 28 30 10 12 Thus CovX Y 2 28 30 80 1230 10 12 96145 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 123 b EXYX XEYX X12 X820 where the above follows since given X there are 12X additional balls to be selected from among 8 blue and 12 nonblue balls Now since X is a hypergeometric random variable it follows that EX 121030 4 and EX 2 1218132329 42 51229 As EY 81230 165 we obtain EXY 2 48 512 29 5 35229 and CovX Y 35229 4165 96145 64 a EX EXtype 1p EXtype 21 p pμ1 1 pμ2 b Let I be the type EXI μI VarXI 2 I σ VarX 2 Var I I E σ μ 2 2 2 2 2 1 2 1 2 1 2 1 1 1 p p p p p p σ σ μ μ μ μ 65 Let X be the number of storms and let GB be the events that it is a good bad year Then EX EXGPG EXBPB 34 56 42 If Y is Poisson with mean λ then EY 2 λ λ2 Therefore EX 2 EX 2GPG EX 2BPB 124 306 228 Consequently VarX 228 422 516 66 EX 2 2 2 2 1 1 2 3 3 E X Y E X Y E X Y 2 2 19 5 7 3 E X E X 2 183 24 2 3 E X E X 2 1443 2 3 E X since EX 15 Hence VarX 443 152 218 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 124 Chapter 7 67 Let F denote the fortune after n gambles EF EELF F1 E22p 1F1p Fit 2p 1 Fri 1 2p1 ELF 1 1 2p 1PELF2 1 2p 1 TELFol 68 a 6 4e b 6e 3 Ae 3 3 3 P30 6ee i dese x c P30 PO 6e 4e 69 a fereax 1 2 0 co 3 co ay XL ler T4 1 b Je Se de ee ydy 3 36 O96 16 oo xo feve 31 2 2 0 Cc SF a 3 81 fe e dx 0 1 70 a pap 12 0 1 b pap 13 0 1 1 n ni TL PXi PX i pjap p l pdp 0 o iWni POO ey i nl Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 125 72 a PN i 1 1 1 0 0 1 1 i P N i p dp p dp i b PN i PN i PN i 1 1 i i 1 c EN 1 1 1 i i P N i i 73 a ER EERS ES μ b VarRS 1 ERS S VarR 1 VarS 1 σ2 c fRr S R S f s F r s ds 2 2 2 2 2 s r s C e e ds μ σ 2 2 2 2 exp 2 1 1 r K S μ σ σ σ σ ds exp ar2 br Hence R is normal d ERS EERSS ESERS ES 2 μ2 σ2 Cov R S μ2 σ2 μ2 σ2 75 X is Poisson with mean λ 2 and Y is Binomial with parameters 10 34 Hence a PX Y 2 PX 0PY 2 PX 1PY 1 PX 2PY 0 2 2 8 2 9 2 10 10 10 3 4 1 4 2 3 41 4 2 1 4 2 1 e e e b PXY 0 PX 0 PY 0 PX Y 0 e2 1410 e21410 c EXY EXEY 2 10 3 4 15 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 126 Chapter 7 77 The joint moment generating function EetXsY can be obtained either by using EetXsY tX sY e f x y dy dx or by noting that Y is exponential with rate 1 and given Y X is normal with mean Y and variance 1 Hence using this we obtain EetXsYY esYEEtXY 2 2 sY e eYt t and so EetXsY 2 2 t s t Y e E e 2 2 1 1 te s t s t 1 Setting first s and then t equal to 0 gives EetX 2 2 1 1 te t t 1 EesY 1 s1 s 1 78 Conditioning on the amount of the initial check gives EReturn EReturnA2 EReturnB2 AFA B1 FA2 BFB A1 FB2 A B B AFB FA2 A B2 where the inequality follows since B A and FB FA both have the same sign b If x A then the strategy will accept the first value seen if x B then it will reject the first one seen and if x lies between A and B then it will always yield return B Hence EReturn of xstrategy if 2 otherwise B A x B A B c This follows from b since there is a positive probability that X will lie between A and B Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 127 79 Let Xi denote sales in week i Then EX1 X2 80 VarX1 X2 VarX1 VarX2 2 CovX1 X2 72 2666 936 a With Z being a standard normal PX1 X2 90 90 80 936 P Z PZ 1034 150 b Because the mean