Download 第三章 随机向量的分布及数字特征

Chapter 3 Random vector and their numerical characteristics
CHAPTER 3 RANDOM VECTORS AND THEIR NUMERICAL
CHARACTERISTICS
A lot of random phenomenon or a randomized trial will need to visit a few random
variables. For example, examine the development of preschool children in a certain
area, it is needed to observe their height and weight; fired a shell that require the
simultaneous study of several point of impact coordinates; study of market supply
model, taking into account the need to supply of goods, consumer income and the
market price and other factors, and so on. In general, these random variables require
them as a whole (or vector) to be studied, so it is inevitably to define in the same
sample space on multiple random variables describe them. Multidimensional random
variable (or random vector) concept is in the context of these real concerned. This
chapter focuses on two-dimensional random vector, mainly including random vector
(function) of the distribution, independence, and numerical characteristics.
§ 3.1 Random vectors and their distribution functions
1. n -dimensional random vector
Definition 3.1 Let X1 , X 2 , , X n be n random variables defined on the same
sample space, it is said that X  ( X1 ,..., X n ) is a n -dimensional random variable or
random vector.
For n-dimensional random vector, each of its components is a one-dimensional
random variable, one can study it separately. But in addition, there are linkages
between the various components, in many problems, it is more important. The method
for studying random vector is analogue to the one used in one-dimensional random
variable study----with distribution function, probability density, or distribution of the
column to describe its statistical law.
2. The joint distribution function of random vector
Definition 3.2 for any x1 , x2 , , xn , define
F ( x1 , x2 ,   ,nx )( P 1X  1 x, 2X  2x,  n, X n  x)
(3.1)
the n  dimensional joint distribution function of random vector ( X 1 , X 2 , , X n ) ,
where ( X1  x1 , X 2  x2 ,
, X n  xn ) stands for
1
n
i 1
( X i  xi ) .
Chapter 3 Random vector and their numerical characteristics
From now on, we just focus on the joint distribution of two-dimensional random
vector, and one can easily find that the same idea can be fitted in more dimensional
case.
Two-dimensional random vector ( X , Y ) is a random point in a plane with
coordinates, the value of distribution function F ( x, y ) on the point ( x, y ) is the
probability that random point fall in area shown in Figure 3-1.
Figure 3-1
Figure3-2
By figure 3-2 one can easily find that the probability of ( X , Y ) fall in
{( x, y) : x1  x  x2 , y1  y  y2 } is
P( x1  X  x2 , y1  Y  y2 )  F ( x2 , y2 )  F ( x1 , y2 )  F ( x2 , y1 )  F ( x1, y1 ).
(3.2)
3．Characteristics of joint distribution
Theorem 3.1 Joint distribution function F ( x, y ) has the following
characteristics.
(1)Monotonic F ( x, y ) are non-decreasing with respect to x or y respectively,
i.e.,
for any fixed y, x1  x2 implies F ( x1 , y)  F ( x2 , y) ;for any fixed x, y1  y2 时有
F ( x, y1 )  F ( x, y2 ) ;
(2)Normalized
For any x and y , 0  F ( x, y )  1
2
Chapter 3 Random vector and their numerical characteristics
F (, y )
F (, )
lim F ( x, y )  0, F ( x, )
lim F ( x, y)  0,
x 
y 
lim F ( x, y )  1;
x , y 
(3) Right continuous F ( x, y ) are right continuous with respect to x or y ,
F ( x 0 , y ) F ( x , y ) ,F (x ,y  0 ) F ; x( y, )
(4) Nonnegative For all a  b, c  d ,
P(a  X  b, c  Y  d )  F (b, d )  F (a, d )  F (b, c)  F (a, c)  0
Characteristic (1)-(4) is necessary for any bivariate joint distribution. The
following example show us even condition (1)-(3) is satisfied, F ( x, y ) is not a joint
distribution yet.
0, x  y  0;
Example3.1 Let G ( x, y )  
Then, G ( x, y ) fulfill characteristics
1, x  y  0.
(1) (2) (3), but not fulfill (4) and thus G ( x, y ) not a distribution function. In fact,
(1) Suppose x  0, if x  y  0, since x  x  y  0,
G ( x, y )  G ( x  x, y )  1. Then G ( x, y )  0 if x  y  0 ; G ( x  x, y )  0 , if
x  x  y  0 ; G ( x  x, y )  1 if x  x  y  0 .
That is to say
G ( x, y )  G ( x  x, y ).
G ( x, y ) is non-decreasing w.r.t. x , similarly, one can find that
G ( x, y ) is
non-decreasing w.r.t. y .
(2)
If x  y  0 , lim G( x  x, y)  lim G( x, y  y)  0  G( x, y).
x0
y 0
And lim G( x  x, y)  lim G( x, y  y)  1  G( x, y) if x  y  0 . This proves
x0
y0
that G ( x, y ) is right continuous w.r.t. x or y respectively.
3
Chapter 3 Random vector and their numerical characteristics
(3) Obviously, 0  G ( x, y )  1 and G (, y )  G ( x, )  0, G(, )  1.
One can also find that
G (1,1)  G (1, 1)  G (1,1)  G(1, 1)  1  1  1  0  1  0, thus, G ( x, y ) does not
fulfill (4) and thus not a joint distribution function.
4. Marginal distribution function
If the joint d.f. of X , Y is known, then the d.f. of X , Y can be derived from
the joint d.f. and denote by FX ( x), FY ( y ) the d.f. of X , Y respectively, i.e.,
FX ( x)  P( X  x)  P( X  x, Y  )  F ( x, )  lim F ( x, y);
y 
FY ( y )  P(Y  y )  P( X  , Y  y )  F (, y )  lim F ( x, y ).
x 
Generally, if the joint d.f. of n -dimensional r.v. ( X1 , X 2 ,
F ( x1 ,
, X n ) is given by
, xn ) ,then the marginal k (1  k  n) -dimensional d.f. of ( X1 , X 2 ,
, X n ) is
derived easily. For example, the marginal d.f. of X1 ,( X1 , X 2 ),( X1 , X 2 , X 3 ) is
FX1 ( x1 )  F ( x1 , ,
, ),
FX1 , X 2 ( x1 , x2 )  F ( x1 , x2 , ,
, ),
FX1 , X 2 , X 3 ( x1 , x2 , x3 )  F ( x1 , x2 , x3 , ,
 ).
Example 3.2 Assume that the joint d.f. of ( X , Y ) is
x 

