Download Figure 3.1.1 The distribution function in Example 3.1.3

3. Random Variables
3.1 Definition of Random Variables
In engineering or scientific problems, we are not only interested in
the probability of events, but also interested in some variables depending
on sample points. （定义在样本点上的变量）
For example, we maybe interested in the life of bulbs produced by a
certain company, or the weight of cows in a certain farm, etc. These ideas
lead to the definition of random variables.
1. random variable definition
Definition 3.1.1 A random variable is a real valued function defined on
a sample space; i.e. it assigns a real number to each sample point in the
sample space.
Here are some examples.
Example 3.1.1
A fair die is tossed. The number X shown is a random
variable, it takes values in the set {1, 2, 6} .
Example 3.1.2
The life t of a bulb selected at random from bulbs
produced by company A is a random variable, it takes values in the
interval (0 , ) .
Since the outcomes of a random experiment can not be predicted in
advance, the exact value of a random variable can not be predicted before
the experiment, we can only discuss the probability that it takes some
27
value or the values in some subset of R.
2. Distribution function
Definition 3.1.2
Let X be a random variable on the sample space S .
Then the function
F ( X )  P( X  x) .
xR
is called the distribution function of X
Note The distribution function F ( X ) is defined on real numbers, not on
sample space.
Example 3.1.3
Let X be the number we get from tossing a fair die.
Then the distribution function of X is (Figure 3.1.1)
0,
n

F ( x)   ,
6
1,
if x  1;
if n  x  n  1, n  1, 2,
,5;
if x  6.
Figure 3.1.1 The distribution function in Example 3.1.3
3. Properties
The distribution function F ( x) of a random variable X has the
following properties:
(1) F ( x) is non-decreasing.
28
In fact, if x1  x2 , then the event { X  x1} is a subset of the event
{ X  x2 } ,thus
F ( x1 )  P( X  x1 )  P( X  x2 )  F ( x2 )
F ( x)  0 ,
(2) F ()  xlim

F (  )  l i m
F x( ).
x  
(3)For any x0  R ,
1
lim F ( x)  F ( x0  0)  F ( x0 ) .This is to say, the
x  x0 0
distribution function F ( x) of a random variable X is right continuous.
Example 3.1.4 Let X be the life of automotive parts produced by
company A , assume the distribution function of X is (in hours)
x


2000
,
F ( x)  P( X  x)  1  e
0,
x  0;
x  0.
Find P( X  2000) , P(1000  X  3000) .
Solution
By definition,
P( X  2000)  F (2000)  1  e1  0.6321 .
P(1000  X  3000)  P( X  3000)  P( X  1000)
 F (3000)  F (1000)  (1  e1.5 )  (1  e0.5 )  0.3834
Question： What are the probabilities P( X  2000) and P( X  2000) ?
Example 3.1.5
A player tosses two fair dice, if the total number shown
is 6 or more, the player wins $1, otherwise loses $1. Let X be the
amount won, find the distribution function of X .
Solution
29
Let X1 be the total number shown, then the events { X 1  k} contains
k 1 sample points, k  2,3, 4,5 . Thus
P( X 1  k ) 
k 1
,
36
k  2,3, 4,5
And
{ X  1} 
5
{ X 1  k}
k 2
so
5
P( X  1)   P ( X  k ) 
k 2
P( X  1)  1  P( X  1) 
5
18
13
18
Thus
0 ,
5

