Papoulis a. - Probability.random.variables.and.Stochastic.processes.problem.solutions - 4th(2002)

7/29/2019 Papoulis a. - Probability.random.variables.and.Stochastic.processes.problem.solutions - 4th(2002)

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Solutions Manual

to accompany

Probability,Random Variables

andStochastic Processes

Fourth Edition

Athanasios PapoulisPolytechnic University

S. Unnikrishna PillaiPolytechnic University


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Solutions Manual to accompany

PROBABILITY, RANDOM VARIABLES AND STOCHASTIC PROCESSES, FOURTH EDITION

ATHANASIOS PAPOULIS

Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,

New York, NY 10020. Copyright 2002 by The McGraw-Hill Companies, Inc. All rights reserved.

The contents, or parts thereof, may be reproduced in print form solely for classroom use with PROBABILITY, RANDOM

VARIABLES AND STOCHASTIC PROCESSES, FOURTH EDITION, provided such reproductions bear copyright notice, but may

not be reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc.,

including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.

www.mhhe.com


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CHAPTER 2

2 - 1 We use D e M o r g a n ' s l a w :- -a) X + 6 + I + B AB + A% = A ( B + % ) = Abecause = (01 BB = I01

2-2 If A = { 2 < x ; 5 ) B = { 3 < x < 6 1- S = {-=-


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2-6 Any su bs et of S con tai ns a coun table number of e lements , hence, i tcan be wri t t en a s a countable union of e lementa ry events . I t i st h e r e f o r e a n e v e n t.

2-7 Forming a l l unions , in te rs ec t i on s , and complements of the s e t s E l )and {2,3), we o b t a i n t h e f o l l o w i n g sets:

(01, C11, (41, {2,31, {1,41, {1,2,31, {2,3,41, {1,2,3,41

2-8 I f ACB,P(A) = 114, and P(B) = 113, then

2-10 We use ind uct io n. The formula i s t r u e f o r n = 2 becauseP(A1A2) - P ( A ~ I A ~ ) P ( A ~ ) .uppose that i t i s t r u e f o r n . S in cewe conclude th a t i t must be t r u e f o r n + l .

2-11 F i r s t so l u t i on . The t o t a l number of m element su bs ets equ als (") (s eemPro bl. 2-26). The t o t a l number of m e lement subse ts conta in ing 5 e qua l s0n- l(m-l) Hence

Second solut ion. Clearly, P{C, I A ~ ) = mln i s t h e p r o b ab i l i t y t h a t 50

i s i n a s p e c i f i c Am . Hence ( t o t a l p r o b a b i l i t y )

where the summation i s over a l l sets A .


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22-12 ( a) P E 6 < t < 8 1 = -- - 10P E 6 r t s 8 1 2(b) ~ { 6t - 81t > 51 = P ( t , 51 = -

2-13 From (2-27) it fo l lows tha t

Equat ing th e two s ide s and se t t in g t l= tO+ t w e o b ta in

for every to . Hence,

D i f f er e n t ia t i n g t h e s e t t i n g c = a ( O ) , we conclude that

2-14 I f A and B are independent , th en P (AB) = P (A)P (B) . I f t h e y a r emutually ex clu sive , the n P(AB) = 0 , Hence, A and B ar e mutual lyexc lusi ve and independent i f f P(A)P(B) = 0.


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Clear ly , A1 = A1A2 + ~ ~ i i ~ence

I f t h e ev e n ts A and ar e independent , then1 2

hence , t he events A and A ar e independent . Furthermore, S i s1 2independent with any A because SA = A. T h is y i e l d sP(SA) = P(A) = P(S)P(A)

Hence, t h e theorem i s t r u e f o r n = 2 . To p ro ve i t i n g e ne r al we useindu ct ion : Suppose th at A i s independent of A1, ..., . Cle a r ly ,- n+l nAn+l and An+l are independent of B1, ... B . There foren

2.16 The desired probabilit,ies are given by (a)

(TI)


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2.17 Let Al IA2 and Ad represent tire eventsAl = "ball numbered less tha,n or equal to rn is drawn?A2= ('ball numbered rn i s drawn"AS= ('ball numbered greater tillan rn is drawn"