F ( x ,y ) A  B  a r c t anC 
2 

y
a r c t, a nwhere A, B, C are
2
constants and x  (, ), y  (, ) , (1) give the value of A,B,C;(2) give the
marginal d.f. of X and Y;(3) determine P( X  2).
Answer
4
Chapter 3 Random vector and their numerical characteristics
 


(1) F (+, )  lim F ( x, y)  A  B   C    1 and
x 
2 
2

y 
 

 



F ( , ) A  B    C    0, F (, )  A  B    C    0
2 
2
2 
2


Yields
A
1

2
,B 

2
(2) FX ( x)  F ( x, ) 
FY ( y ) F 
( y, )
,C 

2
and
F ( x, y ) 
1 
x  
y
 arctan   arctan  .
2 
 2
2  2
2
1 
x
1 1
x
 arctan      arctan , x  (, )
2 
 2
2
2 
2
1    



 2  2 2  2
y 1 1
a r c tan 
2
2
y
a r yc
t a
n 
,
2
(
1 1   1
(3) P( X  2)  1  P( X  2)  1  FX (2)  1      
2  4 4
5. Discrete bivariate distribution law
Definition 3.3 If ( X , Y ) assumes only finite or countable ( xi , y j ) , ( X , Y ) is
said to be a discrete bivariable.
Definition 3.4 Let ( X , Y ) assumes value ( xi , y j ) , i, j  1, 2,
. and define
pij  P( X  xi , Y  y j ), i, j  1, 2, 
The joint distribution law of ( X , Y ) and the following characteristics holds:
(1) pij  0;
 
(2)
 p
i 1 j 1
ij
 1.
Usually, the joint d.f. of ( X , Y ) is specified by the following table.
Table 3.1
5
,
)
Chapter 3 Random vector and their numerical characteristics
Y
y1
y2
yj
p11
p12
p1 j
p21
p22
p2 j
X
x1
x2
xi
pi1
pi 2
pij
Example 3.3
Put one number from 1,2,3,4 and write it down with X , and
put one from 1,
, X and denote it by Y ,try to determine the joint distribution
law of ( X , Y ) and P ( X  Y ) .
Answer The d.f. of X is P( X  i ) 
1
, i  1, 2,3, 4. Y possibly assumes value
4
1,2,3,4 and denote j the value of Y , then if j  i , P( X  i, Y  j )  P( )  0. If
1  j  i  4 , P( X  i, Y  j )  P( X  i ) P(Y  j | X  i ) 
1 1
 . So the joint
4 i
distribution law is
Table 3.2
Y
1
2
3
4
1
1/4
0
0
0
2
1/8
1/8
0
0
3
1/12
1/12
1/12
0
4
1/16
1/16
1/16
1/16
X
And
6
Chapter 3 Random vector and their numerical characteristics
1 1 1
1 25
  

 0.5208.
4 8 12 16 48
P( X  Y )  p11  p22  p33  p44 
Marginal distribution law
Suppose that the joint distribution law of ( X , Y ) is given by
P( X  xi , Y  y j )  pij , i, j  1, 2,
,
One problem is how to determine the marginal d.f. of X , Y . Since X possibly
assume value x1 , x2 ,
( X  xi )  ,
i
, and Y possibly assume value y1 , y2 ,
, and
(Y  y j )  , Thus
j
+


 +
 
P( X  xi )  P  X  xi , (Y  y j )   P  ( X  xi , Y  y j )  =  P ( X  xi , Y  y j )
j 1


 j 1
 j 1

  pij
j 1
pi i =1,2,
 

 
 
P(Y  y j )  P  ( X  xi ), Y  y j   P  ( X  xi , Y  y j )    P ( X  xi , Y  y j )
 i 1

 i 1
 i 1

  pij
i 1
p j , j  1, 2,
Where pi and p
j
denote


j 1
i 1
 pij , pij respectively and represent the
probability of evens ( X  xi ) and (Y  y j ) , which is shown in Table 3.3
Table 3.3
Y
X
x1
y1
y2
yj
p11
p12
p1 j
7
P( X  xi )
p1
Chapter 3 Random vector and their numerical characteristics
p21
x2
xi
pi 2
pi1
P (Y  y j )
p2 j
p22
pi
pij
p2
p1
p2
p
1
j
Example 3.4 The joint d.f. of ( X , Y ) is given by Table 3.4, try to determine the
marginal d.f. of X and Y .
Table 3.4
Y
-1
0
2
0
0.1
0.2
0
1
0.3
0.05
0.1
2
0.15
0
0.1
X
Answer
表 3.5
Y
-1
0
2
pi
0
0.1
0.2
0
0.3
1
0.3
0.05
0.1
0.45
2
0.15
0
0.1
0.25
0.55
0.25
0.2
1
X
p
j
8
Chapter 3 Random vector and their numerical characteristics
Continuous bivariate joint distribution
Definition 3.4 It is said that ( X , Y ) have a continuous distribution if there
exists a nonnegative function p( x, y ) such that for any x and y ,
F ( x, y)  
x

y
 
p(u, v)dudv,
And p( x, y ) is called the density function of ( X , Y ) .
By the definition, p( x, y ) fulfills the following characteristics.
(1)
(2)
p ( x, y )  0;




 
p( x, y) 1.
(3) for any continuous point of p( x, y ) ,
2
F ( x, y)  p( x, y) .
xy
In fact,
(3.3)
y
x
  y
F ( x, y )   
     p(u, v)dv  du    p( x, v)dv

x
x    
  
y

 2 F ( x, y )  
   p( x, v) d v

x y
 y 

(4) P(( X , Y )  G)   p( x, y )dxdy .
p( ,x )y
(3.4)
G
Example 3.5 Suppose that the joint density function of ( X , Y ) is specified as
follows:
ke ( x  y ) ,
p( x, y)  
0,
x  0, y  0
其它
(1) Determine constant k ; (2) determine F ( x, y ) , the joint d.f. of ( X , Y ) ; (3)
9
Chapter 3 Random vector and their numerical characteristics
（X  1, Y  1）
.
determine P
Answer (1) Since 





f ( x, y)dxdy  1 ,
1 

0

 k
0
x
y
(2)F ( x, y ) = 



0
ke ( x  y ) dxdy
e x dx 

0
 
e y dy  k (e x |0 )2  k
f ( x, y ) dxdy
 x y e  ( x  y ) dxdy  (1  e  x )(1  e  y ), x  0, y  0
  0 0

