管理统计学
Statistics In Management
西南交通大学经济管理学院
王建琼 刁明碧
2001年 6月
第一部分 预备知识
Primary Knowledge
第 1章 概率论
Basic Probability
Probability Distributions
本章概要
? Basic Probability Concepts:
Sample Spaces and Events,Simple
Probability,and Joint Probability(联合概率 )。
? Conditional Probability(条件概率 )
? Bayesian Theorem（贝叶斯定理）
? The Probability of a Discrete Random Variable
? Binomial,Poisson,and Hypergeometric Distributions
? Covariance(协方差 ) and its Applications in Finance
Sample Spaces
（样本空间）Collection of all Possible Outcomes
e.g,All 6 faces of a die:
e.g,All 52 cards of a bridge deck:
Events（事件）
? Simple Event,Outcome from a Sample Space
with 1 Characteristic
e.g,A Red Card from a deck of cards.
? Joint or Compound Event,Involves 2 Outcomes
Simultaneously（同时发生称为,Joint”）
e.g,An Ace which is also a Red Card from a deck of cards.
An Ace given that it is a Red Card,
Visualizing Events（形象化事件）
?Contingency Tables（列联表）
?Tree Diagrams
（树状图）
Ace Not Ace Total
Red 2 24 26
Black 2 24 26
Total 4 48 52
Simple Events（ 简单 事件）
The Event of a Happy Face
There are 5 happy faces in this collection of 18 objects
Joint Events（ 联合事件 - - -交 ）
The Event of a Happy Face AND Light Colored
3 Happy Faces which are light in color
12 Items,5 happy faces and 7 other light objects
Compound Events（ 复合事件 ---并）
The Event of Happy Face OR Light Colored
??
Special Events（特殊事件）
Null Event（无效事件）
Club & Diamond on
1 Card Draw
Complement（补集） of
Event For Event A,
All Events Not In A:
Null Event
A
除非它是次品
补集的文氏图
The event,not A” happens whenever A does not
Venn diagram,A (in circle),“not A” (shaded)
Prob(not A) = 1 – Prob(A)
? If Prob(Succeed) = 0.7,then Prob(Fail) = 1–0.7 = 0.3
A
not A
3 Items,3 Happy Faces Given they are Light Colored
Dependent or Independent Events
相依或独立事件 给定的
The Event of a Happy Face GIVEN it is Light Colored
E = Happy Face???Light Color
Contingency Table（列联表）
A Deck of 52 Cards（ 去掉大、小王 ）
Ace(幺点 ) Not anAce Total
Red
Black
Total
2 24
2 24
26
26
4 48 52
Sample Space(样本空间 )
Red Ace
Tree Diagram（树图）
Event Possibilities
Red
Cards
Black
Cards
Ace
Not an Ace
Ace
Not an Ace
Full
Deck
of Cards
Probability（概率）
?Probability is the numerical
measure of the likelihood
that the event will occur.
?Value is between 0 and 1.
?Sum of the probabilities of
all mutually exclusive（互斥
的） events is 1.
Certain
必然的
Impossible
不可能的
.5
1
0
计算事件的概率 ----关键确定有效事件与全部事件
?The Probability of an Event,E:
?Each of the Outcome in the Sample Space
equally likely to occur(等概率发生 ),
S p a c eS a m p l einO u t c o m e sT o t a l
O u t c o m e sE v e n tofN u m b e r
)E(P ?
T
X?
e.g,P( ) = 2/36
(There are 2 ways to get one 6 and the other 4)
有效事件数
样本空间事件总数
Joint Probability（联合概率）
The Probability of a Joint Event,A and B:
e.g,P(Red Card and Ace) =
C a r d sofN um b e rT o ta l
A c e sdRe
52
2
26
1?
P(A and B) =
Number of Event Outcomes from both A and B
Total Outcomes in Sample Space
A与 B同时发生
交集的文氏图
Intersection happens whenever both events happen
Venn diagram,Intersection,A and B” shaded)
? e.g.,A =,sign contract”,B =,get financing”
? Did the intersection happen? Great! Project has been
launched!
