Capítulo 6: Problemas e Análises

81 Chapter 6 Problems 2 a p0 0 8 7 13 12 1439 p0 1 p1 0 8 5 13 12 1039 p1 1 5 4 13 12 539 b p0 0 0 8 7 6 13 12 11 28143 p0 0 1 p0 1 0 p1 0 0 8 7 5 13 12 11 70429 p0 1 1 p1 0 1 p1 1 0 8 5 4 13 12 11 40429 p1 1 1 5 4 3 13 12 11 5143 3 a p0 0 1013912 1526 p0 1 p1 0 1013312 526 p1 1 313212 126 b p0 0 0 1013912811 60143 p0 0 1 p0 1 0 p1 0 0 1013912311 45286 pi j k 3132121011 5143 if i j k 2 p1 1 1 313212111 1286 4 a p0 0 8132 p0 1 p1 0 513813 p1 1 5132 b p0 0 0 8133 pi j k 8132513 if i j k 1 pi j k 8135132 if i j k 2 5 p0 0 1213311123 p0 1 p1 0 121331 11123 p1 1 213113 1213113 1113213113 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 82 Chapter 6 8 fYy 2 2 y y y c y x e dx 3 4 3 y cy e 0 y 0 Yf y dy 1 c 18 and so fYy 3 6 y e y 0 y fXx 2 2 1 8 y x y x e dy 1 1 4 e x x upon using 2 2 2 2 y y y y y e y e ye e 9 b fXx 2 2 2 0 6 6 2 7 2 7 xy x dy x x c PX Y 1 2 0 0 6 15 7 2 56 x xy x dydx d PY 12X 12 PY 12 X 12PX 12 2 1 2 2 1 2 0 1 2 2 0 2 2 xy x dxdy x x dx 10 a fXx ex fYy ey 0 x 0 y PX Y 12 b PX a 1 ea 11 5 212 45215402 12 e5 5e5 2 3 5 5 5 5 2 3 e e Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 83 14 Let X and Y denoted respectively the locations of the ambulance and the accident of the moment the accident occurs PY X a PY X Y a PX Y X a min 2 0 2 y a L L y dxdy L 2 0 2 L a y a L L y L a y dxdy dxdy L 1 2 2 L a a a a L a L L L L 0 a L 15 a 1 x y R f x y dydx c dydx cAR where AR is the area of the region R b fx y 14 1 x y 1 fxfy where fv 12 1 v 1 c PX 2 Y 2 1 1 4 c dydx area of circle4 π4 16 a A Ai b yes c PA P Ai n12n1 17 1 3 since each of the 3 points is equally likely to be the middle one 18 PY X L3 2 3 4 y x L dydx L 2 L y L 0 x 2 L 6 2 2 0 2 6 3 4 L L L L L L x L dydx dydx L 2 2 2 2 4 5 7 12 24 72 L L L L 79 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 84 Chapter 6 19 1 1 0 0 0 x 1 dydx dx x 1 a 1 1 ln y dx y x 0 y 1 b 0 x 1 x dy 1 0 y 1 c 1 2 d Integrating by parts gives that 1 1 0 0 ln 1 ln y y dy y y y dy yielding the result EY 1 0 yln y dy 14 20 a yes fXx xex fYy ey 0 x 0 y b no fXx 1 21 x f x y dy x 0 x 1 fYy 0 2 y f x y dx y 0 y 1 21 a We must show that f x y dxdy 1 Now f x y dxdy 1 1 0 0 24 y xy dxdy 1 2 012 1 y y dy 1 2 3 012 2 y y y dy 1212 23 14 1 b EX 1 0 xfX x dx 1 1 0 0 24 x x xy dydx 1 2 2 012 1 x x dx 25 c 25 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 85 22 a No since the joint density does not factor b fXx 1 0 x y dy x 12 0 x 1 c PX Y 1 1 1 0 0 x x y dydx 1 2 0 1 1 2 x x x dx 13 23 a yes fXx 12x1 x 1 0 6 1 ydy x x 0 x 1 fYy 12y 1 0 1 x x dx 2y 0 y 1 b EX 1 2 0 6 1 x x dx 12 c EY 1 2 0 2y dy 23 d VarX 1 3 0 6 1 1 4 x x dx 120 e VarY 1 3 0 2 y dy 49 118 24 PN n 1 0 0 1 pn p b PX j pj1 p0 c PN n X j 1 0 n j p p 25 1 e i by the Poisson approximation to the binomial 26 a FABCa b c abc 0 a b c 1 b The roots will be real if B2 4AC Now PAC x 1 1 0 0 0 0 1 0 1 x x a c x a x a c dadc dcda dcda x x log x Hence FACx x x log x and so fACx log x 0 x 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 86 Chapter 6 PB24 AC 2 1 4 0 0 log b xdxdb 1 2 2 2 0 log 4 4 4 b b b db log2 5 6 36 where the above uses the identity 3 3 2 log log 3 9 x x x x xdx 27 PX1X2 a 2 1 2 0 0 1 ay y x e e dxdy λ λ λ λ 1 2 2 0 1 ay y e e dy λ λ λ 1 2 1 2 1 1 2 a a a λ λ λ λ λ λ PX1X2 1 1 1 2 λ λ λ 28 a 1 2 t e since t e is the probability