《应用随机过程》第9版教材解答全英版

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《应用随机过程：概率模型导论》第九版是Sheldon M. Ross撰写的一本权威教材，专为那些希望深入理解随机过程在实际问题中的应用和理论背景的学生和专业人士设计。该书不仅介绍了随机过程的基本概念，如随机变量序列、马尔可夫链、布朗运动等，还涵盖了更高级的主题，如泊松过程、高维随机向量和随机积分。作为教材的配套教师手册，它提供了详尽的答案解析，旨在帮助教师在教学过程中解答学生的疑问，确保学生能够准确掌握课程内容。每一章节后可能包含习题的答案，这些习题旨在检验学生对所学理论的理解程度，例如如何计算随机过程的概率分布，如何解决与随机过程相关的统计推断问题，以及如何应用随机过程理论来模拟和分析实际世界中的随机现象。第九版的特点可能包括对现代技术的更新，比如可能介绍了与计算机科学、金融工程、信号处理等领域相关的随机过程应用实例。此外，书中可能还强调了随机过程在诸如通信系统、生物统计、气候建模等领域的实际应用，突出了理论与实践相结合的重要性。版权方面，本书受到严格保护，未经Elsevier公司的书面许可，任何形式的复制或传播，无论是电子的还是机械的，包括复印、录音或通过任何信息存储和检索系统，都是非法的。读者如果需要获取许可，可以直接联系Elsevier的科技与许可部门，或者通过Elsevier的官方网站进行在线申请。《应用随机过程：概率模型导论》第九版是一本实用且深入的教育资源，对于学习者来说，它是提升随机过程理论技能和应用能力的强大工具，而对于教育工作者，则是教学支持和评估的重要辅助资料。

12 Answers and Solutions

we have

[X]=

∑

i=1

E[X

]

r(r − 1)

(4n − 2)

Because

r(r − 1)(r − 2)(r − 3)

2n(2n − 1)(2n − 2)2n − 3)

For Var (X) use

Var

(X)=

∑

Var(X

)+2

∑

i< j

Cov(X

, X

)

nVar(X

)+n (n − 1) Cov(X

, X

)

where

Var

)=E[X

](1 − E[X

])

Cov(X

, X

r(r − 1)(r − 2)(r − 3)

2n(2n − 1)(2n − 2)(2n −3)

−(

E[X

])

59. (a) Use the fact that F(X

) is a uniform (0, 1) ran-

dom variable to obtain that

= P{F(X

) < F(X

) > F(X

) < F(X

)}

P{U

< U

> U

< U

}

where the U

, i = 1, 2, 3 , 4, are independent

uniform

(0, 1) random variables.

(b) p



(1 − x

)dx



− x

/2)dx



(1/3 − x

/2 + x

/6)dx

= 1/3 −1/6 + 1/24 = 5/24

that X

< X

> X

< X

. They are as follows:

> X

60. E[e



dx =

−1

−e

+ 1

E[e

[

(te

+ e

−e

) − 2t(te

−e

+ 1)]

−2(te

− e

+ 1)

To evaluate at t

= 0, we must apply l’Hospital’s

rule.

This yields

[X]= lim

t = 0

+ e

−e

= lim

t = 0

]=lim

t = 0

2te

+ t

−2te

−2e

+ 2e

= lim

t = 0

Hence, Var

(X)=

−





61. φ

(t)=

+ e

62. E

αλX



αλx

λe

−

λx

dx =

1 −α

Therefore,

= −

αλ

(1−α).

The inequality ln

(1 −x) ≤−x shows that

≥ 1/λ.

63. φ

(t)=

∞

∑

n=1

(1 − p)

n−1

= pe

∞

∑

n=1

((1 − p)e

)

n−1

1 − (1 − p)e

64. (See Section 2.3 of Chapter 5.)

65. Cov

, X

)=Cov(

∑

k=1

∑

t=1

)

∑

t=1

∑

k=1

Cov(a

, a

)

14 Answers and Solutions

73. As N

is a binomial random variable with para-

meters

(n, P

), we have (i) E[N

]=nP

(ii) Va r( X

)

=(1 − P

); (iii) for i = j, the covariance of

and N

can be computed as:

Cov

, N

)=Cov



∑



where X

) is 1 or 0, depending upon whether

or not outcome k is type i

(j). Hence,

Cov

, N

∑



Cov(X

, Y



Now for k

= , Cov(X

, Y



)=0 by independence

of trials and so

Cov

, N

∑

Cov(X

, Y

)

∑

(E[X

] − E[X

]E[Y

])

= −

∑

E[X

]E[Y

] (since X

= 0)

= −

∑

= −nP

(iv) Letting



1, if no type i’s occur

0, otherwise,

we have that the number of outcomes that never

occur is equal to

∑

and thus,



∑



∑

E[Y

]

∑

P{outcomes i does not occur}

∑

(

1 − P

)

74. (i) As the random variables are independent,

identically distributed, and continuous, it fol-

lows that, with probability 1, they will all

have different values. Hence the largest of

,...,X

is equally likely to be either X

...orX

. Hence, as there is a record at time

n when X

is the largest value, it follows that

{a record occurs at n} =

(ii) Let I



1, if a record occurs at j

0, otherwise.

Then



∑



∑

E[I

∑

(iii) It is easy to see that the random variables

, I

,...,I

are independent. For instance, for

< k

= 1/I

= 1} = P{I

= 1},

since knowing that X

is the largest of

,...,X

clearly tells us nothing

about whether or not X

is the largest of

,...,X

. Hence,s

Var

∑

Var(I

∑

j=1





−1



(iv) P

{N > n}

P{X

is the largest of X

,...,X

} =

Hence,

[N]=

∞

∑

n=1

P{N > n} =

∞

∑

n=1

= ∞.

75. (i) Knowing the values of N

,...,N

is equiva-

lent to knowing the relative ordering of the

elements a

,...,a

. For instance, if N

= 0,

= 1, N

= 1 then in the random permu-

tation a

is before a

, which is before a

. The

independence result follows for clearly the

number of a

,...,a

that follow a

i+1

does not

probabilistically depend on the relative order-

ing of a

,...,a

(ii) P

= k} =

, k

= 0,1,...,i − 1

which follows since of the elements

,...,a

i+1

the element a

i+1

is equally

likely to be ﬁrst or second or ...or

(i + 1)

(iii) E

i−1

∑

k=0

k =

i −1

i−1

∑

k=0

(

i −1)(2i −1)

and so

Var

(

i −1)(2i −1)

−

(

i −1)

−1

76. E

[XY]=µ

E[( XY )

]=(µ

)(

)

Var(XY )=E[(XY )

] − (E[XY])

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