F ( x ) P X(  x ) 
18
1,
x 1;
,   1x 
1;
x  1.
Figure 3.1.2 The distribution function in Example 3.1.5
The distribution function of random variables is a connection between
probability and calculus. By means of distribution function, the main
tools in calculus, such as series, integrals are used to solve probability and
statistics problems.
3.2 Discrete Random Variables 离散型随机变量
30
In this book, we study two kinds of random variables.
Definition 3.2.1
A random variable X is called a discrete random
variable, if it takes values from a finite set or, a set whose elements can
be written as a sequence {a1 , a2 , an , }
Assume a discrete random variable X takes values from the set
X  {a1 , a2 ,
an , } . Let
P( X  an )  pn , n  1, 2,
Then we have pn  0 , n  1, 2, ,
p
n
.
(3.2.1)
 1.
n
the probability distribution of the discrete random variable X (概率分
布)
X
a1
probability
p1
a2
an
p2
pn
注意随机变量 X 的分布所满足的条件
(1)
Pi ≥0
(2) P1+P2+…+Pn=1
离散型分布函数
And the distribution function of X is given by
F ( x)  P( X  x)   pn
(3.2.2)
an  x
In general, it is more convenient to use (3.2.1) instead of (3.2.2).
Equation (3.2.1) is called the probability distribution of the discrete
random variable X .
31
Example 1
For an experiment in which a coin is tossed three times (or 3 coins are
tossed once), construct the distribution of X. (Let X denote the number of
head occurrence)
Solution
n=3, p=1/2
X
pr
0
1/8
1
3/8
2
3/8
3
1/8
Example 2
在一次试验中，事件 A 发生的概率为 p, 不发生的概率为 1－p, 用
X=0 表示事件 A 没有发生，X=1 表示事件 A 发生，求 X 的分布。
two-point distribution(两点分布)
X 0
1
P 1-p p
某学生参加考试得 5 分的概率是 p, X 表示他首次得 5 分的考试次数，
求 X 的分布。
geometric distribution （几何分布）
X 1 2
P
3
4
p q1p q2p q3p
… k
…
qk－1p …
32
Example 3 (射击 5 发子弹)
某射手有 5 发子弹，射一次命中率为 0.9,如果命中目标就停止射
击，如果不命中则一直射到子弹用尽，求耗用子弹数 x 的概率分布。
*****
Example 3.2.1
A die is tossed, by X we denote the number shown, Assume that the
probability P( X  k ) is proportional to k , k  1, 2, , 6 . Find the
probability distribution of X .
Solution Assume that
P( X  k )  ck , c  constant, k  1, 2,
,6.
Since the events P( X  k ) , k  1, 2, , 6 are mutually exclusive and
their union is the certain event, i.e., the sample space S , we have
6
1   P ( X  k )  21c ,
k 1
thus c 
1
. The probability distribution of X is (Figure 3.2.1)
21
k
k  1, 2, , 6 .
P( X  k ) 
,
21
p
2/7
x
1
2
3
4
5
6
Figure 3.2.1 Probability distribution in Example 3.2.1
Question. What is the difference between distribution functions and
33
probability distributions
例 2 有一种验血新方法：把 k 个人的血混在一起进行化验，如果结
果是阴性，那么对这 k 个人只作一次检验就够了，如果结果是阳性，
那么必须对这 k 个人再逐个分别化验，这时 k 个人共需作 k+1 次检
验。
假设对所有人来说，化验是阳性反应的概率为 p，而且这些人反
映是独立的。设 X 表示每个人需要化验的次数，求 X 的分布（construct
the distribution of X ）
Binomial distribution（二项分布）
Example 3.2.2
A fair die is tossed 4 times. Let X be the number of
six got. Find the probability distribution of X .
Solution. The possible values of X are 0,1, 2,3, 4 .
First we find the probability P( X  0) .
Since X  0 means that no six occur in 4 tosses.
The probability that six fails to occur in a single toss is 5/ 6 , and all
trials are independent, so
4
5
P ( X  0)    .
6
Now consider the probability P( X  k ) , k  1, 2,3, 4 .