0, 其它

1
1 1
(3) P( X  1, Y  1)   dx  e  ( x  y ) dy  1   .
1
0
e e
Marginal density
Suppose p( x, y ) the joint density function of ( X , Y ) . Then
FX ( x) P( X
x)

,x 
Y  )
P( X
x




[
p( u, y) , d ]y d u
And thus
p X ( x)  

pY ( y)  



p( x, y)dy .
(3.5)
p( x, y)dx .
(3.6)
p X ( x) and pY ( y ) are called the marginal density function of ( X , Y ) .
Example 3.6 Suppose the joint density function of ( X , Y ) is specified by
8 xy, 0  x  y  1
p ( x, y )  
其他
 0,
Try to determine the marginal density function of X and Y .
Answer By (3.5) and (3.6), we have
10
Chapter 3 Random vector and their numerical characteristics
pX ( x) 


p( x, y) ,dy
1
if 0  x  1 , p X ( x)   8 xydy  4 x(1  x 2 ) ;
x
if x  0 or x  1 , pX ( x)  0 . Thus,
4 x ( 1 x 2 ，
)
0 x  1
.
p X ( x)  
0
，
其它

4 y3， 0  y  1
pY ( y)  
.
0， 其它
TWO IMPORTANT MULTIDIMENSIONAL DISTRIBUTIONS
1. Multivariate distribution
Suppose D a bounded area in R n and S D its measurement, if the joint
density function of multivariate ( X 1 , X 2 , , X n ) is specified by
 1
 , ( x1 , x2 ,   , xn )  D,
p( x1 , x2 ,   , xn )   S D

0, 其它.

(3.7)
Then ( X 1 , X 2 , , X n ) is called follows a multivariate-uniform distribution and
denoted it by ( X1 , X 2 , , X n ) ~ U（D).
Example 3.7 Suppose D is a circle with r the radar and defined on plane,
( X , Y ) is distributed uniformly on D and the density function is give by
 1
, x2  y 2  r 2 ,

p ( x, y )    r 2
 0, x 2  y 2  r 2 .
r
Try to determine the value of P ( X  ).
2
Answer
11
Chapter 3 Random vector and their numerical characteristics
r
P (| X | ) 
2
r
2

r 2  x2
r
 
2
1
1
2 2  r 2 dydx   r 2
r x

1
x
[ x r 2  x 2  r 2 arcsin ] |r/r2/ 2
2
r
r

1
r2
1
2
[
r
r

 2r 2 arcsin ]
2
r
4
2

1

[

r/2
r / 2
2 r 2  x 2 dx
3 
 ]  0.609.
2 3
2. Two-dimensional normal distribution
If the density function of ( X , Y ) is given by
p ( x, y ) 
1
( x  1 )2
( x  1 )( y  2 ) ( y  2 ) 2
1
[

2


]},   x, y  
2(1   2 )
 12
 1 2
 22
exp{
2 1 2 1   2
(3.8)
It is said that ( X , Y ) follows the binormal distribution, and denoted it with
( X , Y ) ~ N ( 1 , 2 ,  12 ,  22 ,  ), where   1 , 2  ;1 ,  2  0; 1    1.
Example 3.8 Suppose ( X , Y ) ~ N ( 1 , 2 ,  12 ,  22 ,  ) ,try to determine the  X ( x)
and Y ( y) .
Answer  X ( x)  


p( x, y)dy. 令 s 


1

2 1 1   2
 X ( x)  

1

e
2 1
i.e. X ~ N (1 ,  12 ) .
( x  1 )2
212
e

x  1
1
1
2(1  2 )


,t 
y  2
,得
2
( s 2  2  st  t 2 )
dt
1
2 1   2

e
( t   s )2
2(1 
2

1
)
dt 
e
2 1
imilarly, one can find that Y ~ N (  2 ,  22 ) .
12
( x  1 )2
212
,
Chapter 3 Random vector and their numerical characteristics
§ 3.2 INDEPENDENCE OF RANDOM VARIABLE
The concept of independence of random variables is very important in probability
theory, which is the promotion of independent random events.
Definition 3.5 Suppose that ( X , Y ) has joint d.f. F ( x, y ) ,and marginal d.f.
FX ( x) and FY ( y ) respectively, if for any x, y  R , the following equation holds
F ( x ,y ) FX x( FY) y (, )
(3.9)
i.e. P( X  x, Y  y )  P( X  x) P(Y  y ) , it is said that X
and
are
Y
independent.
We give the following theorem without proof.
Theorem 3.2 A sufficient and necessary condition for X
and Y to be
independent is for any real number set A and B such that { X  A},{Y  B} are
well-defined, the following equations holds.
P( X  A, Y  B)  P( X  A) P(Y  B) .
(3.10)
Theorem 3.3 If X and Y are independent, then for any continuously or
piece-wise continuous function f ( x), g ( x) , f ( X ) and g (Y ) are also independent .
Definition 3.6 Suppose F ( x1 , x2 , , xn ) is the d.f. of n -dimensional random
vector ( X 1 , X 2 , , X n ) , Fi ( xi ) is the marginal d.f. of X i , if for any real number
x1 , x2 , , xn ,
n
F ( x1 , x2 , , xn )   Fi ( xi )
(3.11)
i 1
X1 , X 2 , , X n is said to be mutual independent.
Independent discrete random vector
13
Chapter 3 Random vector and their numerical characteristics
Theorem
3.4
Suppose
values ( xi , y j ), i, j  1, 2,
( X ,Y )
are
discrete
distributed
and
assume
, then the sufficient and necessary condition for X and Y
to be independent is for all i, j  1, 2,
the following equations holds.
P( X  xi , Y  y j )  P( X  xi ) P(Y  y j ) .
(3.12)
Example 3.10 Let the d.f. of ( X , Y ) is given by the Talbe 3.6,
Table 3.6
Y
1
2
1
1/3
1/6
2
a
1/9
3
b
1/18
X
Try to determine the value of constant a , b such that X
and Y
independent.
Answer The marginal d.f. of ( X , Y ) is
Table 3.7
X
1
P
2
a  1/ 9
1/2
3
b  1/18
Table 3.8
Y
P
1
2
a  b  1/ 3
1/ 3
To make sure X and Y independent, by pij  pi.  p. j we have
P( X  2, Y  2)  P( X  2) P(Y  2) , P( X  3, Y  2)  P( X  3) P(Y  2) ,i.e.
14
are
Chapter 3 Random vector and their numerical characteristics
 1 
1 1
2

