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Solutions Manual
to pany
Probability,
Random Variables
and
Stochastic Processes
Fourth Edition
Athanasios Papoulis
Polytechnic University
S. Unnikrishna Pillai
Polytechnic University
Solutions Manual to pany
PROBABILITY, RANDOM VARIABLES AND STOCHASTIC PROCESSES, FOURTH EDITION
ATHANASIOS PAPOULIS
Published by McGraw-Hill Higher Education, an imprint of The McGraw-panies, Inc., 1221 Avenue of the Americas,
New York, NY 10020. Copyright © 2002 by The McGraw-panies, 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-panies, Inc.,
including, but not limited to, in work or other electronic storage or transmission, or broadcast for distance learning.
CHAPTER 2
2-1 We use- De Morgan's - law:
(a) X+6+ I+B= AB + A% = A(B+%) = A
because = (01 BB = I01
2-2 If A = {2<x;5)- B = {3<x<61- - S = {-=-<x<=-) then
A+B = {2<x<6)- - AB = {3<x<5)- -
(A+B)(E)= {2<x<61- - [{x<31 + Ex>51]
= {2<x<3)- + {5<x<61-
2-3 If AB = (0 1 then A c;; hence
P (A) -< P (i)
2-4 (a) P(A) = P(AB) + P(A~) P(B) = P(AB) + P(XB)
If, therefore, P(A) =P(B) = P(AB) then
P(G)= 0 ~(h)= 0 hence
P(XB+AIB) = P(XB) + P(A%) = o
(b) If P(A) = P (B) = 1 then 1 = P (A) 5 P (A + B) hence
1 = P(A+B) = P(A) + P(B) - P(AB) = 2 - P(AB)
This vields P(AB) = 1
2-5 From (2-1 3) it follows that
P(A+B+C) = P(A) + P(B+C) - P[A(B+c)]
P(B+C) = P(B) + P(C) - P(BC)
P [A(B + C) ] = P (AB) + P(AC) - P(ABC)
because ABAC = ABC. Combining, we obtain the desired result.
Using induction, we can show similarly t