Since X  k means that six occurs exactly k times, they may occur
in any k tosses of 4 tosses.
The event that they occur in a special order (for example, the first k
tosses), has probability (5 / 6)4k (1/ 6)k ,
34
and we have C4k such combinations. Thus
P( X  k )  C4k (5 / 6) 4 k (1/ 6) k
i.e.
P( X  0) 
625
125
25
, P( X  1) 
, P( X  2) 
1296
324
216
P( X  3) 
4
1
, P( X  4) 
.
325
1296
X 0 1 2 3 4
P
Binomial Distributions
An experiment often consists of repeated trials, each with two
possible outcomes “success” and “failure”. The most useful application
deals with the testing of items as they come off an assembly line, where
each test or trial may indicate a defective or a non-defective item. We
may choose to define either outcome as a success. The process is referred
to a Bernoulli process. Each trial is called a Bernoulli trial.
Consider an experiment consists of n independent repeated trials,
each trials result in two outcomes “success” and “failure”, and the
probability of success, denote by p , remains constant. Then this process
is called a Bernoulli process.
Definition 3.4.1 The number X of successes in n Bernoulli trials is
called a binomial random variable. The probability distribution of this
discrete random variable is called the binomial distribution with
35
parameters n and p , denoted by B(n, p) .
The random variable in Example 3.2.2 is an example of binomial
random variable.
Theorem 3.4.1 The probability distribution of the binomial distribution
with parameters n and p is given by
k
P( X  k)  b( k, n, 
p) nk C
p
( 1 n , kp)k  0,1, 2,
Proof
,n
(3.4.1)
First, consider the probability of obtaining k consecutive
successes, followed by n  k consecutive failures. These n events are
independent, therefore the desired probability is p k (1  p)nk .
Since the k successes and n  k failures may occur in any order,
and for any specific order, the probability is again p k (1  p)nk . We must
now determine the total number of sample points in the experiment that
have k successes and n  k failures. This number is equal to the
number of partition of n outcomes into two groups with k in one
group and n  k in the other, i.e. Cnk . Because the partitions are mutually
exclusive, thus we have
k
P( X  k)  b( k, n, 
p) nk C
p
( 1 n , kp) k  0,1, 2,
,n
Let q  1  p , the binomial expansion of the expression (q  p)n
gives
1  (q  p) n  Cn0 q n  Cn1 pq n 1 
 b(0; n, p)  b(1; n, p) 
 Cnn p n
 b(n; n, p) .
36
Each term correspond to various values of binomial distribution, this is
the reason that we called it “binomial distribution”.
Example 2
For an experiment in which 9 coins are tossed, Let X denotes the
number of head occurrence, construct the binomial distribution of X
what is the probability of obtaining between 3 and 6 successes.
poisson distribution（泊松分布）
Definition 3.5.1 A discrete random variable X is called a Poisson
random variable, if it takes values from the set {0,1, 2, } , and if
P( X  k )  p (k ;  ) 
k
k!
 e
 0
,
k  0,1, 2,
(3.5.1)
Distribution (3.5.1) is called the Poisson distribution with
parameter  , denoted by P( ) .
Note that