of the normal X1 X2 is less than 90 the probability that it exceeds 90 is increased as the variance of X1 X2 increases Thus this probability is smaller when the correlation is 2 c In this case PX1 X2 90 90 80 72 2266 P Z PZ 1076 141 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 128 Chapter 7 Theoretical Exercises 1 Let μ EX Then for any a EX a2 EX μ μ a2 EX μ2 μ a2 2Ex μμ a EX μ2 μ a2 2μ aEX μ EX μ2 μ a2 2 EX a x a x a a x f x dx x a f x dx aFa 1 x a x a xf x dx xf x dx a F a Differentiating the above yields derivative 2afa 2Fa afa afa 1 Setting equal to 0 yields that 2Fa 1 which establishes the result 3 EgX Y 0 P g X Y a da 0 0 g x y x y g x y a f x y dydxda daf x y dydx g x y dydx 4 gX gμ gμX μ gμ 2 2 X μ gμ gμX μ gμ 2 2 X μ Now take expectations of both sides 5 If we let Xk equal 1 if Ak occurs and 0 otherwise then X 1 n k k X Hence EX 1 1 n n k k k k E X P A But EX 1 1 n n k k k P X k P C Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 129 6 X 0 X t dt and taking expectations gives EX 0 0 E X t dt P X t dt 7 a Use Exercise 6 to obtain that EX 0 0 P X t dt P Y t dt EY b It is easy to verify that X st Y and Y st X Now use part a 8 Suppose X st Y and f is increasing Then PfX a PX f 1a PY f 1a since x st Y PfY a Therefore fX st fY and so from Exercise 7 EfX EfY On the other hand if EfX EfY for all increasing functions f then by letting f be the increasing function fx 1 if 0 otherwise x t then PX t EfX EfY PY t and so X st Y Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 130 Chapter 7 9 Let Ij th 1 if a run of size begins at the flip 0 otherwise k j Then Number of runs of size k 1 1 n k j j I ENumber of runs of size k 1 1 n k j j E I PI1 1 1 2 1 1 n k j n k j P I P I pk1 p n k 1pk1 p2 pk1 p 10 1 1 1 1 1 1 1 n n n n n i i i i i E X X E X X nE X X Hence 1 1 k n i i E X X k n 11 Let Ij 1 outcome never occurs 0 otherwise j Then X 1 r jI and EX 1 1 r n j j p 12 For X having the Cantor distribution EX 12 VarX 18 13 Let Ij 1 record at 0 otherwise j 1 1 1 1 1 is largest of 1 n n n n j j j j E I E I P X X X j Var 1 1 1 1 1 Var 1 n n n j j I I j j Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 131 15 μ 1 n i i p by letting Number 1 n i i X where 1 is success 0 i i X VarNumber 1 1 n i i i p p maximization of variance occur when pi μn minimization of variance when pi 1 i 1 μ pμ1 μ μ To prove the maximization result suppose that 2 of the pi are unequalsay pi pj Consider a new pvector with all other pk k i j as before and with 2 i j i j p p p p Then in the variance formula we must show 2 1 2 2 i j i j p p p p pi1 pi pj1 pj or equivalently 2 2 2 2 i j i j i j p p p p p p 0 The maximization is similar 16 Suppose that each element is independently equally likely to be colored red or blue If we let Xi equal 1 if all the elements of Ai are similarly colored and let it be 0 otherwise then 1 r i i X is the number of subsets whose elements all have the same color Because 1 1 1 21 2 i r r r A i i i i i E X E X it follows that for at least one coloring the number of monocolored subsets is less than or equal to 1 11 2 i r A i 17 2 2 2 2 1 2 1 2 Var 1 1 X X λ λ λ σ λ σ 2 2 2 2 1 2 2 2 1 2 2 21 0 d d σ λσ λ σ λ λ σ σ As VarλX1 1 λX2 2 1 2 1 E X X λ λ μ we