P ( A 1 o cc u,r s nl = k - 1 , A2 o ccu rs n2= 1 and A3 o ccu rs n3 = 0 )

2.18 All cars are equally likely so that the first car is selected withprobability p = 113. This gives the desired probabil ity to be

2.19 P{ 'dr aw ing a w hi te bad1 " } =&&P(" a t l ea t o n e wh i t e b all i n k triu , ls " )= 1 - P( "a l l b lack ba l l s in k t r ia l s" )

2.20 Let D = 2r represent the penny diameter. So long as the centerof the penny is at a distance of r away from any side of the square,th e penny will be entirely inside th e square. Th is gives th e desiredprobabili ty to b e


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2.21 Refer to Exanlple 3.14.( a ) Using (3.391,we get

(h )

P ( " t wo one -d ig i t and f ou r two-d ig i t numbers1 ' )=

n2-22 The number of equations of the form P(AiAk) = P(Ai)P(Ak) equals ( * I .The number of equations involving r sets equals (:). Hence the totalnumber N of such equations equals

And since

we conclude that

2-23 We denote by B1 and B2 respectively the balls in boxes 1 and 2 andby R the set of red balls. We have (assmption)

P(B1) = P(BZ) '0.5 P(R\B~) 0 . 999 ~ ( ~ 1 8 ~ )0.001Hence (Bayes' theorem)


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2-24 W denote by B1 and B r e s p ec t i v e ly t h e b a l l i n b oxes 1 and 2 and by2D a l l pa ir s of defect ive par ts . We have (assumption)

To f in d P(D IB1) we proceed a s i n Example 2-10:F i r s t s o l u t i o n . In box B1 t h e r e a r e 1 0 00 x 99 9 p a i r s . The number ofpa ir s with both e lements defect ive equals 100x9 9. Hence,

Second so lu t ion . The p rob ab i l i ty tha t the f i r s t bu lb se lec ted f romB1 i s defective equals 100/1000. The probabi l i ty that the second i sdefe ct iv e assuming th e f i r s t was ef fe ct iv e equals 99/999. Hence,

W s im i l a r l y f i n d

(a) P (D ) = P ( D I B ~ ) P ( B ~ )P ( D ~ B ~ ) P ( B ~ )0.0062e (D B,)P (sl )

(b) pm l ID) = P(Dl = 0.80

2-25 Reasoning as i n Example 2-13, we conclude th a t t h e pr ob ab il it y th at th ebus and the t r a i n meet eq uals

Equating wi th 0.5, we fi nd x = 60 - 1 0 6 1 .

2-26 We wish t o show t h a t th e number N (k) of th e element sub set s of Snequals

This i s t r ue fo r k = l because th e number of l -e lement subsets equals n .Using inducti on i n k, w e sh a l l show tha t

n - kN ( k + 1 ) = Nn(k) Ir+lnWe at ta ch t o each k-element subs et of S one of the remaining n - k elemenof S. We, then, form Nn(k)(n -k) k+l- ele men t sub set s. However, the sesubsets ar e not a l l d i f f er en t . They form groups each of which has k + lide n t i ca l e lements. We mus t, th e re fo re , d iv ide by k + l .


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2 -2 7 In this experiment we have 8 outcomes. Each outcome is a selection of a particular coinand a specific sequence of heads or tails; for example fhh is the outcome "we selected thefair coin and we observed hh". The event F = (the selected coin is fai r) consists of thefour outcomes fhh, fht, fth and fhh. Its complement F s the selection of the two-headead coin. The event HH = (heads at both tosses) consists of two outcomes. Clearly,

Our problem is to find P(F(HH). From (2-41) and ( 2 - 43) it follows that


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CHAPTER 3

3.1 (a ) P ( A occurs atleast twice in n t r ia l s )= 1- P ( A never occurs in n t r ia l s )- P ( A occurs once in n t r ials]= 1 - ( 1 - p ) " - p(1- )"-l

( b ) P ( A occurs atleast thrice in n t r ia l s )= 1- P ( A never occurs in n trials)- P ( A occurs once in n trials)

-P(A occurs twice in n trials)- 1- I - )" - p(1- )"-l -E k p g ( l- )n-2

P("doub1e six atleast three tim es in n trials")