? e.g.,Did I have eggs and cereal for breakfast?,No”
A B
P(A2 and B2)
P(A1 and B2)
P(A2 and B1)
P(A1 and B1)
Event
Event Total
Total 1
交集事件的联合概率
Joint Probability Marginal (Simple) Probability
边际概率
P(A1)A1
A2
B1 B2
P(A2)
P(B1) P(B2)
Compound Probability
（并集事件概率）,或”的关系
The Probability of a Compound Event,A or B:
S p a ceS a m p l einO u t c o m esT o t a l
BorAei t h erf ro mO u t c o m esE ve n tofN u m b e r)BorA(P ?
e.g,
P(Red Card or Ace)
C a r dsN um be r ofTo t a l
A c e sdReC a r dsdReA c e s
52
2264 ???
13
7
52
28
??
并集的文氏图
Union happens whenever either (or both) happen
Venn diagram,Union,A or B” shaded)
? e.g.,A =,get Intel job offer”,B =,get GM job offer”
? Did the union happen? Congratulations! You have a job
? e.g.,Did I have eggs or cereal for breakfast?,Yes”
A B
A+B
重复部分
理应减去
P(A1 and B1)
P(B2)P(B1)
P(A2 and B2)P(A2 and B1)
Event
Event Total
Total 1
Compound Probability
Addition Rule（可加性准则）
P(A1 and B2) P(A1)A1
A2
B1 B2
P(A2)
P(A1 or B1 ) = P(A1) +P(B1) - P(A1 and B1)
For Mutually Exclusive Events,P(A or B) = P(A) + P(B)
互斥的或不相容或不相交事件
Conditional Probability
（条件概率）
The Probability of the Event,
Event A given that Event B has occurred
P(A ?B) =
e.g,
P(Red Card given that it is an Ace) =
)B(P
)Ba n dA(P
2
1
4
2 ?
A c e s
A c e sdRe
若将 A,B对换，有
效与全部事件数会怎
样变化？
条件概率（续）
Given the extra information that B happens for sure,how must
you change the probability for A to correctly reflect this new
knowledge?
? This is a (conditional) probability about A
? The event B gives information
Unconditional
? The probability of A
Conditional
? A new universe,since B must happen
Prob (A given B) = Prob (A and B)Prob (B)
A B A and B B
现实中的条件概率
? Prob (Win given Ahead at halftime)
? Higher than Prob (Win) evaluated before the game
began
? Prob (Succeed given Good results in test market)
? Higher than Prob (Succeed) evaluated before
marketing study
? Prob (Get job given Poor interview)
? Lower than Prob (Get job given Good interview)
? Prob (Have AIDS given Test positive)
? Higher than Prob (Have AIDS) for the population-at-
large
Black
ColorType
Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
条件概率的列联表分析
Conditional Event,Draw 1 Card,Note Kind & Color
26
2
5226
522 ??
/
/
P ( R e d )
R e d )A N D P ( A c e = R e d ) |P ( A c e
Revised
Sample
Space
条件概率及事件的独立性
)B(P
)Ba n dA(P
Conditional Probability,P(A?B) =
P(A and B) = P(A ?B) P(B)
Events are
Independent:
P(A? B) = P(A)
Or,P(A and B) = P(A) P(B)
Events A and B are Independent when the probability
of one event,A is not affected by another event,B.
Multiplication Rule:
贝叶斯公式
后验概率
)B(P)BA(P)B(P)BA(P
)B(P)BA(P
kk
ii
???????
?
11
)A(P
)Aa ndB(P i?
P(Bi ?A) =
Adding up
the parts
of A in all
the Bi事件 B
j之间独立且构成
必然事件的完全划分；
分母利用全概率公式。
What are the chances of repaying a loan,
given a college education?
贝叶斯公式的列联表分析
Loan Status
Education Repay Defaul
t
Prob.
College,2,05,25
No College
Prob,1
P(Repay College) = 08.)D e f au l tan dC ol l e ge(P)payRean dC ol l e ge(P
)payRean dC ol l e ge(P ?
?
?
拖欠
Discrete Random Variable
(离散型随机变量 )? Random Variable,represents outcomes of an
experiment.
e.g,Throw a die twice:
Count the number of times 4 comes up (0,1,or 2 times)
? Discrete Random Variable,
? Obtained by Counting (0,1,2,3,etc.)
? Usually finite by number of values
e.g,Toss a coin 5 times,Count the number of tails.
(0,1,2,3,4,or 5 times)
离散事件实例
Probability Distribution
Values Probability
0 1/4 =,25
1 2/4 =,50
2 1/4 =,25
Event,Toss 2 Coins,Count of Tails.
