概率模型导论：第十一版简介

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"《Introduction to Probability Models》是概率模型领域的一本经典教材，适用于本科及研究生阶段的学习。该书已更新至第11版，由学术出版社（Academic Press，Elsevier的印记）出版。自1972年首次出版以来，每隔几年就会进行修订，以保持内容的最新性和相关性。本书旨在介绍概率建模的基本概念和方法，并涵盖了随机变量、概率分布、统计推断等多个核心主题。" 《Introduction to Probability Models》的第11版于2014年发布，前一版则在2010年出版。该书在过去的几十年里经历了多次更新，反映了概率论与统计学领域的不断发展和变化。内容可能包括但不限于以下方面： 1. 概率基础：书中首先会讲解概率论的基本原理，如样本空间、事件的概率、条件概率、独立事件等，为后续的建模工作奠定基础。 2. 随机变量：介绍离散型和连续型随机变量，包括它们的概率分布，如二项分布、泊松分布、均匀分布、正态分布等，以及它们的期望值和方差。 3. 联合分布与条件分布：探讨多个随机变量之间的相互关系，如何描述它们的联合分布，以及如何从已知信息中推断条件分布。 4. 概率分布的极限定理：如大数定律和中心极限定理，这些定理对于理解统计推断和实际问题的模拟具有重要意义。 5. 随机过程：可能会涉及马尔科夫链、布朗运动等随机过程理论，这对于理解和分析动态系统的随机行为至关重要。 6. 统计推断：包括参数估计、假设检验等，这部分内容将帮助读者学习如何基于观测数据对未知参数进行合理估计或做出统计决策。 7. 贝叶斯统计：可能会介绍贝叶斯定理及其在概率建模中的应用，这提供了一种从先验信息出发更新知识的框架。 8. 蒙特卡洛模拟：利用计算机进行随机抽样，解决那些难以解析求解的概率问题。 9. 实际应用：通过实例和案例研究，展示概率模型在工程、科学、经济和金融等领域的应用。此外，这本书可能还包括习题和解答，以帮助读者巩固所学知识并提升解决实际问题的能力。对于希望深入理解和应用概率模型的读者来说，这是一本非常有价值的参考书。在获取和使用本书内容时，需要遵守版权规定，未经许可不得擅自复制或传播。若需获得使用许可，应直接联系出版社或通过Elsevier的在线平台申请。

8 Introduction to Probability Models

Solution: If we let C be the event that Bev takes chemistry and A denote the event

that she receives an A in whatever course she takes, then the desired probability is

P(AC). This is calculated by using Equation (1.5) as follows:

P(AC) = P(C)P( A|C)



Example 1.7 Suppose an urn contains seven black balls and ﬁve white balls. We

draw two balls from the urn without replacement. Assuming that each ball in the urn

is equally likely to be drawn, what is the probability that both drawn balls are black?

Solution: Let F and E denote, respectively, the events that the ﬁrst and second

balls drawn are black. Now, given that the ﬁrst ball selected is black, there are six

remaining black balls and ﬁve white balls, and so P(E |F) =

.AsP(F ) is clearly

, our desired probability is

P(EF) = P(F)P(E |F)

132



Example 1.8 Suppose that each of three men at a party throws his hat into the center

of the room. The hats are ﬁrst mixed up and then each man randomly selects a hat.

What is the probability that none of the three men selects his own hat?

Solution: We shall solve this by ﬁrst calculating the complementary probability

that at least one man selects his own hat. Let us denote by E

, i = 1, 2, 3, the event

that the ith man selects his own hat. To calculate the probability P(E

∪ E

we ﬁrst note that

P(E

) =

, i = 1, 2, 3

P(E

) =

, i = j (1.6)

P(E

) =

To see why Equation (1.6) is correct, consider ﬁrst

P(E

) = P(E

)P(E

)

Now P(E

), the probability that the i th man selects his own hat, is clearly

since

he is equally likely to select any of the three hats. On the other hand, given that the

ith man has selected his own hat, then there remain two hats that the j th man may

select, and as one of these t wo is his own hat, it follows that with probability

will select it. That is, P(E

) =

and so

P(E

) = P(E

)P(E

) =

To calculate P(E

) we write

P(E

) = P(E

)P(E

)

P(E

)

10 Introduction to Probability Models

Let E

be the event that the sum of the dice equals seven. Is E

independent of F ?

