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Chapter 3Chapter 3 Random vectors Random vectors
and their numerical and their numerical
characteristicscharacteristics
3.1 Random vectors1.n-dimension variables
n random variables X1 , X2 , ...,Xn compose a n-dimension vector (X1,X2,...,Xn), and the vector is named n-dimension variables or random vector.
2. Joint distribution of random vector
Define function F(x1,x2,…xn)= P(X1≦x1,X2≦x2,...,Xn ≦xn) the joint distribution function of random vector (X1,X2,...,Xn).
Let (X, Y) be bivariable and (x, y)R2, define
F(x,y)=P{Xx, Yy}
the bivariate joint distribution of (X, Y)
Bivariate distribution
00 , yx
00 ,,, yyxxyx
Geometric interpretation : the val
ue of F( x, y) assume the probability
that the random points belong to eara
in dark
For (x1, y1), (x2, y2)R2, (x1< x2 , y1<y2 ), then
P{x1<X x2 , y1<Yy2 }
= F(x2, y2) - F(x1, y2) - F (x2, y1) + F (x1, y1).
(x1, y1)
(x2, y2)
(x2, y1)
(x1, y2)
1x 2x 3x
1y
2y
3ySuppose that the joint distribution of (X,Y) is (x,y), find the probability that (X,Y) stands in erea
G .
Answer
)],(),(),(),([
)],(),(),(),([}),{(
22313221
13323312
yxyxyxFyxF
yxyxyxFyxFGYXP
Joint distribution F(x, y) has the following characteristics:
0),(lim),(
yxFFyx
1),(lim),(
yxFFyx
0),(lim),(
yxFyFx
0),(lim),(
yxFxFy
(1) For all (x, y) R2 , 0 F(x, y) 1,
(2) Monotonically increment
for any fixed y R, x1<x2 yields
F(x1, y) F(x2 , y) ;
for any fixed x R, y1<y2 yields
F(x, y1) F(x , y2).
);y,x(F)y,x(Flim)y,0x(F 0xx
00
).y,x(F)y,x(Flim)0y,x(F 0yy
00
(3) right-hand continuous for xR, yR,
(4) for all (x1, y1), (x2, y2)R2, (x1< x2 , y1<y2 ),
F(x2, y2) - F(x1, y2) - F (x2, y1) + F (x1, y1)0.
Conversely, any real-valued function satisfied the aforementioned 4 characteristics must be a joint distribution of some bivariable.
Example 1. Let the joint distribution of (X,Y) is
)]3
()][2
([),(y
arctgCx
arctgBAyxF
1) Find the value of A , B , C 。2) Find P{0<X<2,0<Y<3}
Answer 1]2
][2
[),(
CBAF
0)]3
(][2
[),( y
arctgCBAyF
0]2
)][2
([),(
Cx
arctgBAxF
2
1
2
ACB
16
1)0,2()3,0()3,2()0,0(}30,20{ FFFFYXP
Discrete joint distribution
If both x and y are discrete random variable, then,(X,
Y) take values in (xi, yj), (i, j = 1, 2, … ), it is said that X
and Y have a discrete joint distribution .
The joint distribution is defined to be a function such t
hat for any points (xi, yj), P{X = xi, Y = yj,} = pij ,
(i, j = 1, 2, … ). That is (X, Y) ~ P{X = xi, Y = yj,}
= pij , (i, j = 1, 2, … ) ,
X Y y
1 y
2 … y
j …
p11 p12 ... P1j ...
p21 p22 ... P2j ...
pi1 pi2 ... Pij ...
......
...
...
...
...
... ...