T
T
T T
离散事件的概率分布
? List of All Possible [ Xi,P(Xi) ] Pairs
Xi = Value of Random Variable (Outcome)
P(Xi) = Probability Associated with Value
? Mutually Exclusive (No Overlap)
? Collectively Exhaustive (Nothing Left
Out)
0 ? P(Xi) ? 1
? P(Xi) = 1
离散随机变量的数字特征
Expected Value（数学期望值）
The Mean of the Probability Distribution Weighted Average
? = E(X) = ?Xi P(Xi)
e.g,Toss 2 coins,Count tails,Compute Expected Value:
??= 0 ?,25 + 1 ?.50 + 2 ?,25 = 1
Variance（方差）
Weighted Average Squared Deviation about Mean
?? = E [ (Xi - ? )2]=? (Xi - ? )2P(Xi)
e.g,Toss 2 coins,Count tails,Compute Variance:
?? = (0 - 1)2(.25) + (1 - 1)2(.50) + (2 - 1)2(.25) =,50
Number of Tails
常见的离散 (变量 )概率分布
Discrete Probability
Distributions
Binomial
二项分布
Hypergeometric
超几何分布
Poisson
泊松分布
二项分布的性质（ 1）
? 2 Identical Trials,e.g,15 tosses of a coin,
10 light bulbs taken from a warehouse
? 2 Mutually Exclusive Outcomes,
e.g,heads or tails in each toss of a coin,
defective or not defective light bulbs
? Constant Probability for each Trial,
e.g,probability of getting a tail is the same
each time we toss the coin and each
light bulb has the same probability of
being defective
二项分布的性质（ 2）
? 2 Sampling Methods:
Infinite Population Without Replacement
Finite Population With Replacement
? Trials are Independent:
The Outcome of One Trial Does Not Affect the
Outcome of Another
二项分布函数 (Function)
P(X) = probability that X successes given a knowledge of n
and p
X = number of successes?in
sample,(X = 0,1,2,...,n)
p = probability of success
n = sample size
P(X) nX ! n X p pX n X!( )! ( )? ? ? ?1
Tails in 2 Toss of Coin
X P(X)
0 1/4 =,25
1 2/4 =,50
2 1/4 =,25
二项分布的数字特征
标准差
n = 5 p = 0.1
n = 5 p = 0.5
Mean（均值）
Standard Deviation
?
?
E X np
np p
? ?
? ?
( )
( )
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
e.g,? = 5 (.1) =,5
e.g,? = 5(.5)(1 -,5)
= 1.118
0
泊松分布
区间
Poisson Process:
? Discrete Events in an interval
? The Probability of One Success in
Interval is Stable
? The Probability of More than One
Success in this Interval is 0
? Probability of Success is
Independent from Interval to
Interval
e.g,# Customers Arriving in 15 min.
# Defects Per Case of Light
Bulbs,
P X x
x
x
( |
!
? ?
??e-
泊松分布函数
P(X ) = probability of X successes given ?
? = expected (mean) number of successes
e = 2.71828 (base of natural logs)
X = number of successes per unit
P X
X
X
( )
!
? ?
??e
e.g,Find the probability of 4
customers arriving in 3 minutes
when the mean is 3.6,
P(X) = e
-3.6
3.6
4!
4
=,1912
泊松分布的数字特征
??= 0.5
??= 6
Mean
Standard Deviation
? ?
?
ii
N
i
E X
X P X
? ?
?
?1/?
?
?
( )
( )
1
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
.2
.4
.6
0 2 4 6 8 10
X
P(X)
超几何分布 试验
? n Trials in a Sample Taken From a
Finite Population of size N
? Sample taken Without Replacement
? Trials are Dependent 总体
? Concerned With Finding the Probability
of X?Successes in the Sample
where there are A?Successes in
the Population
超几何分布函数
P(X) = probability that X successes given n,N,and A
n = sample size
N = population size
A = number of successes
in population
X = number of successes
in sample (X = 0,1,2,...,n)
P X ) ( )( )
A
X( ???
( )
N - A
n - X
N
n
3 Light bulbs were selected
from 10,Of the 10 there
were 4 defective,What is
the probability that 2 of the
3 selected are defective?
P(2) = ( )( )
4
2
6
1
10
3( )
=,30
组合数
超几何分布的数字特征
Mean
Standard Deviation
?
?
E X n
nA N A
? ?
?