a
 9  a  9  3





9


 1   b  1   1 b  1
18  18  3 
9
Independent continuous random vector
Theorem 3.5 Suppose the joint density function of ( X , Y ) is p( x, y ) and is
continuous everywhere, then the sufficient and necessary condition for X and Y to
be independent is
p( x ,y ) pX x(  p) Y y (. )
(3.13)
Example 3.11 If the density functions of ( X , Y ) is
15 x 2 y， 0  x  y  1
p( x, y)  
else
0，
(1)Try to find the marginal distribution of X , Y ;
(2) X is independent with Y or not？
1
15 2
2
4
  15 x ydy， 0  x  1  ( x  x )， 0  x  1
Answer pX ( x)   p( x, y )dy   x
2

0，

else
0，
else


y
2
x 0 y 1 5 y 4， 0 y  1
  1 5x y d，
pY ( x )  p x( y, dx)   0

0， 其它

0，
其它


Since p( x, y ) are not equal to pX ( x) pY ( y ) everywhere, X is not independent
with Y .
Example 3.12 Let ( X , Y ) ~ N (1 ,  2 ,  12 ,  22 ,  ) , then, the sufficient and necessary
condition for X and Y to be independent is   0.
§ 3.3 CONDITIONAL DISTRIBUTIONS
15
Chapter 3 Random vector and their numerical characteristics
The relationships between two random variables ( X , Y ) can mainly be divided
into two types: independent or dependent. In many situations, the values of X or Y
is an influence on each other, which makes the conditional a powerful tool for study
the dependent of two random variables, this is the topic of this section.
Discrete conditional distributions
Suppose the joint distribution law of ( X , Y ) is specified by
pij  P( X  xi , Y  y j ), i, j  1,2, .
By the definition of conditional probability, one can easily give the definition of
conditional distribution law:
Definition 3.7 Suppose
is an discrete two-dimensional
( X ,Y )
distribution, for given j , if P( Y  y j )  0 , then the conditional distribution
law for X given Y  y j can be represented by pi| j which is defined as
P( X  xi Y  y j ) 
P( X  xi , Y  y j )
P(Y  y j )

pij
p• j
,
(3.14)
and (1) pi| j  0; (2)  pi| j  1.
i
Similarly, given X  xi the conditional distribution law of Y is represented by
{ p j|i : j  1, 2, }, where p j|i 
pij
pi.
.
FX |Y ( x | y j )   P( X  xi | Y  y j )   pi| j
We can also similarly define
xi  x
FY | X ( y | xi ) 
or
 P(Y  y
yj y
j
| X  xi ) 
p
yj y
j |i .
xi  x
the conditional d.f. for given X  xi
Y  y j respectively.
Continuous Conditional distributions
16
and
Chapter 3 Random vector and their numerical characteristics
Suppose that now ( X , Y ) has continuous joint distribution density function
p( x, y ) and marginal d.f. pX ( x), pY ( y) .It should be noted that
P( X  x)  0 for
any x and conditional distribution for P( A | B) have not been defined in the case
that P( B)  0 . Therefore, in order that it is possible to derive conditional probability
for X when ( X , Y ) has a continuous joint distribution, the concept of conditional
probability will be extended by considering the conditional density function of Y
given X  x . One natural way is to consider the limit of P(Y  y | x  X  x  x)
when x  0 instead of considering P(Y  y | X  x) directly.
Definition 3.9 Suppose that for any x  0 , P( x  X  x  x)  0 holds, if
P(Y  y, x  X  x  x)
exist, then the limit is
x 0
P( x  X  x  x)
lim P(Y  y | x  X  x  x)  lim
x 0
called the conditional distribution of Y for given X  x and denoted it by
P(Y  y | X  x) or FY | X ( y | x) .
Theorem 3.6 Suppose that ( X , Y ) has continuous density function p( x, y )
and pX ( x)  0, then
FY | X ( y |x

)
y

p( x ,v dv
)
p X ( x)
is a continuous d.f. and its density function is
,
(3.15)
p ( x, y )
, which is called the
p X ( x)
conditional density function of Y given the condition X  x and denoted it by
pY | X ( y | x).
Proof
17
Chapter 3 Random vector and their numerical characteristics
P ( x  X  x  x, Y  y )
P{x  X  x  x}
[ F ( x  x, y )  F ( x, y ) / x
 lim
x  0 [ F ( x  x )  F ( x )] / x
X
X
FY | X ( y | x)  lim
x  0

lim[ F ( x  x, y )  F ( x, y )] / x
x  0
lim[ FX ( x  x)  FX ( x)] / x
x  0

F ( x, y ) F X ( x )
/
x
x


y

p ( x, v ) dv
p X ( x)
Differentiate with respect to y on both side of aforementioned equation yields
pY | X ( y | x) 
p ( x, y )
.
p X ( x)
(3.16)
Similarly, given Y  y , the conditional d.f. and conditional density function of X
is specified as follows:
FX |Y

( x | y) 
x

p X |Y ( x | y ) 
p (u , y ) du
pY ( y )
,
(3.17)
p ( x, y )
.
pY ( y )
(3.18)
Example 3.14 Suppose that ( X , Y ) is uniformly distributed on the area
x 2  y 2  1, try to determine the conditional density function p X |Y ( x | y ).
Answer The joint density function of ( X , Y ) is
1
2
2
 ， x  y 1
p  x, y    
 0，
其它
When  1  y  1 , pY ( y )  


1 y 2
1
 1 y 2

p( x, y )dx  
18
dx 
2 1 y2

.
Chapter 3 Random vector and their numerical characteristics
2
1  y 2， 1  y  1；

Thus pY  y    

0，
el se.
and
1

,  1  y 2  x  1  y 2；

2
pX Y  x y    2 1  y

0,
el se.