3-3 ~f A = {seven), then

If the dice are tossed 10 times, then the probability that w i l l occur10 them equals (5/6)1. Hence, the probability p that {seven} w i l l showat least once equals

1 - (5/6)1


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3-4 If k is the number of heads, then

But= (q + q)n ' n + (;:P qn-l +("2p2q*-2 + * * *(P - 9)' qn - (;lp qn-l + (p2ne2 -

Adding, we obtain1 + (p - q)" 5 2 ~{evenl

N3-5 In this experiment, the total number of outcomes is the number ( , of ways of pickingn out of N objects. The number of ways of picking k out of the K good components

Kequals ( ) and the number of ways of picking n-k out of the N-K defectiveN-Kcomponents equals ( ,-k ). Hence, the number of ways of picking k good components

K N-Kand n-k deafective components equals ( ) ( ,,-k ). From this and (2-25) it follows thatK N-K N~ ' ( k ) ( n - k ) / ( n )

3.6 (a)

(b)

(4


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3.7 (a) Let n represent th e num ber of wins required in 50 games so tha tthe net gain or loss does not exceed $1. This gives the net gain to be

50 17 3 33P(n e t ga in does not exceed $1) = (17) (a ) (a ) = 0.432P(net gain or loss exceeds $1) = 1- .432 = 0.568(b) Le t n represent the number of wins required so that the net gainor loss does not exceed $5. This gives

50-nP ( n e t ga in does not exceed $5) =xn9= 14 (50) (a)" ($) = 0.349P(net gain or loss exceeds $5) = 1- .349 = 0.651


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3.8 Define the eventsA=" r successes in n Bernoulli trials"B="success at the ithBernoulli trial"C= "r- 1 successes in the remaining n - 1Bernoulli trials excluding

the ith rial"

P(C)= (C1 )p'-' qn-'We need

3.9 There are ( ) ays of selecting 13 cards out of 52 cards. Thenumber of ways to select 13 cards of any suit (out of 13 cards) equals( ) 1 Four such (mutually exclusive) suits give the total numberof favorable outcomes to be 4. Thus the desired probability is given by


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3.10 Using the hint, we obtain

LetMkfl =Nkfl - Nk

so that the above iteration gives

This gives

where we have used No= 0. Similarly Na+b= 0 gives

Thus

Ia+ b 1- q / ~ ) ". p # qP - q 1- q/p)a+b P - q'

Ni = i (a+ b - ) , P = q


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which gives for i = a

Arguing as in (3.43),we get the corresponding iteration equation

and proceed as in Example 3.15.3.12 Suppose one bet on k = 1,2, - - ,6.

Then3 2pl = P( k appears o n one dice) = (A) (I)

3 l 2p2 = P ( k appear on t w o dice)= (2) (6) (I)3p3 = P ( k appear o n all th e tree dice) = (A)5po = P( k appear none) = (6)

Thus, we getNet gain = 2p1+ 3p2+ 4p3 - o = 0.343.


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CHAPTERQ

4-1 From the evenness of f(x): 1 - F(x) = F(-x).From the definition of xu: u = F(xu), I - u = F(xl-,). Hence

4-2 From the symmetry of f(x): 1 - F(q+a) = F(q-a). Hence [see (4-8)]

This yieldsI-a = 2F(q+a) - 1 F(q+a) = 1 - 4 2 'I+a = XI-^/^

4-3 (a) In a linear interpolation:

From Table 4-1 page 106

Proceeding simiplarly, we obtain

(b) If z is such that x = q + az then z is N(O, 1 ) and G(z) = Fx(q+az). Hence,- - - -


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4-4 pk - 2G(k) = 1 = 2 erfk(a) From Table 4- 1

(b ) From Table 3-1 with linear interpolation:

(c ) P(q-zuo < x..< q + zuo) = 2G(z,) - I = 7Hence, G(zu) = ( l+7) /2 u = ( l+7) /2..............................................................................