k
k 0
k 0
k!
 P( X  k )  

k
k 0
k!
e   e  
 e    e  1 .
Here are some examples of Poisson random variables:
(a) the number of radioactive particles passing through a counter in
certain time period;
(b) the number of telephone calls received by an office in certain time
period;
(c) the number of bacteria in a given culture;（细菌，培养基）
37
(d) the number of typing errors per page in a certain book.
Example 3.5.1
From a laboratory experiments, it is known that the number X of
radioactive particles passing（放射性粒子） through a counter in a given
millisecond is a Poisson random variable with parameter   4 . What is
the probability that 6 particles enter the counter in a given millisecond?
Solution The probability is
p(6; 4) 
46 4
 e  0.1042 .
6!
Example 3.5.2
The number X of oil tankers arriving each day at a certain port is a
Poisson random variable with parameter 10. What is the probability that
on a given day no more than 3 tankers having arrived?
Solution. The probability is
3
10k 10
e
k 0 k !
3
P   p(k ;10)  
k 0


1
2 1
 e1 01  10   10 
 1000   0.0103 .
2
6


Homework
Chapter 3 (P47) 1， 2， 3， 5，7，
21
二项分布与泊松分布的关系
Theorem 3.5.2
Let X be a sequence of binomial random variables with probability
38
distribution b(k ; n, pn ) . If for some constant  , we have npn   when
n   , and pn  0 , then
b(k ; n, pn )  p(k ;  ) , when n   , and pn  0 .
proof
****
Example 3.5.4 Suppose that, on average, 1 person in 1000 makes a
numerical error in preparing his or her income tax return. If 5000 forms
are selected at random and examined, find the probability that 6, 7 or 8 of
the forms contain an error.
Solution Let X be the number of forms contain an error, then X has
the binomial distribution of parameter n  5000 and p  0.001 .
Using Poisson distribution as approximations, we have
b(6;5000, 0.001)  p(6,5) 
56 5
 e  0.1462 ;
6!
b(7;5000, 0.001)  p(7,5) 
57 5
 e  0.1044 ;
7!
58 5
b(8;5000, 0.001)  p(8,5)   e  0.0653 .
8!
二项分布的应用例子
Example 3.4.1 It is known that 15% of certain articles manufactured are
defective, what is the probability that in a random sample of 5 articles
(a ) exactly 2 are defective.
(b) at least 2 are defective.
39
Solution In this case, n  5, p  0.15 .
(a) The probability is
b(2;5, 0.15)  C52 (0.152 )(0.853 )  0.1382 .
(b)The desired probability is the sum of getting 2, 3, 4, 5 defective
articles, or, we may first find the probability of the complement event, i.e.,
getting 0 or 1 defective article. So, if we denote the number of defective
articles by X , then we have
P( X  2)  1  P( X  1)  1  0.855  C51 (0.15)(0.854 )  0.1648 .
□
Example 3.4.2 A man is able to hit a target 7 times of 10 on the average.
(a) Find the probability that he hits the target exactly 3 times in 6
shots;
(b) In how many shots the probability that he hits the target at least
one time is greater than 0.95?
Solution
(a) The probability is
b(3;6, 0.7)  C63 (0.73 )(0.33 )  0.1852 .
(b) In n shots, the probability that he hits at least one time is
1  b(0; n,0.7)  1  0.3n .
Since when n  3 , we have
1  0.3n  0.95 ,
so in 3 shots, the probability that he hits the target at least one time is
1  0.33  0.973  0.95 .
40
3.3. Expectation and Variance
1．Expectation (mean) 数学期望
Suppose in the final exam, you got 85 in calculus, 90 in algebra and
83 in statistics, then your average score is (85  90  83) / 3  86 .
Assume a player tossed a fair die 20 times. He won $11 when he get
six and lost $1 otherwise. If he gets six 4 times. Then the average amount
he gets per toss is
1
(11 4  (1) 16)  1.4
20
Consider the future games. Since we cannot predict the outcome of
the game, we cannot predict the exact amount he will win in the game.
But we can predict the average amount he will win. Assume he tosses the
die 600 times, in average, six will occur 100 times, thus, the average
amount he will win per toss would be
1
1
5
(11100  (1)  500)  11  (1)    1
600
6
6
We say that in average he will win $1 per toss.
Definition 3.3.1 Let X be a discrete random variable. The expectation
or mean of X is defined as
  E ( X )   xP( X  x)
(3.3.1)
x
Notice
In the case that
X
takes values from an infinite number set, (3.3.1)
becomes an infinite series. If the series converges absolutely（级数绝对
41
收敛）, we say the expectation E ( X ) exists, otherwise we say that the
expectation of X does not exist.
Example 3.3.1
A fair die is tossed . Find the expectation of spots X shown.
Solution
Since X takes values from the set {1, 2,3, 4,5, 6} and the
distribution is
P( X  k ) 
1
,
6
k  1, 2,
,6.
Thus,
6
1 7
E( X )   k   .
6 2
k 1
If a discrete random variable assume each of its values with an equal
probability, we say this probability distribution is a discrete uniform
distribution（离散均匀分布）. The distribution in Example 3.3.1 is a
discrete uniform distribution.
Example 3.3.2
A player tosses a fair coin until a head occurs. If the first head occurs at
n -th time, the player wins n dollars. Find the average amount the player
wins.
Solution Let X be the amount the player wins. Then X takes values
from the set {1, 2, , n, } . The player wins k dollars if and only if he
gets k 1 tails first, and follows by a head. Thus
42
P( X  k ) 
1
2k
So