( )
N n
N 1
Finite
Population
Correction
A
N
N 2 ?
?? )(
随机变量的协方差 (Covariance)
X = discrete random variable X
Xi = ith outcome of X
P(XiYi) = probability of occurrence of the
ith outcome of Y
Y = discrete random variable Y
Yi = ith outcome of Y
i = 1,2,…,N
? ? ? ? )YX(P)Y(EY)X(EX iiiN
i
iXY ???? ??
? 1
?
数字特征应用：平均投资回报
Mean for Investment Returns
Return per $1,000 for two types of investments
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
.2 Recession -$100 -$200
.5 Stable Economy + 100 + 50
.3 Expanding Economy + 250 + 350
Investment
E(X) = ?X = (-100)(.2) + (100)(.5) + (250)(.3) = $105
E(Y) = ?Y = (-200)(.2) + (50)(.5) + (350)(.3) = $90
投资回报的方差
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
.2 Recession -$100 -$200
.5 Stable Economy + 100 + 50
.3 Expanding Economy + 250 + 350
Investment
Var(X) = = (.2)(-100 -105)2 + (.5)(100 - 105)2 + (.3)(250 - 105)2
= 14,725,?X = 121.35
Var(Y) = = (.2)(-200 - 90)2 + (.5)(50 - 90)2 + (.3)(350 - 90)2
= 37,900,?Y = 194.68
2X?
2Y?
投资回报的协方差
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
.2 Recession -$100 -$200
.5 Stable Economy + 100 + 50
.3 Expanding Economy + 250 + 350
Investment
?XY = (.2)(-100 - 105)(-200 - 90) + (.5)(100 - 105)(50 - 90)
+ (.3)(250 -105)(350 - 90) = 23,300
The Covariance of 23,000 indicates that the two investments are
strongly related and will vary together in the same direction.
本章小结
?Discussed Basic Probability Concepts:
Sample Spaces and Events,Simple Probability,
and Joint Probability
?Defined Conditional Probability
?Discussed Bayesian Theorem
?Addressed the Probability of a Discrete Random
Variable
?Discussed Binomial,Poisson,and Hypergeometric
Distributions
? Addressed Covariance and its Applications in Finance
Statistics In Management
西南交通大学经济管理学院
王建琼 刁明碧
2001年 6月
第一部分 预备知识
Primary Knowledge
第 1章 概率论
Basic Probability
Probability Distributions
本章概要
? Basic Probability Concepts:
Sample Spaces and Events,Simple
Probability,and Joint Probability(联合概率 )。
? Conditional Probability(条件概率 )
? Bayesian Theorem（贝叶斯定理）
? The Probability of a Discrete Random Variable
? Binomial,Poisson,and Hypergeometric Distributions
? Covariance(协方差 ) and its Applications in Finance
Sample Spaces
（样本空间）Collection of all Possible Outcomes
e.g,All 6 faces of a die:
e.g,All 52 cards of a bridge deck:
Events（事件）
? Simple Event,Outcome from a Sample Space
with 1 Characteristic
e.g,A Red Card from a deck of cards.
? Joint or Compound Event,Involves 2 Outcomes
Simultaneously（同时发生称为,Joint”）
e.g,An Ace which is also a Red Card from a deck of cards.
An Ace given that it is a Red Card,
Visualizing Events（形象化事件）
?Contingency Tables（列联表）
?Tree Diagrams
（树状图）
Ace Not Ace Total
Red 2 24 26
Black 2 24 26
Total 4 48 52
Simple Events（ 简单 事件）
The Event of a Happy Face
There are 5 happy faces in this collection of 18 objects
Joint Events（ 联合事件 - - -交 ）
The Event of a Happy Face AND Light Colored
3 Happy Faces which are light in color
12 Items,5 happy faces and 7 other light objects
Compound Events（ 复合事件 ---并）
The Event of Happy Face OR Light Colored
??