Characteristics of joint distribution :(1) pij 0 , i, j = 1, 2, … ; (2) 1
1 1
= i j
ijp
x1
x2
xi
The joint distribution can also be specified in the following table
Example 2. Suppose that there are two red balls and three white balls in a bag, catch one ball form the bag twice without put back, and define X and Y as follows:
ball whiteisput second the0
ball red isput second the1
ball whiteisput first the0
ball red isput first the1
Y
X
Please find the joint distribution of (X,Y)
XY
1 0
1 0
10
110
3
10
3
10
3
25
22}1,1{
P
PYXP
25
32}0,1{
PYXP
25
23}1,0{
PYXP
25
23}0,0{
P
PYXP
Continuous joint distributions and density functions
1. It is said that two random variables (X, Y) have a continuous joint distribution if there exists a nonnegative function f (x, y) such that for all (x, y)R2 , the distribution function satisfies
x y
dudvvufyxF ,),(),(
and denote it with
(X, Y) ~ f (x, y) , (x, y)R2
2. characteristics of f(x, y)
(1) f (x, y)0, (x, y)R2;
(2)
);,(),(2
yxfyx
yxF
(3) 若 f (x, y) 在 (x, y)R2 处连续,则有
( , ) 1;f x y dxdy
- -
G
dxdyyxfGYXP .),(}),{(
(4) For any region G R2,
Let
others
yxyxfYX
0
10,101),(~),(
Find P{X>Y}
2
11}{
0
1
0
x
dydxYXP
1
1
x
y
Find (1)the value of A ;
(2) the value of F(1,1) ;
(3) the probability of (X, Y)stand in region D : x0, y0, 2X+3y6
casesother for ,0
0,0,),(~),(
)32( yxAeyxfYX
yx
Answer (1) Since
6 A 1
0
1
0
32)32( )1)(1(6)1,1()2( eedxdyeF yx1
1
- -
-(2x+3y)
0 0
f(x, y)dxdy = Ae dxdy = 1
3. Bivariate uniform distribution Bivariate (X, Y) is said to follow uniform distribution if the density function of is specified by
21 ( , )
( , ) the measure of 0 ,
x y D Rf x y D
,
el se
D
G
S
SGYXP }},{(
By the definition, one can easily to find that if (X, Y) is uniformly distributed, then
Suppose that (X,Y) is uniformly distributed on area D, which is indicated by the graph on the left hand, try to determine:(1)the density function of (X,Y) ;(2)P{Y<2X} ;(3)F(0.5,0.5)
1DS
others
Dyxyxf
0
),(1),()1(
Answer
where , 1 、 2 are constants and 1>0,2>0 、 |
|<1 are also constant , then, it is said that (X, Y)
follows the two-dimensional normal distribution
with parameters 1, 2, 1, 2, and denoted it by
),,,,(~),( 2
22121 NYX
(2)Two dimensional normal distribution
Suppose that the density function of (X, Y) is specified by
,e12
1)y,x(f
])y()y)(x(
2)x(
[)1(2
1
221
22
22
21
2121
21
2
The concept of joint distribution can be easily generalized to n-dimensional random vectors.
nnn bxabxaxxD ,...:,... 111
D nnn dxdxxxfDXXP ...),...,x(...... 1211
Definition. Suppose that (X1,X2,...Xn) is a n-dimensional random vector , if for any n-dimensional cube
there exists a nonnegative f(x1,x2,...xn) such that
It is said that (X1,X2,...Xn) follows continuous n-dimensional distribution with density function f(x1,x2,...xn)
Definition. Suppose that (X1,X2,...Xn) assume finite or countable points on Rn . We call that X1,X2,...X
n) is a discrete distributed random vector and P{X1=x1,X2=x2,...Xn=xn}, (x1,x2,...xn) R∈ n
is the distribution law of (X1,X2,...Xn)
Multidimensional random varibables
Discrete d.f.
Distribution function
Continuous d.f.