E( X )   k 
k 1
1
2
2k
二项分布的数学期望
Now we give the expectation and variance of binomial distribution.
Theorem 3.4.3
The expectation and variance of a binomial random
variable X with parameters n and p are given by
E ( X )  np,
D( X ) n p(1
.p)
(3.4.2)
Proof Consider the identity
n
n
k 0
k 0
( px  q) n   Cnk ( px) k q n k   Cnk p k q n k x k .
Regard p, q as constants, x as a variable. Differentiate both sides of
this identity with respect to x , we have
n
np( px  q) n 1   kCnk p k q n  k x k 1 .
(3.4.3)
k 1
Put x  1 , we get
n
 kC
k 1
k
n
p k q n  k  np .
But P( X  k )  Cnk p k q n k , thus
n
n
k 0
k 0
E ( X )   kP( X  k )   kCnk p k q n k  np
43
Take differentiation on both sides of (3.4.3),
n
n(n  1) p 2 ( px  q) n  2   k (k  1)Cnk p k q n  k x k  2 .
k 2
Put x  1 , we get
n
 k (k  1)C
k
n
k 2
p k q n  k  n(n  1) p 2 .
(3.4.4)
Add (3.4.3) and (3.4.4) to get
n
D( X )  E ( X ) 2   E ( X )    k 2 P( X  k )  (np) 2
2
k 0
n
  k 2Cnk p k q n  k  (np ) 2
k 1
 n(n  1) p 2  np  (np)2  np  np 2  np(1  p)
泊松分布的期望
Theorem 3.5.1 The expectation and variance of a Poisson random
variable X with parameter  are, respectively,
E ( X )   and D( X )   .
(3.5.2)
Proof By the definition,

k
k 0
k!

E ( X )   k  P( X  k )   k 
k 0

 k 1
k 0
k!

e


k
k 0
k!
 