that AJ is still in service when MJ arrives and 12 is the conditional probability that MJ then finishes first b Using that the time at which MJ finishes is gamma with parameters 2 1 yields the result 2 1 3e 29 a If W X1 X2 is the sales over the next two weeks then W is normal with mean 4400 and standard deviation 2 2230 32527 Hence with Z being a standard normal we have PW 5000 5000 4400 32527 P Z PZ 18446 0326 b PX 2000 PZ 2000 2200230 PZ 87 PZ 87 8078 Hence the probability that weekly sales exceeds 2000 in at least 2 of the next 3 weeks p3 3p21 p where p 8078 We have assumed that the weekly sales are independent Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 87 30 Let X denote Jills score and let Y be Jacks score Also let Z denote a standard normal random variable a PY X PY X 0 PY X 5 2 2 2 2 160 170 5 160 170 20 15 20 15 Y X P PZ 42 3372 b PX Y 350 PX Y 3505 2 2 2 2 330 205 20 15 20 15 X Y P PZ 82 2061 31 Let X and Y denote respectively the number of males and females in the sample that never eat breakfast Since EX 504 VarX 376992 EY 472 VarY 360608 it follows from the normal approximation to the binomial that X is approximately distributed as a normal random variable with mean 504 and variance 376992 and that Y is approximately distributed as a normal random variable with mean 472 and variance 360608 Let Z be a standard normal random variable a PX Y 110 PX Y 1095 976 1095 976 7376 7376 X Y P PZ 13856 0829 b PY X PY X 5 32 5 32 7376 7376 Y X P PZ 3144 3766 32 a e2 b 1 e2 2e2 1 3e2 The number of typographical errors on each page should approximately be Poisson distributed and the sum of independent Poisson random variables is also a Poisson random variable Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 88 Chapter 6 33 a 1 e22 22e22 e222222 b 1 4 44 0 44 i i e i c 1 5 66 0 66 i i e i The reasoning is the same as in Problem 26 34 Use the distribution of the sum of independent geometric random variables to obtain the result 12 12 47 36 35 a PX1 1X2 1 513 1 PX1 0X2 1 b same as in a 36 a PY1 1Y2 1 212 1 PY1 0Y2 1 b PY1 1Y2 0 312 1 PY1 0Y2 0 37 a PY1 1Y2 1 p1 11 12133 1 PY1 0Y2 1 b PY1 1Y2 0 p1 012133 1 PY1 0Y2 0 where p1 1 and p1 0 are given in the solution to Problem 5 38 a PX j Y i 1 1 5 j j 1 j i 1 j b PX jY i 5 5 1 1 15 1 5 k i k i k k j j 5 j i c No Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 89 39 For j i PY iX i 1 36 P Y i X i P X i P X i For j i PY jX i 2 36 P X i Hence 1 1 2 1 1 36 36 i j i P Y j X i P X i P X i and so PX i 2 1 36 i and PY jX i 1 2 2 2 1 i i i j i j i 41 a fXYxy 1 1 x y x y xe xe dx y 12xexy1 0 x b fYXyx 1 1 x y x y xe xe dy xexy 0 y PXY a 1 0 0 a x xe x y dydx 0 1 a x e e dx 1 ea fXYa ea 0 a 42 fYXyx 2 2 2 2 x x x x x y e x y e dx 2 2 3 3 4 x y x x y x FYXyx 2 2 3 3 4 y x x y dy x 2 3 3 3 3 3 2 3 4 x y y x x x y x Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 90 Chapter 6 43 fλn P N n g P N n λ λ C1eλλnαeαλαλs1 C2eα1λλns1 where C1 and C2 do not depend on λ But from the preceding we can conclude that the conditional density is the gamma density with parameters α 1 and n s The conditional expected number of accidents that the insured will have next year is just the expectation of this distribution and is thus equal to n sα 1 44 PX1 X2 X3 PX2 X1 X3 PX3 X1 X2 3PX1 X2 X3 1 2 3 1 2 3 0 1 1 2 3 3 x x x xi i dx dx dx take a 0 