That is to say when 1  y  1 , the conditional d.f. of X for given Y  y
is an uniform distribution on interval   1  y 2

Example 3.15 Suppose ( X , Y ) ~ N (1 ,  2 ,  12 ,  22 ,  ) ,
1 y2  .

TRY TO DETERMINE
p X |Y ( x | y ).
Answer Since ( X , Y ) ~ N (1 ,  2 ,  12 ,  22 ,  ) , Y ~ N (  2 ,  22 ) and
p X |Y ( x | y ) 
p ( x, y )
pY ( y )
1

2 1 2 1   2
 exp{
( x  1 ) 2
( x  1 )( y  2 ) ( y  2 ) 2
1
[

2


]}
2(1   2 )
 12
 1 2
 22
( y  2 ) 2
1
 exp{
}
2 22
2 2
2

 
  
1
1

 exp  2
 x   1    y  2    
2
2 1 1   2   
2 1 1   2
  

1
   x   
Thus for given Y  y ,
X



N  1   1  y  2 ，  12 1   2   .
2


Similarly, for given X  x ,
Y



N  2   2  x  1 ，  22 1   2   .
1


19
Chapter 3 Random vector and their numerical characteristics
§3.4 Functions of random variables
1. Discrete distributed random vectors
Example 3.16 Suppose the joint d.f. of ( X , Y ) is specified as follows:
Y
1
2
1
1/5
1/5
2
0
1/5
3
1/5
1/5
X
Try to find the d.f. of Z1  X  Y , Z 2  max{X , Y } .
Answer
Z1 assumes values 2,3,4,5 and
P(Z1=2)=P(X+Y=2)=P(X=1,Y=1)=1/5
P(Z1=3)=P(X+Y=3)=P(X=1,Y=2)+ P(X=2,Y=1) =1/5
P(Z1=4)=P(X+Y=4)=P(X=2,Y=2)+ P(X=3,Y=1) =2/5
P(Z1=5)=P(X+Y=5)=P(X=3,Y=2)=1/5
3
4
5 
2
Thus the distribution law of Z1 is 
.
1/ 5 1/ 5 2 / 5 1/ 5 
Z 2  max{ X , Y } assumes 1,2,3 and
P(Z2=1)=P(X=1,Y=1)=1/5
P(Z2=2)=P(X=1,Y=2)+P(X=2,Y=1)+P(X=2,Y=2) =1/5+0+1/5=2/5
20
Chapter 3 Random vector and their numerical characteristics
P(Z2=3)=P(X=3,Y=1)+P(X=3,Y=2) =1/5+1/5=2/5
2
3 
1
and Z 2 is distributed as 
.
1/ 5 2 / 5 2 / 5 
Generally, suppose that the joint d.f. of ( X , Y ) is specified by
P( X  xi，Y  yi )  pij （i，j  1，，
2 ）
And denote zk , (k  1, 2, ) the values that Z  g ( X , Y ) assumed. Then, the
d.f. of Z is give by the following formula:
P Z  zk   P  g ( X , Y )  zk  

i , j:g ( xi，y j )  zk
P  X  xi , Y  y j ， k  1，，
2
(3.19)
Particularly, if Z  X  Y , then
P( Z  zk )   P( X  xi , Y  zk  xi )
i
  P ( X  zk  y j , Y  y j )
j
When X and Y are independent of each other, then
P( Z  zk )   P( X  xi ) P(Y  zk  xi )
i
  P( X  zk  y j ) P(Y  y j )
j
Example 3.17 Suppose X ~ P(1 ), Y ~ P(2 ), and X is independent of Y
then Z  X  Y ~ P(1  2 ).
Proof Z  X  Y possibly assume 0,1, 2,3,
21
Chapter 3 Random vector and their numerical characteristics
k
P( Z  k )  P( X  Y  k )   P( X  i ) P(Y  k  i )
i 0
k

i 0

i
1
i!
e 1 

k i
2
(k  i )!
e 2
(1  2 )k  ( 1 2 )
1 k
k!
i k i



e

1 2
k ! i 0 i !(k  i )!
k!
 e ( 1 2 )
where k  0,1, 2,3,
Remark If X 1 ,
n
 Xi
i 1
P(1  2 ) .
that is to say Z
, X n are mutually independent and X i
P(i ),1  i  n, then
n
P( i ).
i 1
Example 3.18 Suppose X ~ b(n, p), Y ~ b(m, p), and X is independent of Y ,
then Z  X  Y ~ b(m  n, p).
1，，
2 ,n  m
Proof Z  X  Y possibly assume value 0，
k
P( Z  k)  P( X Y )k

(P X ) i (P Y k) i
i 0
By considering the coefficient of x k in equation (1  x)nm  (1  x)n (1  x)m , it is
b
C C
not difficult to find that
i a
k i
m
i
n
 Cnk m .
Then
b
P( Z  k )   P( X  i ) P(Y  k  i )
i a
b
  Cni pi q n i Cmk i p k i q m( k i )
i a
 pk q
n m
b
k
C C
i a
C
k
nm
k
p q
n
i
m
k
i
nmk
Where k  0,1, 2,3,
, n  m , namely, Z
22
B(n  m, p) .
Chapter 3 Random vector and their numerical characteristics
Remark If X 1 ,
, X k are mutually independent, and X i ~ B(ni , p),1  i  k ,
then
k
X
i 1
~ B(n1    nk , p ) .
i
Particularly, if X 1 ,
, X n are mutually independent and
n
X i ~ B(1, p),1  i  n , then  X i ~ B ( n, p ) .
i 1
2. Variables with continuous distributions
(1) Distribution of Z  X  Y
Let p( x, y ) be the joint density function of ( X , Y ) , then the density function
of Z  X  Y can be given by the following way.
Fz ( z )  P( Z  z )  P( X  Y  z )    p ( x, y )dxdy
D


x y z

zx


p ( x, y )dxy   [ 
p( x, y )dy ]dx

z
z

令u  y  x  dx  p( x, u  x)du     p( x, u  x)dx  du



 
 