 e  
k
k 1 (k  1)!
 e
 e    .
Homework chapter 3
8, 9, 10, 22, 27, 30
2008-3-19
验血问题
44
验血次数 X 的数学期望为
E( X ) 
1 k
1
1
q  (1  )(1  q k )  1  q k  .
k
k
k
1
k
N 个人平均需化验的次数为 N (1  q k  ) . 由此可知，
只要选择 k
使
1  qk 
1
 1,
k
则 N 个人平均需化验的次数  N .
1
k
当 p 固定时,我们选取 k 使得 L  1  q k  小于 1 且取到最小值,这
时就能得到最好的分组方法.
1
k
例如, p  0.1 ,则 q  0.9 ,当 k  4 时, L  1  q k  取到最小值. 此时
得到最好的分组方法.若 N  1000 ,此时以 k  4 分组,则按第二方案平均
只需化验
1000(1  0.94 
1
)  594(次) .
4
这样平均来说,可以减少 40%的工作量.
补充例
问题提出
某工厂需要在五周内采购 1000 吨原料，估计原料价格为 500 元的概
率为 0.3，600 元的概率为 0.3，700 元的概率为 0.4，试求最佳采购策
略，使采购价格的期望值最小。
思考题
如果你能预先知道 5 周的原料价格，当然是按最低价购买全部原料,
则此时价格的期望值是多少？
The expectation of discrete random variables has the following
45
properties
数学期望性质
Theorem 3.3.1
Let X be a discrete random variable, then (assume all
expectations exist):
(a) If P( X  a)  1 for some constant a  IR , then E ( X )  a .
(b) Let g ( X ) be a function of X , then
E ( g ( X ))   g ( X ) P( X  x)
(3.3.2)
x
(c) If X1 , X 2 , , X n are discrete random variables, then
E(X1  X 2 
 X n )=E(X1 )  E(X 2 ) 
 E(X n )
(3.3.3)
(d) If P( X  0)  1 , then E ( X )  0 .
(e) If P( X  0)  1 and E ( X )  0 , then P( X  0)  1 .
(f) For any constant b , E (bX )  bE ( X )
(3.3.4)
(g) Schwarz’s inequality. （许瓦慈不等式）Let X,Y be random
variables, then
 E ( XY ) 
2
 E ( X 2 ) E (Y 2 )
(3.3.5)
The equality holds iff P( X  0)  1 or P(Y  aX )  1 for some constant a.
Proof (a) By the definition.
E ( X )   xP( X  x) .
x
In the summation, for the term x  a, P( X  a)  1, and for the other
terms x  a, P( X  x)=0, thus E(X)= a.
(b), (c) The proofs are given in advanced probability theory, so is
omitted here.
46
(d) If P( X  0)=1 , then in the summation
E ( X )   xP( X  x)
x
each term is non-negative, thus E ( X )  0 .
(e).If P( X  0)=1 and E ( X )  0 , then in the summation
E ( X )   xP( X  x)
x
the left side is 0 and each term in right side is non-negative, so each
term is 0 . Thus, for the terms x  0 , we must have P( X  x)  0 . This
means P( X  0)  1 .
(f).Set g ( X )  bX in (b), we have
E (bX )   bxP( X  x)  b bxP( X  x)  bE ( X )
x
x
then E ( X 2 )  0 . Consider the variable
(h) Assume P( X  0)  1 ,
Z  Y  sX ,
s is a constant, then
0  E ( Z 2 )  E  (Y  sX ) 2   E (Y 2  2sXY  s 2 X 2 )
 E (Y 2 )  2sE ( XY )  s 2 E ( X 2 )
put s  E ( XY ) / E( X 2 ), we have
 E ( XY ) 
2
 E ( X 2 ) E (Y 2 )
If P( X  0)  1, then E ( XY )  E ( X 2 )  0 , the equality holds in (3.3.5). If
P(Y  aX )  1 , then
 E ( XY ) 
2
  E (aX 2 )   a 2  E ( X 2 )   E ( X 2 ) E )(Y 2 )
2
2
so the equality holds. On the other hand, if the equality holds in (3.3.5),
but P( X  0)  1 , then we must have
P(Y  sX )  1 ,
s  E ( XY ) / E ( X 2 )
47
2．Variance 方差
Except the expectation of a random variable, we are interested in
some other quantities related to a random variable. Let’s consider an
example.
Example
A cigarette manufacturer tests tobaccos grown from two districts for
nicotine contents, obtains the following results
District 1 24, 27, 25, 22, 22 (in milligrams)
District 2 28 ,27, 25, 20, 20
The average nicotine content for both district are the same：
24 milligrams.
But the manufacturer prefer the tobaccos from district 1, because it has
smaller dispersion than district 2, i.e., it is more stable.
To measure the dispersion of a data set X1 , X 2 ,
X
1
( X1  X 2 
n
X n , whose average is
 X n ) , we use the quantity “variance”, denote by  2 , is
defined as
2 
1 n
 ( X i  X )2
n i 1
For example, if  12 and  22 are the variances of nicotine constant for the
district 1and 2, resp., then
1
(24  24) 2  (27  24) 2  (25  24) 2  (22  24) 2  (22  24) 2   3.6

5
1
 12   (28  24)2  (27  24) 2  (25  24) 2  (20  24) 2  (20  24) 2   11.6 .
5
 12 
48
For many purpose it is desirable that a measure of dispersion be
expressed in the same unit as the original data, thus the square root of the
variance, called standard deviation is used. Thus for the data set
X1 , X 2 ,
X n , the standard deviation （标准差）is
1
1 n
2
    ( X i  X )2 
 n i 1

To measure the dispersion of random variables, we also use the
quantity variance and standard deviation.
Definition 3.3.2 Let
X
be a discrete random variable, having
expectation E ( X )   . Then the variance of X , denote by D ( X ) is
defined as the expectation of the random variable ( X   )2
D( X )  E  ( X   ) 2 
(3.3.6)
The square root of the variance D ( X ) , denote by
called
the
standard
deviation
of
X
:
D( X ) , is

D( X )  E  X   

1
2 2
(3.3.7)
The variance of random variable has the following properties.
（方差性质）
Theorem 3.3.2 For a discrete random variable X ,
(a) D( X )  E ( X 2 )   E ( X ) 
(b) D(aX  b)  a 2 D( X ) ,
2
(3.3.8)
a , b are constants.
(3.3.9)
(c ) If D ( X )  0 , then P( X  E ( X ))  1 , i.e. X is a constant.
Proof (a) Let   E ( X ), D( X )  E  ( X   )2   E ( X 2  2 X    2 )
49
 E ( X 2 )  2 E ( X )   2
 E ( X 2 )  2 2   2  E ( X 2 )   E ( X ) 
2
(b) E (aX  b)   (ax  b) P( X  x)
x
  axP( X  x )  bP X(  x )
x
x
 aE ( X )  b  a  b
Thus D(aX  b)  E  ((ax  b)  (a   b))2 
 E (a 2 ( X   ) 2 )  a 2 E  ( X   ) 2   a 2 D( X )
(c) Since P  ( X   ) 2  0   1 , by Theorem 3.3.1(e)
P( X    0 
) ,1i.e. P  X  E( X )   1
Example 3.3.3 A die is tossed. Find the variance and standard deviation
of the spots X shown, if
(a)this die is a fair die, i.e. the probability distribution of X is
P( X  k ) 
1
,
6
k  1, 2,
, 6.
(b)the probability distribution of X is
X
1
P(X=k) 0.1
2
3
4
5
6
0.1
0.1
0.2
0.2
0.3
7
2
Solution (a) By Example 3.3.1. E ( X )  . Thus
D( X )  E ( X )   E ( X ) 
2
D( X ) 
2
2
 7  35
  k  P( X  k )    
 2  12
k 1
6
2
35
12
(b) E ( X )  1 0.1  2  0.1  3  0.1  4  0.2  5  0.2  6  0.3  4.2
50
6
E ( X 2 )   k 2 P( X  k )  20.4
K 1
D( X )  E ( X 2 )   E ( X )   20.4  4.22  2.76
2
D( X )  1.66
Example 3.3.5 Find the variance of the amount won by the player in
Example 3.3.2.
（ P( X  k ) 
1
）
2k
Solution We have E ( X )  2 , and

k2
 22  12  4  8 .
k
k 1 2
D( X )  E ( X 2 )   E ( X )   
2
泊松分布的期望和方差
Theorem 3.5.1 The expectation and variance of a Poisson random
variable X with parameter  are, respectively,
E ( X )   and D( X )   .
(3.5.2)
Proof By the definition,


k
k 0
k 0
k!
E ( X )   k  P( X  k )   k 

 k 1
k 0
k!


k
k 0
k!
 e    

 e  
k 1
k
(k  1)!
 e
 e    .
k 2 k   
k
 e    k (k  1)  k   e 
k!
k 1 k !
k 1


E ( X 2 )   k 2 P( X  k )  
k 0


k 2
k
(k  2)!
k

e  

k 1
(k  1)!
 e    2  
D ( X )  E ( X 2 )   E ( X )   ( 2   )   2  
2
Variance of Binomial distribution
51