Transcript E[(X+Y)Z]
4.3 Covariance ﹠Correlation
1.Covariance
Definition 4.3
suppose that ( X , Y )are dependent tw o dim ensio nal random vector,
if E [( X E X )(Y E Y )]exsits, it is called the covariance of X and Y .
That is to say
C ov ( X , Y ) E [( X E ( X ))(Y E (Y ))]
P articu larly , w e h ave C o v ( X , X ) D ( X ).
If X and Y are discrete random variables,
C ov( X , Y )
[x
i
i
E ( X )][ y j E (Y )] p ij ,
j
If X and Y are continuous random variables,
C ov ( X , Y )
[ x E ( X )][ y E (Y )] f ( x , y ) dxdy .
C ov( X , Y ) E ( X Y ) E ( X ) E (Y );
Proof
C ov( X , Y ) E {[ X E ( X )][Y E (Y )]}
E [ XY YE ( X ) XE (Y ) E ( X ) E (Y )]
E ( XY ) 2 E ( X ) E (Y ) E ( X ) E (Y )
E ( XY ) E ( X ) E (Y ).
Example Suppose that (X, Y) is uniformly
distributed on D={(X, Y)：x2+y21} .Prove that X and
Y are uncorrelated but not independent.
Proof
1
f ( x, y)
0
x y 1
2
1 x
2
1
xdx
1
1
E ( XY )
1
1
others
1
E(X )
2
1 x
1 x
2
xy
dx
1 x
2
2
dy 0
dy 0
1
1
1
Cov ( X , Y ) E ( XY ) E ( X ) E (Y ) 0
Thus X and Y are uncorrelated. Since
fX (x)
fY ( y )
1 x
2
1
1 x
2
dy
2
1 x
2
0
1 y
1 x 1
others
2
1
1 y
2
dy
2
1 y
0
2
1 y 1
others
f ( x , y ) f X ( x ) fY ( y )
Thus, X is not independent of Y.
2. Properties of covariance:P82
(1) Cov(X, Y)=Cov(Y, X);
(2) Cov(aX, bY)=abCov(X, Y), where a, b are constants
Proof
Cov(aX, bY)=E(aXbY)-E(aX)E(bY)
=abE(XY)-aE(X)bE(Y)
=ab[E(XY)-E(X)E(Y)]
=abCov(X,Y)
(3) Cov(X+Y，Z)=Cov(X, Z)+Cov(Y, Z);
Proof Cov(X+Y，Z)= E[(X+Y)Z]-E(X+Y)E(Z)
=E(XZ)+E(YZ)-E(X)E(Z)-E(Y)E(Z)
=Cov(X,Z)+Cov(Y,Z)
(4) D(X+Y)=D(X)+D(Y)+2Cov(X, Y).
Remark D(X-Y)=D[X+(-Y)]=D(X)+D(Y)-2Cov(X,Y)
Example 4.15----P84
3.Correlation Coefficients
Definition4.4 Suppose that r.v. X，Y has finite variance,
dentoed by DX>0,DY>0，respectively, then,
XY
C o v( X , Y )
D(X )
D (Y )
is name the correlation coefficients of r.v. X and Y .
Properties of coefficients
(1) |XY|1；
(2) |XY|=1There exists constants a, b such that P {Y=
aX+b}=1；
(3) X and Y are uncorrelated XY;
1. Suppose that (X,Y) are uniformly distributed on D:0<x<1,0<y<x,
try to determine the coefficient of X and Y.
Answer
2
f ( x, y)
0
( x, y) D
others
x=y
D
1
1
E(X )
x
2 xdx
0
2 dx ydy
0
1
E ( XY )
1
2x
1
D (Y )
3
2
dx dy
ydy
0
0
x
2 dx
0
x
2 xdx
0
D( X )
x
0
x
0
1
3
0
1
E (Y )
dy
2
y dy
2
4
9
18
1
1
9
0
1
4
C O V ( X , Y ) E ( X Y ) E ( X ) E (Y )
1
36
XY
COV ( X , Y )
D ( X ) D (Y )
1
2
1
D
1
18
1) X ~ U (0,1), Y X , determ ine X Y
2
2） X ~ U ( 1,1), Y X , determ ine X Y
2
Answer 1)
E(X )
1
2
, E (Y )
1
, E ( XY )
3
1
4
, D( X )
1
, D (Y )
12
4
45
1
XY
12
0 . 968
1
4
12 45
2)
E ( X ) 0 , E ( XY ) 0
XY 0
What does Example 2 indicate？
Exam ple 4.18
Suppose ( X , Y ) ~ N ( 1 , 2 , 1 , 2 , ), t hen XY .
Proof
f ( x, y )
2
1
exp
2
2(1 ρ )
2
1
2 π σ 1σ 2 1 ρ
2
2
( x μ1 ) 2
( x μ1 )( y μ 2 ) ( y μ 2 )
2ρ
2
2
σ1
σ 1σ 2
σ2
fX (x)
fY ( y )
1
e
( x μ1 )
2
2 σ1
2
, x ,
2 πσ 1 ( y μ )2
2
2
1
2σ2
e
, y .
2 πσ 2
E ( X ) μ 1 , E (Y ) μ 2 , D ( X ) σ 1 , D (Y ) σ 2 .
2
2
Suppose ( X , Y ) ~ N ( 1 , 2 , 1 , 2 , ), t hen XY .
2
Exam ple 4.18
2
E ( X ) μ 1 , E (Y ) μ 2 , D ( X ) σ 1 , D (Y ) σ 2 .
2
C ov( X , Y )
Let t
( x μ1 )( y μ 2 ) f ( x , y ) d x d y
1
2 π σ 1σ 2 1 ρ
y μ2
x μ1
ρ
,
2
σ1
1 ρ σ2
1
C ov( X , Y )
1
2π
2
2
( x μ1 )( y μ 2 ) e
x μ1
u
σ1
2
2
2 σ1
y μ2
x μ1
ρ
2
σ
σ1
2 (1 ρ )
2
2
1
e
d y d x.
,
( σ 1 σ 2 1 ρ tu ρ σ 1 σ 2 u )e
2
( x μ1 )
2
u
2
2
t
2
2
dtdu
u
t
σ σ 1 ρ 2 u
t
ρσ σ
ρσ 1 σ 2 2 2
1 2
2
2
2
1 2
du e dt
d u te d t
u e
ue
2 π
2
π
2
2
SO C ov( X , Y ) ρσ 1 σ 2 .
2
Hence X Y
2
C ov( X , Y )
D(X )
2
2
.
D (Y )
Note P86
Thus, if （X，Y）follow two-dimensional distribution,
then “X and Y are independent” is equvalent to “X and Y are uncorrelated
2 ,
Example 4.16—4.18 (P86)
Exercise:P90—11 Find Cov(X,Y),12
Homework:P91—16,17