Special Events（特殊事件）
Null Event（无效事件）
Club & Diamond on
1 Card Draw
Complement（补集） of
Event For Event A,
All Events Not In A:
Null Event
A
除非它是次品
补集的文氏图
The event,not A” happens whenever A does not
Venn diagram,A (in circle),“not A” (shaded)
Prob(not A) = 1 – Prob(A)
? If Prob(Succeed) = 0.7,then Prob(Fail) = 1–0.7 = 0.3
A
not A
3 Items,3 Happy Faces Given they are Light Colored
Dependent or Independent Events
相依或独立事件 给定的
The Event of a Happy Face GIVEN it is Light Colored
E = Happy Face???Light Color
Contingency Table（列联表）
A Deck of 52 Cards（ 去掉大、小王 ）
Ace(幺点 ) Not anAce Total
Red
Black
Total
2 24
2 24
26
26
4 48 52
Sample Space(样本空间 )
Red Ace
Tree Diagram（树图）
Event Possibilities
Red
Cards
Black
Cards
Ace
Not an Ace
Ace
Not an Ace
Full
Deck
of Cards
Probability（概率）
?Probability is the numerical
measure of the likelihood
that the event will occur.
?Value is between 0 and 1.
?Sum of the probabilities of
all mutually exclusive（互斥
的） events is 1.
Certain
必然的
Impossible
不可能的
.5
1
0
计算事件的概率 ----关键确定有效事件与全部事件
?The Probability of an Event,E:
?Each of the Outcome in the Sample Space
equally likely to occur(等概率发生 ),
S p a c eS a m p l einO u t c o m e sT o t a l
O u t c o m e sE v e n tofN u m b e r
)E(P ?
T
X?
e.g,P( ) = 2/36
(There are 2 ways to get one 6 and the other 4)
有效事件数
样本空间事件总数
Joint Probability（联合概率）
The Probability of a Joint Event,A and B:
e.g,P(Red Card and Ace) =
C a r d sofN um b e rT o ta l
A c e sdRe
52
2
26
1?
P(A and B) =
Number of Event Outcomes from both A and B
Total Outcomes in Sample Space
A与 B同时发生
交集的文氏图
Intersection happens whenever both events happen
Venn diagram,Intersection,A and B” shaded)
? e.g.,A =,sign contract”,B =,get financing”
? Did the intersection happen? Great! Project has been
launched!
? e.g.,Did I have eggs and cereal for breakfast?,No”
A B
P(A2 and B2)
P(A1 and B2)
P(A2 and B1)
P(A1 and B1)
Event
Event Total
Total 1
交集事件的联合概率
Joint Probability Marginal (Simple) Probability
边际概率
P(A1)A1
A2
B1 B2
P(A2)
P(B1) P(B2)
Compound Probability
（并集事件概率）,或”的关系
The Probability of a Compound Event,A or B:
S p a ceS a m p l einO u t c o m esT o t a l
BorAei t h erf ro mO u t c o m esE ve n tofN u m b e r)BorA(P ?
e.g,
P(Red Card or Ace)
C a r dsN um be r ofTo t a l
A c e sdReC a r dsdReA c e s
52
2264 ???
13
7
52
28
??
并集的文氏图
Union happens whenever either (or both) happen
Venn diagram,Union,A or B” shaded)
? e.g.,A =,get Intel job offer”,B =,get GM job offer”
? Did the union happen? Congratulations! You have a job
? e.g.,Did I have eggs or cereal for breakfast?,Yes”
A B
A+B
重复部分
理应减去
P(A1 and B1)
P(B2)P(B1)
P(A2 and B2)P(A2 and B1)
Event
Event Total
Total 1
Compound Probability
Addition Rule（可加性准则）
P(A1 and B2) P(A1)A1
A2
B1 B2
P(A2)
P(A1 or B1 ) = P(A1) +P(B1) - P(A1 and B1)
For Mutually Exclusive Events,P(A or B) = P(A) + P(B)
互斥的或不相容或不相交事件
Conditional Probability
（条件概率）
The Probability of the Event,
Event A given that Event B has occurred
P(A ?B) =
e.g,
P(Red Card given that it is an Ace) =
)B(P
)Ba n dA(P
2
1
4
2 ?
A c e s
A c e sdRe
若将 A,B对换，有
效与全部事件数会怎
样变化？
条件概率（续）
Given the extra information that B happens for sure,how must
you change the probability for A to correctly reflect this new
knowledge?
? This is a (conditional) probability about A
? The event B gives information
Unconditional
? The probability of A
Conditional
? A new universe,since B must happen
Prob (A given B) = Prob (A and B)Prob (B)
A B A and B B
现实中的条件概率
? Prob (Win given Ahead at halftime)
? Higher than Prob (Win) evaluated before the game
began
? Prob (Succeed given Good results in test market)
? Higher than Prob (Succeed) evaluated before
marketing study
? Prob (Get job given Poor interview)
? Lower than Prob (Get job given Good interview)
? Prob (Have AIDS given Test positive)
? Higher than Prob (Have AIDS) for the population-at-
large
Black
ColorType
Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
条件概率的列联表分析
Conditional Event,Draw 1 Card,Note Kind & Color
26
2
5226
522 ??