b 1 3 3 2 3 1 1 1 1 1 1 2 3 2 3 2 3 0 0 0 0 3 3 1 x x x x dx dx dx x x dx dx 1 2 3 3 0 1 3 1 2 2 x dx 45 3 2 2 0 5 22 x x x x X x f x xe dx xe xe dx 30x 12e2xxex1 exx 12 46 3 2 L d L 47 3 3 4 3 4 2 2 14 14 5 1 22 Xf x dx x x dx 48 a Pmin Xi a 1 Pmin Xi a 1 5 1 a P Xi a e λ b Pmax Xi a 5 1 a P Xi a e λ 49 It is uniform on 1 ns 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 91 50 Start with 2 2 1 2 1 2 2 1 2 1 2 z z fz z z z e π Making the transformation using that its Jacobian is 1 yields that 2 2 1 2 2 1 2 x y x X Y Z Z f x y f x y x e π 51 1 4 X X f x y 2 42 2 2 2 Y X x zdz y x y 48xyy2 x2 PX4 X1 a 1 2 2 0 0 48 a a x xy y x dydx 1 1 2 2 1 0 48 a xy y x dydx 52 1 1 2 2 R r f r r θ π π 0 r 1 0 θ 2π Hence R and θ are independent with θ being uniformly distributed on 0 2π and R having density fRr 2r 0 r 1 53 fRθrθ r 0 r sin θ 1 0 r cos θ 1 0 θ π2 0 r 2 54 J 1 2 1 2 1 1 cos 2 sin 2 2 2 2 sin 2 cos x u z u z u z u cos2 u sin2 u 1 fuzu z 1 2 e z π But x2 y2 2z so fXYx y 2 2 2 1 2 x y e π Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 92 Chapter 6 55 a If u xy v xy then J 2 1 y x x y y 2 x y and y u v x vu Hence b fuvu v 2 1 1 2 2 fX Y vy u v v vu u 1 1 u v u fuu 2 2 1 1 1 log 2 u u dv u vu u u 1 For v 1 fVv 2 2 1 1 2 2 v du vu v v 1 For v 1 fVv 2 1 2 1 1 2 2 du vu 0 v 1 56 a u x y v xy y 1 u v x 1 uv v J 2 2 2 2 1 1 1 1 1 1 x v x y y u y y y x y fuv u v 2 1 u v 0 uv 1 v 0 u 1 v 58 y1 x1 x2 y2 1xe J 1 1 1 0 xe 1xe y2 x1 log y2 x2 y1 log y2 2 1 2 1 2 log log 1 2 2 1 y y y fY Y y y e e y λ λ λ λ 1 2 2 1 y e y λ λ 1 y2 y1 log y2 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 93 59 u x y v x z w y z z 2 2 2 v w u v w u w v u x y J 1 1 0 1 0 1 0 1 1 2 fu v w 1 1 exp 2 2 u v w u v w u w v v w u 60 PYj ij j 1 k 1 PYj ij j 1 k PYk1 ik1Yj ij j 1 k 1 1 1 1 k i k j j i k n k P n Y i Y i j k n kn kn if 1 1 1 k j j i n 0 otherwise Thus the joint mass function is symmetric which proves the result 61 The joint mass function is PXi xi i 1 n 1 n k xi 0 1 i 1 n 1 n i i x k As this is symmetric in x1 xn the result follows Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 94 Chapter 6 Theoretical Exercises 1 PX a2 Y b2 Pa1 X a2 b1 Y b2 PX a1 b1 Y b2 Pa1 X a2 Y b1 PX a1 Y b1 The above following as the left hand event is the union of the 4 mutually exclusive right hand events Also PX a1 Y b2 PX a1 b1 Y b2 PX a1 Y b1 and similarly PX a2 Y b1 Pa1 X a2 Y b1 PX a1 Y b1 Hence from the above Fa2 b2 Pa1 X a2 b1 Y b2 Fa1 b2 Fa1 b1 Fa2 b1 Fa1 b1 Fa1 b1 2 Let Xi denote the number of type i events i 1 n PX1 r1 Xn rn 1 1 1 events n n n i P X r X r r 1 1 n i n r i e r λλ 1 1 1 1 1 1 n i n n r i r r n n n i r e P p r r r λλ 1 i i i n P r p i i e r λ λ 3 Throw a needle on a table ruled with equidistant parallel lines a distance D apart a large number of times Let L L D denote the length of the needle Now estimate π by 2L fD where f is the fraction of times the needle intersects one of the lines Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 95 5 a For a 0 FZa PX aY 0 0 a y X Y f x f y dxdy 0 X Y F ay f y dy fZa 0 X Y f ay yf y dy b FZa PXY a 0 0 a y X Y f x f y dxdy 0 X Y F a y f y dy fZa 0 1 X Y f a y f y dy y If X is exponential with rate λ and Y is exponential with rate μ then a and b reduce to a FZa 0 ay y e y e dy λ λ μ λ μ b FZa 0 a y 1 y e e dy y λ μ λ μ 6 X Y X Y x y t t x X Y F t