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Probability Random Variables and Stochastic Processes - Papoulis

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Solutions Manual

to accompany

Probability,

Random Variables

and

Stochastic Processes

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,

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

2-1

use

De Morgan's law:

(a)

X+6+

I+B

A(B+%)

because

(01

I01

2-2

{2<x;5)

{3<x<61

- -

{-=-<x<=-)

then

A+B

{2<x<6)

- -

{3<x<5)

(A+B)(E)

{2<x<61

- -

[{x<31

Ex>51]

{2<x<3)

{5<x<61

2-3

then

c;;

hence

P (A)

(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)

~(h)

hence

P(XB+AIB)

P(XB)

P(A%)

(b)

P(A)

P (B)

then

P (A)

hence

P(A+B)

P(A)

P(B)

P(AB)

This

vields P(AB)

2-5 From (2-1

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

]

P (AB)

P(AC)

P(ABC)

because ABAC

ABC. Combining, we

obtain

the

desired

result.

Using

induction,

can

show similarly

that

?(A +A2+*-+A

)

P(A1)

P(A2)+*** +P(An)

P(A A

)

...

'An-lAn'

P (A1A2A3)

P (An-2An)

*..*.I................*.,.*

kP(A A

An)

___

_..__---I

----

2-6

Any subset of

contains a countable number of elements, hence,

can be written as a countable union of elementary events.

therefore an event.

2-7 Forming all unions, intersections, and complements of the sets

El)

and {2,3),

obtain the following

sets:

(01,

C11,

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

2-8 If ACB,P(A)

114, and P(B)

113, then

2-10 We use induction. The formula

true for n=2 because

P(A1A2)

P(A~IA~)P(A~).

Suppose that

true for n. Since

we conclude that

must be true forn+l.

2-11 First solution. The total number of

element subsets equals (") (see

Probl. 2-26). The total number of

element subsets containing

equals

(m-l) Hence

Second solution.

Clearly,

P{C,

IA~)

mln

the probability that

specific

Am.

Hence (total probability)

where the summation

over

all

sets

2-12 (a) PE6<t<81=-

- -

PE6rts81 2

(b)

~{6

81t

P(t,

2-13 From (2-27)

follows that

Equating the two sides and setting

tl=tO+

obtain

for every

to.

Hence,

Differentiating the setting

c=a(O),

conclude that

2-14

and

are independent,

then P (AB)

(A)P (B)

If they are

mutually exclusive, then

P(AB)

Hence,

and

are mutually

exclusive and independent iff

P(A)P(B)

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