want this value to be small Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 132 Chapter 7 18 a Binomial with parameters m and Pi Pj b Using a we have that VarNi Nj mPi Pj1 Pi Pj and thus mPi Pj1 Pi Pj mPi1 Pi mPj1 Pj 2 CovNi Nj Simplifying the above shows that CovNi Nj mPiPj 19 CovX Y X Y CovX X CovX Y CovY X CovY Y VarX CovX Y CovY X VarY VarX VarY 0 20 a CovX YZ EXY EXZY XEYZ EXZEYZ Z EXYZ EXZ EYZ EXZEYZ EXZEYZ EXYZ EXZEYZ where the next to last equality uses the fact that given Z EXZ and EYZ can be treated as constants b From a ECovX YZ EXY EEXZEYZ On the other hand CovEXZ EYZ EEXZEYZ EXEY and so ECovX YZ CovEXZ EYZ EXY EXEY CovX Y c Noting that CovX XZ VarXZ we obtain upon setting Y Z that VarX EVarXZ VarEXZ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 133 21 a Using the fact that f integrates to 1 we see that cn i 1 1 0 1 i n i x x dx i 1n in From this we see that EXi cn 1 i 1cn i in 1 2 E X i cn 2 i 2cn i 1 2 1 i i n n and thus VarXi 2 1 1 2 i n i n n b The maximum of in 1 i is obtained when i n 12 and the minimum when i is either 1 or n 22 CovX Y b VarX VarY b2 VarX 2 Var Var b X b X Y b b X ρ 26 Follows since given X gX is a constant and so EgXYX gXEYX 27 EXY EEXYX EXEYX Hence if EYX EY then EXY EXEY The example in Section 3 of random variables uncorrelated but not independent provides a counterexample to the converse 28 The result follows from the identity EXY EEXYX EXEYX which is obtained by noting that given X X may be treated as a constant 29 x EX1 XnX1 Xn x 1 i n i E X X x E X X x 1 i nE X X x Hence EX1X1 Xn x xn Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 134 Chapter 7 30 ENiNjNi NiENjNi Nin Ni1 j i p p since each of the n Ni trials no resulting in outcome i will independently result in j with probability pj1 pi Hence ENiNj 2 2 2 2 1 1 1 j j i i i i i i i i p p nE N E N n p n p np p p p nn 1pi pj and CovNi Nj nn 1pi pj n2pi pj npi pj 31 By induction true when t 0 so assume for t 1 Let Nt denote the number after stage t ENtNt 1 Nt 1 Enumber selected Nt 1 Nt 1 r b w r ENtNt 1 Nt 1 b w b w r ENt t b w w b w r 32 c c A A A E XI E XI A P A E XI A P A E X A P A 34 a ETrTr1 Tr1 1 1 pETr b Taking expectations of both sides of a gives ETr ETr1 1 1 pETr or ETr 1 1 1 E Tr p p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 135 c Using the result of part b gives ETr 1 1 1 E Tr p p 2 1 1 1 1 E Tr p p p p 1p 1p2 1p2ETr2 1p 1p2 1p3 1p3ETr3 0 1 1 1 r i r i p p E T 1 1 r i i p since ET0 0 35 PY X j j P Y X X j p j j P Y j X j p j j P Y j p 1 j j j p p 36 Condition on the first ball selected to obtain Mab 1 1 a b a b a b M M a b a b a b 0 Ma0 a M0b b Mab Mba M21 4 3 M31 7 4 M32 32 37 Let Xn denote the number of white balls after the nth drawing EXn1Xn 1 1 1 1 1 n n n n n X X X X X a b a b a b Taking expectations now yields a To prove b use a and the boundary condition M0 a c Pn 1st is white EPn 1st is whiteXn n n X M E a b a b Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 136 Chapter 7 40 Let I equal 1 if the first trial is a success and 0 if it is a failure Now if I 1 then X 1 because the variance of a constant is 0 this gives Var 1 0 X I On the other hand if I 0 then the conditional distribution of X given that I 0 is