/
/
P ( R e d )
R e d )A N D P ( A c e = R e d ) |P ( A c e
Revised
Sample
Space
条件概率及事件的独立性
)B(P
)Ba n dA(P
Conditional Probability,P(A?B) =
P(A and B) = P(A ?B) P(B)
Events are
Independent:
P(A? B) = P(A)
Or,P(A and B) = P(A) P(B)
Events A and B are Independent when the probability
of one event,A is not affected by another event,B.
Multiplication Rule:
贝叶斯公式
后验概率
)B(P)BA(P)B(P)BA(P
)B(P)BA(P
kk
ii
???????
?
11
)A(P
)Aa ndB(P i?
P(Bi ?A) =
Adding up
the parts
of A in all
the Bi事件 B
j之间独立且构成
必然事件的完全划分；
分母利用全概率公式。
What are the chances of repaying a loan,
given a college education?
贝叶斯公式的列联表分析
Loan Status
Education Repay Defaul
t
Prob.
College,2,05,25
No College
Prob,1
P(Repay College) = 08.)D e f au l tan dC ol l e ge(P)payRean dC ol l e ge(P
)payRean dC ol l e ge(P ?
?
?
拖欠
Discrete Random Variable
(离散型随机变量 )? Random Variable,represents outcomes of an
experiment.
e.g,Throw a die twice:
Count the number of times 4 comes up (0,1,or 2 times)
? Discrete Random Variable,
? Obtained by Counting (0,1,2,3,etc.)
? Usually finite by number of values
e.g,Toss a coin 5 times,Count the number of tails.
(0,1,2,3,4,or 5 times)
离散事件实例
Probability Distribution
Values Probability
0 1/4 =,25
1 2/4 =,50
2 1/4 =,25
Event,Toss 2 Coins,Count of Tails.
T
T
T T
离散事件的概率分布
? List of All Possible [ Xi,P(Xi) ] Pairs
Xi = Value of Random Variable (Outcome)
P(Xi) = Probability Associated with Value
? Mutually Exclusive (No Overlap)
? Collectively Exhaustive (Nothing Left
Out)
0 ? P(Xi) ? 1
? P(Xi) = 1
离散随机变量的数字特征
Expected Value（数学期望值）
The Mean of the Probability Distribution Weighted Average
? = E(X) = ?Xi P(Xi)
e.g,Toss 2 coins,Count tails,Compute Expected Value:
??= 0 ?,25 + 1 ?.50 + 2 ?,25 = 1
Variance（方差）
Weighted Average Squared Deviation about Mean
?? = E [ (Xi - ? )2]=? (Xi - ? )2P(Xi)
e.g,Toss 2 coins,Count tails,Compute Variance:
?? = (0 - 1)2(.25) + (1 - 1)2(.50) + (2 - 1)2(.25) =,50
Number of Tails
常见的离散 (变量 )概率分布
Discrete Probability
Distributions
Binomial
二项分布
Hypergeometric
超几何分布
Poisson
泊松分布
二项分布的性质（ 1）
? 2 Identical Trials,e.g,15 tosses of a coin,
10 light bulbs taken from a warehouse
? 2 Mutually Exclusive Outcomes,
e.g,heads or tails in each toss of a coin,
defective or not defective light bulbs
? Constant Probability for each Trial,
e.g,probability of getting a tail is the same
each time we toss the coin and each
light bulb has the same probability of
being defective
二项分布的性质（ 2）
? 2 Sampling Methods:
Infinite Population Without Replacement
Finite Population With Replacement
? Trials are Independent:
The Outcome of One Trial Does Not Affect the
Outcome of Another
二项分布函数 (Function)
P(X) = probability that X successes given a knowledge of n
and p
X = number of successes?in
sample,(X = 0,1,2,...,n)
p = probability of success
n = sample size
P(X) nX ! n X p pX n X!( )! ( )? ? ? ?1
Tails in 2 Toss of Coin
X P(X)
0 1/4 =,25
1 2/4 =,50
2 1/4 =,25
二项分布的数字特征
标准差
n = 5 p = 0.1
n = 5 p = 0.5
Mean（均值）
Standard Deviation
?