The answer is yes since

P(E

F) = P({(4, 3)}) =

while

P(E

)P(F) =

We leave it for you to present the intuitive argument why the event that the sum of

the dice equals seven is independent of the outcome on the ﬁrst die. 

The deﬁnition of independence can be extended to more than two events. The events

, E

,...,E

are said to be independent if for every subset E



, E



,...,E



, r  n,

of these events

P(E



···E



) = P(E



)P(E



) ···P(E



)

Intuitively, the events E

, E

,...,E

are independent if knowledge of the occurrence

of any of these events has no effect on the probability of any other event.

Example 1.10 (Pairwise Independent Events That Are Not Independent) Let

a ball be drawn from an urn containing four balls, numbered 1, 2, 3, 4. Let E =

{1, 2}, F ={1, 3}, G ={1, 4}. If all four outcomes are assumed equally likely, then

P(EF) = P(E)P(F ) =

P(EG) = P(E)P(G) =

P(FG) = P(F)P(G) =

However,

= P(EFG) = P(E)P(F)P(G)

Hence, even though the events E, F, G are pairwise independent, they are not jointly

independent. 

Example 1.11 There are r players, with player i initially having n

units, n

>0, i =

1,...,r. At each stage, two of the players are chosen to play a game, with the winner

of the game receiving 1 unit from the loser. Any player whose fortune drops to 0 is

eliminated, and this continues until a single player has all n ≡



i=1

units, with

that player designated as the victor. Assuming that the results of successive games are

independent, and that each game is equally likely to be won by either of its two players,

ﬁnd the probability that player i is the victor.

Solution: To begin, suppose that there are n players, with each player initially

having 1 unit. Consider player i . Each stage she plays will be equally likely to

result in her either winning or losing 1 unit, with the results from each stage being

independent. In addition, she will continue to play stages until her fortune becomes

either 0 or n. Because this is the same for all players, it follows that each player has the

Introduction to Probability Theory 11

same chance of being the victor. Consequently, each player has player probability

1/n of being the victor. Now, suppose these n players are divided into r teams,

with team i containing n

players, i = 1,...,r. That is, suppose players 1,...,n

constitute team 1, players n

+ 1,...,n

+ n

constitute team 2 and so on. Then the

probability that the victor is a member of team i is n

/n. But because team i initially

has a total fortune of n

units, i = 1,...,r, and each game played by members

of different teams results in the fortune of the winner’s team increasing by 1 and

that of the loser’s team decreasing by 1, it is easy to see that the probability that

the victor is from team i is exactly the desired probability. Moreover, our argument

also shows that the result is true no matter how the choices of the players in each

stage are made. 

Suppose t hat a sequence of experiments, each of which results in either a “success”

or a “failure,” is to be performed. Let E

, i  1, denote the event that the i th experiment

results in a success. If, for all i

, i

,...,i

P(E

···E

) =



j=1

P(E

)

we say that the sequence of experiments consists of independent trials.

1.6 Bayes’ Formula

Let E and F be events. We may express E as

E = EF ∪ EF

because in order for a point to be in E, it must either be in both E and F ,oritmustbe

in E and not in F. Since EF and EF

are mutually exclusive, we have that

P(E) = P(EF) + P(EF

)

= P(E |F)P(F) + P(E|F

)P(F

)

= P(E |F)P(F) + P(E|F

)(1 − P(F)) (1.7)

Equation (1.7) states that the probability of the event E is a weighted average of the

conditional probability of E given that F has occurred and the conditional probability

of E given that F has not occurred, each conditional probability being given as much

weight as the event on which it is conditioned has of occurring.

Example 1.12 Consider two urns. The ﬁrst contains two white and seven black balls,

and the second contains ﬁve white and six black balls. We ﬂip a fair coin and then

draw a ball from the ﬁrst urn or the second urn depending on whether the outcome

was heads or tails. What is the conditional probability that the outcome of the toss was

heads given that a white ball was selected?

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