Differentiate w.r.t. z on both side one can obtain the density function of Z ,
which is denoted by pZ ( z )
and
pZ ( z)  


pZ ( z )  


p( x, z  x)dx .
p( z  y, y)dy .
Example 3.20 Suppose that X
and Y
are independent and uniformly
distributed on interval (0,1) , let Z  X  Y , try to determine the density function of
Z.
x
Answer Since both X and Y are independent and uniformly
x=z distributed on
x=z-1
1
23
O
1
2
z
Chapter 3 Random vector and their numerical characteristics
interval (0,1) ,
1 0  x  1
pX  x   
其它
0
and pZ  z  
1 0  y  1
pY  y   
其它
0

 p  x  p  z  x  dx
X
Y

Note that the integrand is 1 when 0  x  1, 0  z  x  1 and 0 for else case, that
is
0  x  1, and z 1  x  z .
(1) When z  0 or z  2 , pZ ( z )  0;
z
(2) When 0  z  1, pZ ( z )   1dx  z;
0
1
(3) When 1  z  2, pZ ( z )   1dx  2  z.
z 1
0  z 1
 z

Summarily, pZ  z   2  z 1  z  2 .
 0
else

Example
3.21
Suppose X ~ N ( 1 ,  12 ), Y ~ N ( 2 ,  22 ) and
independent, prove that Z  X  Y ~ N ( 1  2 ,  12   22 ).
Proof
PZ ( z )  








p X ( x) pY ( z  x)dx
( x  1 ) 2
( z  x  2 ) 2
1
1
exp{
}

exp{

}dx
2 12
2 22
2 1
2 2
1 ( x  1 ) 2 ( z  x  2 ) 2
exp{ [

]}dx
2 1 2
2
 12
 22
1
24
are
mutually
Chapter 3 Random vector and their numerical characteristics

1

2 1 2



exp{
1
2  12   22
1
2  12   22
 12   22
 12 ( z  1  2 ) 2 ( z  1   2 ) 2
(
x



) 
}dx
1
2 12 22
 12   22
2( 12   22 )
exp{
exp{
( z  1  2 ) 2 
}
2( 12   22 )  2
1
 1 2
 12   22
( x  1 
exp{
2(
 12 ( z  1  2 ) 2
)
 12   22
dx
 1 2 2
 12   22
)
( z  1  2 ) 2
},
2( 12   22 )
That is to say Z  X  Y ~ N ( 1  2 ,  12   22 ).
Remark If X i (i  1, 2,
Xi
N ( i ,  i ) , then
2
, n) are mutually independent random variables and
n
c X
i 1
i
i
n
n
i 1
i 1
~ N ( ci i ,  ci2 i2 )
(2) Transformation of a two-dimensional probability density
function
u  g1 ( x, y),
Suppose p( x, y ) is the density function of ( X , Y ) and 
has
v  g2 ( x, y).
 x  x(u , v),
continuous partial derivatives, the inverse transformation 
exists, the
 y  y (u, v)
Jacobian determinant J is defined as follows:
x
( x, y ) u
J

(u, v) x
v

y
1

u  (u, v) 





y  ( x, y ) 

v

 U  g1 ( X , Y )
Define 
V  g 2 ( X , Y ),
u
x
v
x
u
y
v
y
1


  0.



then the joint density function of (U ,V ) can be
determined as p(u, v)  p( x(u, v), y(u, v)) J .
Example 3.22 Suppose that X
and Y
25
are independent, identically and
Chapter 3 Random vector and their numerical characteristics
U  X  Y ,
N (  ,  2 ) distributed, denote 
V  X  Y .
u  x  y,

v  x y
Since
x
u
J
x
v
with
inverse
transformation
 x  (u  v) / 2,

 y  (u  v) / 2,
and
y
1
u 1/ 2 1/ 2

 .
y 1/ 2 1/ 2
2
v
Thus the joint density function of (U ,V ) is
1
p(u, v)  p( x(u, v), y (u, v)) | J | p X ((u  v) / 2) pY ((u  v) / 2) |  |
2
2
1
[(u  v) / 2   ]
1
[(u  v) / 2   ]2

exp{
}
exp{

}
2 2
2 2
2 2
2

1
4
exp{
2
(u  2 ) 2  v 2
},
4 2
Which shows that (U ,V )
N (2 ,0, 2 2 , 2 2 ,0) and the marginal distribution is
U ~ N (2 ,2 2 ), V ~ N (0,2 2 ) ,thus
p(u, v)  pU (u)  pV (v) .That is to say that
U and V are independent.
Remark To find the function of ( X , Y ) ,denoted by U  g ( X , Y ) , one way is to
implement a variable V  h( X , Y ) (for example, let V  X or V  Y ) and applying
the formula mentioned in section 3, one can derive the joint density of (U ,V ) ,denote
it by p (u , v) . The last step is to integrate p (u , v) with respect to v , and then the
density of U is derived.
Example 3.23
Suppose that X is independent of Y with marginal density
p X ( x) and pY ( y ) respectively, then the density function of U  XY is specified by
pU (u )  


pX (u v) pY (v)
1
dv.
v
26
Chapter 3 Random vector and their numerical characteristics
u  xy
x  u / v
In fact, denote V  Y and 
, the inverse transformation is 
and
v y
 yv
the Jacobian determinant is
1
J v
0

u
1
v 2  , thus the joint density of (U ,V ) is
v
1
p(u, v)  p X (u / v)  pY (v) | J | p X (u / v)  pY (v)
1
.
|v|
Integrate p (u , v) w.r.t. v , one can obtain the density function of U  XY :
pU (u )  


p X (u v) pY (v)
1
dv.
v
Remark Under the conditions of Example 3.23, one can easily find the density
function of U  X Y is pU (u)  


pX (uv) pY (v) v dv.
§3.5 Numerical characteristics of random vector
Expectations of the functions of random vectors
Theorem 3.7 Suppose that P( X  xi , Y  y j ) (or p( x, y ) ) is the joint density
function of ( X , Y ) , then the expectation of Z  g ( X , Y ) is defined as
 g ( xi , y j ) P( X  xi , Y  y j ), i n di scr et e di st r i but ed case，
 i j
E (Z )  
 

g ( x, y ) p( x, y )dxdy, i n cont i nuous di st r i but ed case.
  