4-5 (a) F(x) = x for 0 I x 5 1; hence, u = F(xu) = xu(b) F(x) = 1-e-2Xfor x 2 0; hence, u = I -e-2Xu

4-6 Percentage of units between 96 and 104 ohms equals lOOp where p = P(96 < R < 104) =-F(104) - F(96)(a) F(R) = O.l(R-95) for 95 I R I 105. Hence,

p = 0.1(104-95) - 0.1(96-95) = 0.8

(b ) p = G(2.5) - G(-2.5) = 0.9876. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4-7 From (4-34), with a = 2 and P= l /hwe get f(x) = c2 ~ e ' ~ ~ U ( x )


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for 8 < x c 12 and zero other wise

-ax -a cF(x) (1 - e )U (x-c) f (x ) 5 (1 - e ) 6 (x-c) + e-aX~(x-c )2 14-10 (a) P{ l - x - 2 ) = G(?) - G(?) - 0.1499

because (1 5 x 2, 2 1.) = ( 1 < x < 2 )- - -

If x ( t l ) 5 xthenti 5 y = G(x)Hence,


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4-12 (a) P{z < 1024) = ~ ( ~ ~ ~ ~ ~ 0OoO) = G(1.2) = 0.8849

(b) P{X < 1024 1x > 961) = -P{x. > 961)

14-14 (a ) 1. f x ( x ) = - 9001 ( )s(x-k)zgoO k = ~

10 15(b) Pi435 < x < 4601 = G(T5) - G(- 5) 0.5888- -4-15 I f x > b then {z 2 x ) = S F(x) = 1

If x < a then {xcx)=($41 F ( x ) = O4-16 If y(ci ) 2 w, then x (c ) < w because x(; ) < y ( i i ) .- - i - - i - -

Hence,

Theref or e F (w) 5 Fx(w>Y


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4-17 From (4-80)

4-18 It follows frnn (2-41) with

A1 = {X 5 X) A2 = {x > x)" "

4-19 It follows from

4-20 We replace in (4-80) all probabilities with conditional probabilitiesassuming {x i x o I . This yields

But f ( x l x < x ) = O- for x > x and0 0{ x = x , 5 2 x 1 { x = = x ) for X < X . Hence,

0 - 0

Writing a similar equation for P(B~X < x ) we conclude that, if P(A~X = x) = P ( B ~ ? =- Gfor x 5 x then P(A(X Io) = P(B/X 5 xo)0'


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4-21 (a) Clearly, f(p) = 1 for 0 I p 5 1 and 0 otherwise; hence

(b) We wish to find the conditional probability P(0.3 5 p 5 0.71A) where A = (6 heads in-10 tosses). Clearly P(A)p=p)= ~ ' ( l - ~ ) l . ence, [see (4-81)],

This yields

4-22 (a) In this problem, f(p) = 5 for 0.4 I p I 0.6 and zero otherwise; hence [see(4-82)],..

P(H) = 5 1''' pdp = 0.50 .4

(b) With A = (60 heads in 100 tosses) it follows from (4-82) that

for 0.4 I p 5 0.6 and 0 otherwise. Replacing f(p) by f(p)A) in (4-82), we obtain


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4 - 2 4 For a f a i r c o in6 &/2. Ifkl = 0.49n and k = 0.52n then2

P{kl 2 k 5 k2} = ~ ( 0 . 0 4 G ) ~ (0 .0 2& ) - 1 - 0 .9From Table 4-1 ( p a g e l 0 6 ) i t fo l lows that0 . 0 2 6 > 1 . 3 n > 65


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4-25(a) Assume n = 1,000 (Note correction to the problem)

P(A) = 0.6 np = 600 npq = 240 k2 = 650 k1 = 550

0 Oln(b) ~(0.59n 5 k 2 0.61nl - ZG(-) - 1m- 2 6 ( / % ) - 1 - 0.476Hence, (Table 3-1) n = 9220- - - . . .

4-26 With a = 0, b = T I 4 it follows thatp = 1-e = 0.22 np = 220 npq = 171.6 k2 = 100

k2 - "P= - 9.16 and (4-100) yields6

4 27 The eventA = {k heads show at the first n tossings but not earlier)

occurs iff the following two events occurB = {k-1 heads show at the first n-1 tossing)C = {heads show at the nth tossing3

