Probability Random Variables and Stochastic Processes - Papoulis.pdf

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Probability,

Random Variables

and

Stochastic Processes

Fourth Edition

Athanasios Papoulis

S. Unnikrishna Pillai

Solutions Manual to accompany Probability, Random Variables Stochastic Processes and Fourth Edition Athanasios Papoulis Polytechnic University S. Unnikrishna Pillai Polytechnic University

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

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 + % ) = A because = (01 BB = I01 2-2 2-3 - If A = { 2 < x ; 5 ) - - A + B = { 2 < x < 6 ) - - (A+B)(E) = { 2 < x < 6 1 - = { 2 < x < 3 ) B = { 3 < x < 6 1 AB = { 3 < x < 5 ) - - - - [ { x < 3 1 + E x > 5 1 ] + { 5 < x < 6 1 - S = {-=-

2-6 Any s u b s e t of S c o n t a i n s a countable number o f elements, hence, i t can be w r i t t e n a s a countable union of elementary e v e n t s . It i s t h e r e f o r e a n event. 2-7 Forming a l l unions, i n t e r s e c t i o n s , and complements of t h e s e t s E l ) and {2,3), we o b t a i n t h e following 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, t h e n 2-10 We use i n d u c t i o n . The formula i s t r u e f o r n = 2 because P(A1A2) - P ( A ~ I A ~ ) P ( A ~ ) . Suppose t h a t i t i s t r u e f o r n. Since we conclude t h a t i t must be t r u e f o r n + l . 2-11 F i r s t s o l u t i o n . The t o t a l number of m element s u b s e t s e q u a l s (") m ( s e e The t o t a l number of m element s u b s e t s c o n t a i n i n g 5 e q u a l s 0 Probl. 2-26). n- l (m-l) Hence Second s o l u t i o n . C l e a r l y , P{C, I is 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 ) A ~ ) = mln i s t h e p r o b a b i l i t y t h a t 5 0 where t h e summation i s over a l l sets A . m

2-12 2 (a) P E 6 < t < 8 1 = - 10 (b) ~ { 6 - - - < t - < 81t > 51 = P ( t , 51 P E 6 r t s 8 1 2 = - 5 2-13 From (2-27) it follows t h a t Equating t h e two s i d e s and s e t t i n g t l = t O + A t w e o b t a i n f o r every to. Hence, D i f f e r e n t i a t i n g t h e s e t t i n g c = a ( O ) , we conclude t h a t 2-14 I f A and B a r e independent, then P (AB) = P (A)P (B) . I f they a r e mutually exclusive, then P(AB) = 0 , Hence, A and B a r e mutually exclusive and independent i f f P(A)P(B) = 0.

C l e a r l y , A1 = A1A2 + ~ ~ hence i i ~ I f t h e e v e n t s A and 1 a r e independent, t h e n 2 hence, t h e e v e n t s A and A a r e independent. Furthermore, S i s independent w i t h any A because SA = A. This y i e l d s 1 2 P(SA) = P(A) = P(S)P(A) Hence, t h e theorem i s t r u e f o r n = 2 . To prove i t i n g e n e r a l we use i n d u c t i o n : Suppose t h a t A i s independent of A1, ..., A . C l e a r l y , a r e independent of B1, ... ,B . Therefore n+l n - An+l and An+l n 2.16 The desired probabilit,ies are given by (a) (TI ;)

2.17 Let Al I A2 and Ad represent tire events Al = "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 occu,rs nl = k - 1, A2 occurs n2 = 1 and A3 occurs n3 = 0 ) 2.18 All cars are equally likely so that the first car is selected with probability p = 113. This gives the desired probability to be 2.19 P{'drawing a white bad1 " } = && P("atleat one white ball i n k triu,ls ") = 1 - P("all black balls in k trials") 2.20 Let D = 2r represent the penny diameter. So long as the center of the penny is at a distance of r away from any side of the square, the penny will be entirely inside the square. This gives the desired probability to be

2.21 Refer to Exanlple 3.14. ( a ) Using (3.391, we get (h) P ( " t w o one-digit and four two-digit numbers1') = 2-22 The number of equations of the form P(AiAk) = P(Ai)P(Ak) n equals (*I. The number of equations involving r sets equals (:). number N of such equations equals Hence the total And since we conclude that 2-23 We denote by B1 and B2 respectively the balls in boxes 1 and 2 and by R the set of red balls. We have (assmption) P(B1) = P(BZ) '0.5 Hence (Bayes' theorem) P(R\B~) - 0.999 = 0.001 ~ ( ~ 1 8 ~ )

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