?
E X np
np p
? ?
? ?
( )
( )
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
e.g,? = 5 (.1) =,5
e.g,? = 5(.5)(1 -,5)
= 1.118
0
泊松分布
区间
Poisson Process:
? Discrete Events in an interval
? The Probability of One Success in
Interval is Stable
? The Probability of More than One
Success in this Interval is 0
? Probability of Success is
Independent from Interval to
Interval
e.g,# Customers Arriving in 15 min.
# Defects Per Case of Light
Bulbs,
P X x
x
x
( |
!
? ?
??e-
泊松分布函数
P(X ) = probability of X successes given ?
? = expected (mean) number of successes
e = 2.71828 (base of natural logs)
X = number of successes per unit
P X
X
X
( )
!
? ?
??e
e.g,Find the probability of 4
customers arriving in 3 minutes
when the mean is 3.6,
P(X) = e
-3.6
3.6
4!
4
=,1912
泊松分布的数字特征
??= 0.5
??= 6
Mean
Standard Deviation
? ?
?
ii
N
i
E X
X P X
? ?
?
?1/?
?
?
( )
( )
1
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
.2
.4
.6
0 2 4 6 8 10
X
P(X)
超几何分布 试验
? n Trials in a Sample Taken From a
Finite Population of size N
? Sample taken Without Replacement
? Trials are Dependent 总体
? Concerned With Finding the Probability
of X?Successes in the Sample
where there are A?Successes in
the Population
超几何分布函数
P(X) = probability that X successes given n,N,and A
n = sample size
N = population size
A = number of successes
in population
X = number of successes
in sample (X = 0,1,2,...,n)
P X ) ( )( )
A
X( ???
( )
N - A
n - X
N
n
3 Light bulbs were selected
from 10,Of the 10 there
were 4 defective,What is
the probability that 2 of the
3 selected are defective?
P(2) = ( )( )
4
2
6
1
10
3( )
=,30
组合数
超几何分布的数字特征
Mean
Standard Deviation
?
?
E X n
nA N A
? ?
?
( )
N n
N 1
Finite
Population
Correction
A
N
N 2 ?
?? )(
随机变量的协方差 (Covariance)
X = discrete random variable X
Xi = ith outcome of X
P(XiYi) = probability of occurrence of the
ith outcome of Y
Y = discrete random variable Y
Yi = ith outcome of Y
i = 1,2,…,N
? ? ? ? )YX(P)Y(EY)X(EX iiiN
i
iXY ???? ??
? 1
?
数字特征应用：平均投资回报
Mean for Investment Returns
Return per $1,000 for two types of investments
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
.2 Recession -$100 -$200
.5 Stable Economy + 100 + 50
.3 Expanding Economy + 250 + 350
Investment
E(X) = ?X = (-100)(.2) + (100)(.5) + (250)(.3) = $105
E(Y) = ?Y = (-200)(.2) + (50)(.5) + (350)(.3) = $90
投资回报的方差
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
.2 Recession -$100 -$200
.5 Stable Economy + 100 + 50
.3 Expanding Economy + 250 + 350
Investment
Var(X) = = (.2)(-100 -105)2 + (.5)(100 - 105)2 + (.3)(250 - 105)2
= 14,725,?X = 121.35
Var(Y) = = (.2)(-200 - 90)2 + (.5)(50 - 90)2 + (.3)(350 - 90)2
= 37,900,?Y = 194.68
2X?
2Y?
投资回报的协方差
P(XiYi) Economic condition Dow Jones fund X Growth Stock Y
.2 Recession -$100 -$200
.5 Stable Economy + 100 + 50
.3 Expanding Economy + 250 + 350
Investment
?XY = (.2)(-100 - 105)(-200 - 90) + (.5)(100 - 105)(50 - 90)
+ (.3)(250 -105)(350 - 90) = 23,300
The Covariance of 23,000 indicates that the two investments are
strongly related and will vary together in the same direction.
本章小结
?Discussed Basic Probability Concepts:
Sample Spaces and Events,Simple Probability,
and Joint Probability
?Defined Conditional Probability
?Discussed Bayesian Theorem
?Addressed the Probability of a Discrete Random
Variable
?Discussed Binomial,Poisson,and Hypergeometric
Distributions
? Addressed Covariance and its Applications in Finance