f x y dydx f x y dydx Differentiation yields that t x X Y X Y t x X Y X Y d f t f x y dydx dt d f x y dydx dt f x t x dx Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 96 Chapter 6 7 a PcX a PX ac and differentiation yields fcXa 1 1 a c t Xf a c e a c t c c λ λ λ Γ Hence cX is gamma with parameters t λc b A chisquared random variable with 2n degrees of freedom can be regarded as being the sum of n independent chisquare random variables each with 2 degrees of freedom which by Example is equivalent to an exponential random variable with parameter λ Hence by Proposition 2 X2n is a gamma random variable with parameters n 12 and the result now follows from part a 8 a PW t 1 PW t 1 PX t Y t 1 1 FXt 1 FYt b fWt fXt1 FYt fYt 1 FXt Dividing by 1 FXt1 FYt now yields λWt fXt1 FXt fYt1 FYt λXt λYt 9 PminX1 Xn t PX1 t Xn t eλteλt enλt thus showing that the minimum is exponential with rate nλ 10 If we let Xi denote the time between the ith and i 1st failure i 0 n 2 then it follows from Exercise 9 that the Xi are independent exponentials with rate 2λ Hence 2 0 n i i X the amount of time the light can operate is gamma distributed with parameters n 1 2λ 11 I 1 2 3 4 5 x x x x x fx1 fx5dx1dx5 1 2 3 4 5 u u u u u du1 du5 by ui Fxi i 1 5 0 ui 1 2 2 5 u du du 2 3 1 u 2 du3 3 4 4 4 5 3 2 u u du du 1 2 4 0 3 2 u u du 215 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 97 12 Assume that the joint density factors as shown and let Ci ig x dx i 1 n Since the nfold integral of the joint density function is equal to 1 we obtain that 1 1 n i i C Integrating the joint density over all xi except xj gives that X j j j j i j j j i j f x g x C g x C If follows from the preceding that fx1 xn 1 j n X j j f x which shows that the random variables are independent 13 No Let Xi 1 if trial is a success 0 i Then 1 1 1 1 n m n m n m X X X n m P x x X x f X x x x f x P x x 1 i i x n m x cx x and so given 1 n m i X n the conditional density is still beta with parameters n 1 m 1 14 PX iX Y n PX i Y n iPX Y n 1 1 2 2 1 1 1 1 1 i n i n p p p p n p p 1 1 n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 98 Chapter 6 15 Let X denote the trial number of the kth success and let s s f f s f be an outcome of the first n 1 trials that contains a total of k 1 successes and n k failures Using that X is a negative binomial random we have 1 1 1 1 1 1 1 1 1 1 k n k k n k k n k P s s f f s f X n P s s f f s f X n P X n P s s f f s f s n p p k p p n p p k n k and the result if proven 16 PX kX Y m P X k X Y m P X Y m P X k Y m k P X Y m 2 1 1 2 1 k n k m k n m k m n m n n p p p p k m k n p p m 2 n n k m k n m 17 PX n Y m 2 2 i P X n Y m X i P X i 1 2 3 min 3 1 2 0 1 n m m i n i i i e n m i i λ λ λ λ λ λ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 99 18 Starting with P X i Y j p i j P Y j P X i Y j q j i P X i we see that p i j P X i P Y j q j i which gives that 1 i p i j P Y j q j i and the result follows 19 a PX1 X2X1 X3 1 1 2 3 1 3 max P X X X X P X X 13 1 2 23 b PX1 X2X1 X3 3 1 2 1 3 13 1 2 P X X X P X X 13 c PX1 X2X2 X3 1 2 3 2 3 13 13 1 2 P X X X P X X d PX1 X2X2 X3 2 1 2 3 2 3 min P X X X X P X X 13 1 2 23 20 PU sU a PU sPU a 1 1 s a a s 1 PU sU a PU sPU a sa 0 s a Hence UU a is uniform on a 1 whereas UU a is uniform over 0 a 21 fWNwn W P N n W w f w P N n Cew wn n βeβwβwt1 C1eβ1wwnt1 where C and C1 do not