Example 26 Independently put two point two random point X and Y in a
segment with the length a , try to determine the average length of two points.
Answer X , Y ~ U (0, a) , and X is independent of Y ,then the joint density
27
Chapter 3 Random vector and their numerical characteristics
1
 ， 0  x  a, 0  y  a;
function of ( X , Y ) is p( x, y )   a 2
 0，
其它
Thus the average length of X and Y is
E (| X  Y |)  
a
0

a
0
| x y|
1
dxdy
a2
a a
1 a x
{
( x  y )dydx    ( y  x)dydx}
2 0 0
0 x
a
1 a
a2
a
 2 { ( x 2  ax  )dx}  .
0
a
2
3

Properties of the expectation and variance
Proposition 1 E ( X  Y )  EX  EY .
Proof Suppose that ( X , Y ) has density function p( x, y ) and marginal density
function PX ( x) and PY ( y) respectively, then












E( X  Y )  
 ( x  y) p( x, y)dxdy
   xp( x, y )dxdy   
  xP ( x)dx   yP ( y )dy

yp( x, y )dxdy

X


Y
 EX  EY .
Remark E ( X1  X 2    X n )  EX1  EX 2    EX n .
Proposition 2 If X and Y are independent, then EXY  EXEY .
Proof Since X and Y are independent, then
p( x, y)  pX ( x) pY ( y) .Consequently,








EXY  
 xyp( x, y)dxdy
   xyp ( x) p ( y )dxdy
 [  xp ( x)]  [  yp ( y )]
X


Y

X

Y
 EXEY
28
Chapter 3 Random vector and their numerical characteristics
Remark If X1 , X 2 , , X n are mutually independent,
then
E( X1 X 2  X n )  EX1EX 2  EX n .
Proposition 3 If X and Y are independent, then D( X  Y )  DX  DY .
Proof By the definition of variance, one have
D( X  Y )  E[( X  Y ) 2 ]  [ E ( X  Y )]2
 E[ X 2  Y 2  2XY ]  [ E ( X  Y )]2
 [ EX 2  EY 2  2EXEY ]  [( EX ) 2  ( EY ) 2  2 EXEY ]
 {EX 2  ( EX ) 2 }  {EY 2  ( EY ) 2 }
 DX  DY .
Remark If X1 , X 2 , , X n are mutually independent, then
D( X1  X 2    X n )  DX1  DX 2    DX n
Example 3.29 Let X ~ B (n, p ) , try to determine EX , DX .
Answer Suppose that in n trials Bernoulli experiment, with the probability p
that the outcome is A , let
l ，i s
1，t he out come of i t r i a A
X i 
s. n o t
0，t he out come of i t r i a l i A
EX i  p，DX i  pq，i  1 , 2 , n , Then
.
the number that the outcome is A
n
can be specified by X   X i , X ~ B(n, p) ,
i 1
n
n
i 1
i 1
n
n
EX   EX i   p  np, DX   DX i   pq  npq.
i 1
i
1
Covariance
If X
Otherwise
and Y are independent, then If X
E ( X  EX )(Y  EY )  0 ,
29
and Y are independent.
Chapter 3 Random vector and their numerical characteristics
which indicates that X and Y are no longer independent, but are dependent. In
order to define a numerical characteristic to describe the dependent relations of X
and Y , we give the following definition.
Definition 3.10 suppose that ( X , Y ) are dependent two-dimensional random
vector, if E[( X  EX )(Y  EY )] exsits, it is called the covariance of
Cov( X , Y )  E[( X  EX )(Y  EY )] .
Y .That is to say
X and
Particularly, we have
Cov( X , X )  DX .
X and Y are discrete distributed, then
(i) If
C ov( X , Y )   [ xi  EX ][ y j  EY ]pij , where
i
j
P( X  xi , Y  y j )  pij , i, j  1, 2,
(ii) If
X and Y are continuous distributed
Cov( X , Y )  





[ x  EX ][ y  EY ] p( x, y)dxdy.
Properties of Covariance
1 Cov( X , Y )  EXY  EXEY .
2
If
X and Y are discrete distributed, then Cov( X , Y )  0 .
3 or any given ( X , Y ) , D( X  Y )  DX  DY  2Cov( X , Y )
Remark In fact, for any given X1 , X 2 , , X n ,
n
n
i 1
i 1
n
i 1
D( X i )   DX i  2 Cov( X i , X j ) .
i 1 j 1
Example 3.31 Suppose that the joint density function of ( X , Y ) is specified
by
30
Chapter 3 Random vector and their numerical characteristics
3x, 0  y  x  1,
p ( x, y )  
0, 其他.

Try to determine Cov( X , Y ) .
Answer：
1
3
3
x

3
xdydx

3
x
dx

.
0
0 0
4
3
1 x
1 3x
3
EY    y  3xdydx  
dx  .
0 0
0 2
8
1 x
1 3x 4
3
EXY    xy  3xdydx  
dx  .
0 0
0 2
10
EX  
1
x
Cov( X , Y ) 
3 3 3
3
  
 0.
10 4 8 160
Coefficients
Definition 3.11 Suppose that ( X , Y ) is 2-dimensional random vector and
DX  0, DY  0 , the coefficient is defined as:
 X ,Y 
Cov( X , Y ) Cov( X , Y )