And since these two events are independent and

we conclude thatn-1 k n-kP(A) = P(B)P(C) = (k-l)~


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Multiplying by l / t / 2 n and in t eg ra t i ng from x t o =, w e o b t a i n

because

The f i r s t i n e q u a l i t y f o l lo w s s i m i l a r l y b ec au se

4 -29 I f P(A) - p then P(X) = 1-p. Cl ea rl y P1 = 1-Q1 where Q1 e q u a l s t h e p r o b a b i lit h a t A does not occur a t a l l . I f pn


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4 - 33 P(M) = 2/36 P($ = 34/35. The ev en ts M and form a pa r t i t io n,hence, [s ee (2- 41)l

Clearly , P(A]M) = 1 because, i f M occurs a t f i r s t t r y , X wins. The probabil i tyt h a t X wins a f t e r t h e f i r s t t r y e qu al s P ( A I ~ ) . But i n th e ejrperiment th ats t a r t s a t t h e second r o l l i n g , t h e f i r s t pl a ye r i s Y and th e p ro b ab i l i ty th a the wins equ als P(X) = 1-p. Hence, P ( A I ~ )= P ( x ) = 1-p. And since P (M) = 1/18P@ ) = 1 7/18 ( i ) y i e ld s

4 - 34(a) Each of the n pa rt ic le s can be placed i n any one of th e m boxes. There a r enn pa r t ic le s , hence , t he number of po ss ib i l i t ie s equals N - m . I n t h e mp rese lec ted bo xes, t h e p a r t i c l e s can be placed i n NA - n! ways ( a l l per-nmutations of n objects) . Hence p = n!/m .

A l l po ss ib i l i t ie s a r e ob ta ined by permuting the&-1 ob jec ts consis t ingof the m -1 i n t e r i o r w a l l s w it h a nd n p a r t i c l e s . The (m-l)! permutationsof the wall s and th e n! permutations of th e pa r t i cl es must count a s one.Hence

(b) 'III

(c ) Suppose that S i s a set co n s i s t in g of t h e m boxes. Each placing of thep a r t i c l e s sp ec i f i e s a su b se t of S con si st ing of n elements (box). Thenumber of su ch sub se ts equ al s (m) (see Probe 2-26). Hence,n

X$ , 1 : x n p a r t i c l e sI I I I i m-1 i n t e r i o r w a l ls


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4 -35 If k1 + k,


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CHAPTER J5- 1 9 = 2qx+4 = 14 uy2 = 4uX2= 16

5-2 {y I y) = ( -4x + 3 I y) ( x I (y-3)/4). Hence- - - -Since F,(x) = ( I - e - 2 x ) ~ ( x ) ,his yields

F (y) = e ( ~ - S ) / 2 ~ IY fy(y) = e ( ~ - S ) 1 2 ~

5-3 From Example 5-3 with F, = G(x/c):


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5- 4 If y = x2 and F,(x) = (xt2 c)/4 c fo r Ixli2c, then (see Example 5-2) Fy(y) = f i / 2 c a nd- -fy(y) = 1 / 4 f i f o r 0 < y < 2c.

5-5 From Example 5-4 with F,(x) = G(x/b): For ~ X I S ~,(y) = G(y/b) and

5-6 The equat ion y = -Lnx has a sing le solu tion x = e-"-for y > 0 and no solut ions for y < 0.Furthermore, g8(x) = - I / x = -ey. Hence


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5-7 Clearly, z I z iff the number n(0,z) of the points in the interval (0,z) is at feast one.# -

Hence,

The probability p that a particular point is in the integral (0,z) equals zf 100. With n =200, k = 0, and p = z/100, (3-21) yields P(n(0,z) = 0) = (1 Hence,-.

(b) From (4-107) it follows thata F,(z) = 1 - eW2' for z O

otherwisewhich represents Rayleigh density function (with X = 2cr2).

5-9 For both cases, fy( y) = 0 for y < 0.(a) I f y.0 and 1x1 = y, then xlS y, x 2 - - y . Hence

fy(y) = l fx (y) + fx(-y) IU(Y)(b) I f y > O and e - X ~ ( x ) = y , h e n x= - an y .Furthermore, P{z=O) = P{x


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Papoulis a. - Probability.random.variables.and.Stochastic.processes.problem.solutions - 4th(2002)

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