depend on w Hence given N n W is gamma with parameters n t β 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 100 Chapter 6 22 1 1 i n W X i n f w x x 1 1 n w n f x x w f w f x x n 1 i 1 wxi w t C we e w β β 1 1 n i w x n t Ke w β 23 Let Xij denote the element in row i column j PXij is s saddle point 1 min max min ik kj ij ik k m k k i P X X X X min max min ik kj ij ik k k k i P X X P X X where the last equality follows as the events that every element in the ith row is greater than all elements in the jth column excluding Xij is clearly independent of the event that Xij is the smallest element in row i Now each size ordering of the n m 1 elements under consideration is equally likely and so the probability that the m smallest are the ones in row i is 1 1 n m m Hence PXij is a saddlepoint 1 1 1 1 1 1 m n n m m n m m and so Pthere is a saddlepoint ij is a saddlepoint P i j X ij P Xij is a saddlepoint 1 m n n m 24 For 0 x 1 PX n X X x Pn X n x enλ en xλ enλ1 exλ Because the joint distribution factors they are independent X 1 has a geometric distribution with parameter p 1 e λ and x X is distributed as an exponential with rate λ conditioned to be less than 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 101 25 Let Y max X1 Xn Z minX1 Xn PY x PXi x i 1 n 1 n n P Xi x F x PZ x PXi x i 1 n 1 1 n n P Xi x F x 26 a Let d DL Then the desired probability is n 1 3 2 1 1 1 1 2 1 2 1 1 1 2 1 0 n n n n d n d d d n n x d x d x d x d dx dx dx dx 1 n 1dn b 0 27 1 j n i n i x i j n F x F x F x i 1 1 j n i n i X i j n f x iF x f x F x i 1 1 n i n i i j n F x n i F x f x i 1 1 1 n i n i i j n F x f x F x n i i 1 1 1 1 n k n k k j n F x f x F x n k k by k i 1 1 1 1 j n j n F x f x F x n j j 28 1 2 1 1 n n X n n f x x x n n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 102 Chapter 6 29 In order for Xi xi Xj xj i j we must have i i 1 of the Xs less than xi ii 1 of the Xs equal to xi iii j i 1 of the Xs between xi and xj iv 1 of the Xs equal to xj v n j of the Xs greater than xj Hence i j x X i j f x x 1 1 11 11 i j i i i j i j n F x f x F x F x f x i j i n j 1 Fxjnj 31 Let X1 Xn be n independent uniform random variables over 0 a We will show by induction on n that PXk Xk1 t 0 n a t a if f t a i t a It is immediate when n 1 so assume for n 1 In the n case consider PXk Xk1 tXn s Now given Xn s X1 Xn1 are distributed as the order statistics of a set of n 1 uniform 0 s random variables Hence by the induction hypothesis PXk Xk1 tXn s 1 0 n s t s if f t s i t s and thus for t a PXk Xk1 t 1 1 n a n n n t s t ns a t ds s a a which completes the induction The above used that 1 1 1 n n n X n s ns f s n a a a 32 a PX Xn PX is largest of n 1 1n 1 b PX X1 PX is not smallest of n 1 1 1n 1 nn 1 c This is the probability that X is either the i 1st or i 2nd or jth smallest of the n 1 random variables which is clearly equal to j 1n 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 6 103 35 The Jacobian of the transformation is J 2 2 1 1 0 y x y x y Hence 1 2 J y x Therefore as the solution of the equations u x v xy is x u y uv we see that fuvu v 2 2 2 2 2 2 1 2 u u v X Y u u f u u v e v v π Hence fVu 2 1 1 2 2 2 1 2 u v u e du v π 2 2 2 2 1 2 u e u du v σ π where σ2 v21 v2 2 2 2 2 0 1 ue u du v σ π 2 2 0 1 e y dy v σ π 2 1 1 v π Copyright 2010 Pearson Education Inc Publishing as Prentice Hall