the same as the unconditional distribution of 1 the first trial plus a geometric with parameter p the number of additional trials needed for a success Therefore Var 0 Var1 Var X I X X Consequently Var Var 1 1 Var 0 0 1 Var E X I X I P I X I P I p X By the same reasoning used to compute the conditional variances we have 1 1 E X I 1 0 1 1 E X I E X p which can be written as 1 1 1 E X I I p yielding that 2 2 1 1 1 Var Var 1 p E X I I p p p p p The conditional variance formula now gives Var 1 1 Var Var X E X I Var E X I p p X p or 2 1 Var p X p 41 a No b Yes since fYxI 1 fXx fXx fYxI 0 c fYx 1 1 2 2 X X X f x f x f x d EXY EEXYX EXEYX 0 e No since X and Y are not jointly normal Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 137 42 If EYX is linear in X then it is the best linear predictor of Y with respect to X 43 Must show that EY 2 EXY Now EXY EXEXZ EEXEXZ Z EE 2XZ EY 2 44 Write Xn 1 1 n X i i Z where Zi is the number of offspring of the ith individual of the n 1st generation Hence EXn EEXnXn1 EμXn1 μEXn1 so EXn μEXn1 μ2EXn2 μnEX0 μn c Use the above representation to obtain EXnXn1 μXn1 VarXnXn1 σ2Xn1 Hence using the conditional Variance Formula VarXn μ2 VarXn1 σ2μn1 d π Pdies out dies out i j j P X j p j j j π p since each of the j members of the first generation can be thought of as starting their own independent branching process 46 It is easy to see that the nth derivative of 2 0 2 j j t j will when evaluated at t 0 equal 0 whenever n is odd because all of its terms will be constants multiplied by some power of t When n 2j the nth derivative will equal 2 n n j n d t j dt plus constants multiplied by powers of t When evaluated at 0 this gives that EZ2j 2jj2j Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 138 Chapter 7 47 Write X σZ μ where Z is a standard normal random variable Then using the binomial theorem EX n 0 n i i n i i n i σ E Z μ Now make use of theoretical exercise 46 48 φYt EetY EetaXb etbEetaX etbφXta 49 Let Y logX Since Y is normal with mean μ and variance σ2 it follows that its moment generating function is Mt EetY 2 2 2 t t eμ σ Hence since X eY we have that EX M1 eμ σ2 2 and EX 2 M2 2 2 e μ 2 σ Therefore VarX 2 2 2 2 2 2 2 2 1 e e e e μ σ μ σ μ σ σ 50 ψt log φt ψ t φtφt ψ t 2 2 t t t t φ φ φ φ ψ t 2 2 0 t E X E X VarX 51 Gamma n λ 52 Let φs t EesXtY 2 0 0 0 0 sX tY s s t t s t E XYe E XY s t φ 0 0 0 0 s s t t s t E X s t E Y s t φ φ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 7 139 53 Follows from the formula for the joint moment generating function 54 By symmetry EZ 3 EZ 0 and so CovZZ 3 0 55 a This follows because the conditional distribution of Y Z given that Y y is normal with mean y and variance 1 which is the same as the conditional distribution of X given that Y y b Because Y Z and Y are both linear combinations of the independent normal random variables Y and Z it follows that Y Z Y has a bivariate normal distribution c μx EX EY Z μ 2 x σ VarX VarY Z VarY VarZ σ2 1 ρ CorrX Y 2 2 Cov 1 1 Y Z Y σ σ σ σ d and e The conditional distribution of Y given X x is normal with mean EYX x μ 2 2 1 x x x x σ σ ρ μ μ μ σ σ and variance VarYX x 2 2 2 2 2 1 1 1 σ σ σ σ σ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall