Download P-74 Theorem 4.1

Chapter 4
The Mean and Variance
4.1 Expectations of random variables
Although all the probability information of a random
variable is contained in its cdf (or pmf for discrete
random variables and pdf for continuous random
variables), it is often useful to consider various
numerical characteristics of that random variable.
One such number is the expectation of a random
variable; it is a sort of “weighted average” of the values
that X can take. Here is a more precise definition.
Mathematical expectation
Definition 4.1 If X~P{X=xk}=pk, k=1,2,…n, define
n
E ( X )   x k pk
k 1
the mathematical expectation of r.v. X or mean of X.
Example1 Find EX if X is the outcome of a toss of a fair die. Since
P(X =1)= ... = P(X =6)=1/6, we have
n
1
1
1
1
1
1
E ( X )   xk pk  1  2   3   4   5   6   3.5
6
6
6
6
6
6
k 1
Note: E（X ）is not necessarily a possible outcome of the random
experiment as in the previous example.
Mathematical expectation of several important distributions
1. 0-1 distribution
X
P
1
0
p 1 p

E(X)=p
2. Binomial distribution X~B(n, p)
P{ X  k }  C p (1  p)
k
n
k
n k
k  0.1,...n
n
n!
E( X )   k
p k (1  p)n k
k 1 k ! ( n  k )!
n
n!

p k (1  p) n k
k 1 ( k  1)! ( n  k )!
n
( n  1)!
 np
p k 1 (1  p ) n1 ( k 1)
k 1 ( k  1)! ( n  k )!
n 1
令l  k  1 np C nl 1 p l (1  p) n1 l
l 0
 np
3.Poisson distribution
X ~ P{ X  k } 

k
k 0
k!
E( X )   k
k
k!
X ~ P  
e   , k  0, 1, 2, ...

e    e  
k 1
k  1
(k  1)!
 ;
Definition 4.1 Suppose that X~f(x), -<x<,
then define
E( X )  


xf ( x )dx.
the mathematical expectation of r.v. X
Example :P74- Example 4.3
Mathematical expectation of several important distributions
1. Uniform distribution X~U(a, b)
 1
, a  x  b,

X ~ f ( x)   b  a
 0,
el se,
E( X )  
b
a
x
ab
dx 
;
ba
2
2.Exponential distribution X~Exp()
 1 x
 e
f ( x )  
 0


1

X ~ Exp( )
x0
x0

x
E ( X )   x e  dx    xde

0
  xe

x 

0
0


x
  e dx
0



x

3. Normal distribution X~ N(, 2)
1 e
2 
X ~ f (x) 
E( X ) 



令t 

( x  ) 2

22
,  x 

x
e
2 
x

( x   )2
2
2
dx
t  
 2 e

t2

2
dt ;
Mathematical expectation of the functions of r.v.s
EX 1 Suppose that the distribution law of X
X
-1
Pk 1 3
0
1
1
1
3
3
Try to determine the mathematical expectation of Y=X2
Y
Pk
1
2
3
0
1
3
2
1 2
 E (Y )  1   0  
3
3 3
P-74 Theorem 4.1 let X~P{X=xk}=pk, k=1,2,…,then
the mathematical expectation of Y=g(X) is given by
the following equation and denoted by E(g(X))

E (Y )  E[ g ( X )]   g ( x k ) pk .
k 1
P-74 Theorem 4.1 If X~f(x), -<x<, the
mathematical expectation of Y=g(X) is specified as

E (Y )  E[ g( X )]   g( x ) f ( x )dx.

Suppose that X follows N(0，1) distribution, try to
determine E(X2), E(X3) and E(X4)
f ( x) 

E( X ) 
2






1
e
2
1
e
2
2
x
e
2
x2

2
x2

2
x2

2
dx  1

dx   

x
de
2
x2

2

E( X ) 
3
3



E( X ) 
4


3
4

3
x2

2
x
de
2
dx 0
x2

2
x
e
2


x
e
2
x2

2
dx

 3

2
x
e
2
x2

2
dx
Properties of mathematical expectation
1. E(c)=c, where c is a constant;
2。E(cX)=cE(X), c is a constant.
Proof. Let X~f(x), then

E (cX ) 
 cxf ( x )dx


 c  xf ( x )dx  cE ( X )

Example 2. Some disease will occur with probability 1%，
investigate 1000 people now, it is necessary to check the blood.
The method is clarified these people into ten group with each
group 100 and check the mixed blood sample. If the result is
negative, it is not need to do any test any more, if it is positive,
then ,it is necessary to test each blood sample respectively, try to
determine the average times needed for the test.
Let Xj j  1,...10 is the number to be taken of jth group, and X
the number to be taken in 1000 people, then
Xj
1
101
Pj (99%)100
1  (99%)100
EX j  0.99100  (101)(1  0.99100 )
10
10
j 1
j 1
E ( X )  E ( X j )   E ( X j )
 10[0.99100  (101)(1  0.99100 )]
1
 1000  [1 
 0.99100 ]
100
 644
Homework:
P88: 1(2)(3)(4)(5),10Find the mean