 XY
DX DY
Example 3.32 The coefficient of N ( 1 , 2 ,  12 ,  22 ,  )  .
Proof Obviously, we have
E  X   1 , DX  12 , EY  2 , DY   22 . Since
p( x, y) 
Put s 
1
21 2
x  1
1
,t 

  x  1 2 2  x  1  y  2   y  2 2  
1


exp 




2
2
2
2
1
1 2
 2  
1 

 2 1    

y  2
2
, then
31
Chapter 3 Random vector and their numerical characteristics
 
Cov( X , Y ) 
  ( x   )( y   ) p( x, y)dxdy
1
2
 
=
 
1
2 1 2 1   2

 1 2
2

 te



t2
2
  1 2 ste


1
2(1  )2
[ s 2  2  st  t 2 ]
 1 2 dsdt
 


s
(
2 1   2

e
( s   t )2
2(1  2 )
ds )dt
2
t

1
te 2 (  t )dt
2
  1 2 


2
1 2  t2
t e dt
2
  1 2 

  1 2
Properties of coefficients
(1) | X ,Y |  1.
Proof
D( X *  Y * )  D( X * )  D(Y * )  2Cov( X * , Y * )
 1  1  2Cov( X * , Y * )
 2(1   XY )  0,
|  XY | 1 .
i.e.
Definition 3.12 It is said that X and Y are uncorrelated if  X ,Y  0 , are
negative correlated if 1   X ,Y  0 , are positive correlated if 0   X ,Y  1 .
The following assertions are equivalent:
(1) Cov( X , Y )  0;
(2) EXY  EXEY ;
(3) D( X  Y )  DX  DY ;
A sufficient and necessary condition for  X ,Y  1 holds is tha their
(2)
exists
(4) X 与 Y 不相关.
a
constant
a( 0)
and
b
 X ,Y  1 , a  0 ; when  X ,Y  1 , a  0 .
32
such
that
P(Y  aX  b)  1 .When
Chapter 3 Random vector and their numerical characteristics
Proof Suppose that Y  aX  b , then
EY  aEX  b, DY  a 2 DX , then
Cov( X , Y )  E{[ X  EX ][aX  b  aEX  b]}
 aE{[ X  EX ]2 }  aDX .
That is to say
aDX
 XY 
DX a 2 DX

a
|a|
i.e. when a  0 ,  XY  1 ;when a  0 ,  XY  -1 .
(2) Suppose that  XY  1 , since
 XY  1 implies D( X * Y * )  0 ,
X * Y * assume value 1with probability 1.Thus when  XY  1 时, X * Y * =0
and
X  EX
DX
Y  EY
DY
, b  EY
 0 ,i.e. Y  aX  b , where a  
DX
DY
Example 3.33
DY
EX .
DX
If X ~ N (0,1), Y  X 2 ,is that X and Y are correlated?
2
Answer Since X
even function on
N (0,1) and the density function is p ( x) 
1  x2
e ,which is a
2
EX  EX 3  0 . Then it is
R , thus
easy to find
cov( X , Y )  EXY  EXEY  EX 3  EXEX 2  0 and
 XY 
cov( X , Y )
 0.
DX DY
This indicates that X and Y are uncorrelated, however, since Y  X 2 , they are
obviously not independent.
Remark
If
( X ,Y )
N ( 1 , 2 ,  12 ,  22 ,  )
independent  X , Y are uncorrelated.
33
then
X ,Y
are
Chapter 3 Random vector and their numerical characteristics
Example 3.34 Assume that DX  DY  1 , X and Y are uncorrelated, put
X1   X   Y , X 2   X   Y ( 2   2  0), try to determine  X1 , X 2 .
Answer Since X and Y are uncorrelated, sp do  X and  Y , then
D( X   Y )  D( X )  D( Y )   2   2 ,
D( X   Y )  D( X )  D( Y )   2   2 ,
Cov( X   Y ,  X   Y )
  2Cov( X , X )   Cov( X , Y )   Cov(Y , X )   2Cov(Y , Y )
  2   2,
and
X ,X 
1
2
Cov( X 1 , X 2 )  2   2
 2
.
2
DX1 DX 2   
Exercise 3
1.
Suppose that F ( x, y ) is the joint d.f. of ( X , Y ) ,please represent the
probability of following events with F ( x, y ) ：
(1) P(a  X  b, c  Y  d );
(2) P(a  X  b, Y  y );
(3) P( X  a, Y  y );
(4) P( X  x).
2.
If the density function of ( X , Y ) is specified by
1
 , 0  x  1, 0  y  2;
p( x, y)   2

0, else.

Try to find the probability that one of X or Y is less than
3.
If the density function of ( X , Y ) is specified by
34
1
.
2
Chapter 3 Random vector and their numerical characteristics
ke3 x4 y , x  0, y  0;
p( x, y)  
el se.
 0,
Determine:(1)value of k ;(2) F ( x, y ) ;(3) P(0  X  1, 0  Y  2).
4.
If the density function of ( X , Y ) is specified by
e y , 0 x y,
p( x, y) 
 0 , el se .
Try to determine P( X  Y  1) .
5.
If the density function of ( X , Y ) is specified by
3
 xy, 0 x  4 , 0 y  x；
p ( x ,y )  32

0,
else.
Please determine the marginal density function of X , Y .
 1 0 1 
,
1 1
4 2 4
6. Suppose that X  1

Y
0 1 


1 1
2 2
and P( XY  0)  1 . Determine
(1) The joint d.f. of X and Y ;
(2) Whether X and Y are independent or not? why？
7. If the density function of ( X , Y ) is specified by
e y , x  0, y  x,
p( x, y)  
其他.
 0,
Determine (1) the marginal distribution of X , Y ;
(2) conditional distribution of ( X , Y ) ;
(3) P ( X  2 | Y  4).
35
Chapter 3 Random vector and their numerical characteristics
8. Suppose that X and Y are two independent r.v. with density functions
1, 0  x  1,
p X ( x)  
0, el se，
e y ,
pY ( y)  
 0,
y 0,
y  0.
Try to determine the density function of Z  X  Y .
9．Suppose that X and Y are independent and identically distributed with
e x , x  0,
density function p( x)  
 0, x  0.
Determine the joint distribution of U  X  Y and V 
X
and judge
X Y
whether U and V are independent or not?
10. If the density function of ( X , Y ) is specified by
 xe x (1 y ) , x  0, y  0;
p( x, y)  
0，el se.

Determine the density function of Z  XY .
11．Suppose that X and Y are independent and follow exponential distribution
X
with parameter  =1 ,please determine the density function of Z  .
Y
12. If the distribution law of ( X , Y ) is specified by
-1
Y
0
1
X
1/6
1/3
0
1/6
1
1/6
1/6
0
Determine Cov( X , Y ) and  XY .
13. If the density function of ( X , Y ) is specified by
36
Chapter 3 Random vector and their numerical characteristics
1
 ( x  y ), 0  x  2, 0  y  2,
p ( x, y )   8

0, 其他.
Determine E ( X ), E (Y ), Cov( X , Y ),  XY , D( X  Y ).
37