Probability for area Standardized
P{(X,Y)G}
Determine: ( 1) P{X0},(2)P{X1},(3)P{Y y0}
others
yxeyxf
y
0
0),(
EX: Suppose that ( X , Y ) has density funciton
x
y
DAnswer: P{X0}=0
11
0
1}1{
edyedxXPx
y
00
0}{
0
0
00
00
y
ydyedxyYP
y
x
yy
FY(y) = F (+, y) = = P{Yy}
)y,x(Flimy
)y,x(Flimx
3.2 Marginal distribution and independence
FX(x) = F (x, +) = = P{Xx}
Marginal distribution function for bivariate
Define
the marginal distribution of (X, Y) with respect to X and Y respectively
Example 1. Suppose that the joint distribution of (X,Y) is specified by
1 0
( , ) 1 0
0
x y
y y
e xe x y
F x y e ye y x
else
Determine FX(x) and FY(y) 。Answer:
FX(x)=F(x,)=
00
01
x
xe x
FY(y)=F(,y)=
00
01
y
yyee yy
Marginal distribution for discrete distribution
Suppose that
(X, Y) ~ P{X = xi, Y = yj,} = pij , i, j = 1,
2, … Define P{X = xi} = pi. = , i = 1, 2, … P{Y = yj} = p.j = , j = 1, 2, …
1j
ijp
1i
ijp
the marginal distribution of (X, Y) with respect to X and Y respectively.
Marginal density function
the joint distribution of (X,Y) with respec to X and Y.
dyyxfxf X ),()(
dxyxfyfY ),()(
Suppose that (X, Y) ~ f (x, y), (x, y)R2, define
One canc easily to find that the marginal dist
ribution of N(1, 2, 12, 2
2, ) is N(1, 12) and
and N(2, 22).
Example 3. Suppose that the joint density
function of (X,Y)is specified by
others
xyxcyxf
0),(
2
Determine (1) the value of c;
(2)the marginal distribution of (X,Y) with respect to X
2
1
0
(1) 1x
x
dx cdy 6 c
dyyxfxf X ),()()2(100 xorx
10)(66 2
2
xxxdyx
x
Independence of random vectors
Definition It is said that X is independent of Y for any real nu
mber a<b, c<d,p{a<Xb,c<Yd} =p{a<Xb}p{c<Yd},i.e.
event{a<Xb}is independent of {c<Yd}.
Theorem A sufficient and necessary condition for random variables X and Y to be independent is
F(x,y)=FX(x)FY(y)
Remark
(1) If (X,Y) is continuously distributed, then a suffici
ent condition for X and Y to be independent is f(x,y)=
fX(x)fY(y)
(2) If (X,Y) is discrete distributed with law Pi,j=
P{X=xi,Y=yj},i,j=1,2,...then a sufficient and continousl
y for X and Y to be independent is Pi,j=Pi.Pj
EX : try to determine that whether the (X,Y) in
Example1 , Exmaple2 , Example3 are independent or not?
Example 4. Suppose that the d.f. of (X,Y) is give by the following chart and that X is independent of Y, try to determine a and b.
x 1 20 0.15 0.15
1 a b
0.15 0.15 1a b 0.7a b 0.15 ( 0.15) 0.3a
0.35, 0.35a b
§ 3.3 CONDITIONAL DISTRIBUTION
Definition 3.7 Suppose is an discrete two-dimensional distribution, for given , if, then the conditional distribution law for given can be represented by which is defined as (3.14)
and (1)
(2)
( , )X Yj
( , )( ) ,
( )i j ij
i jj j
P X x Y y pP X x Y y
P Y y p
| 0;i jp
| 1.i ji
p
| 0i jp
jY y |i jp
Definition 3.9 Suppose that for any , holds, if
exist, then the limit is called the conditional distribution of for given and denoted it by or .
0x ( ) 0P x X x x
0 0
( , )lim ( | ) lim
( )x x
P Y y x X x xP Y y x X x x
P x X x x
Y
X x ( | )P Y y X x | ( | )Y XF y x
Theorem 3.6 Suppose that has continuous density function and then (3.15)
is a continuous d.f. and its density function is which is
called the conditional density function of given the condition
and denoted it by
( , )X Y( , )p x y ( ) 0,Xp x
|
( , )( | )
( )
y
Y XX
p x v dvF y x
p x
( , ),
( )X
p x y
p x
Y
X x
| ( | ).Y Xp y x
3.4 Functions of random vectorsFunctions of discrete random vectors
Suppose that
(X, Y) ~ P(X = xi, Y = yj) = pij , i, j = 1, 2, …
then Z = g(X, Y) ~ P{Z = zk} = = pk ,
k = 1, 2, …
kji zyxgkiijp
),(:,
(X,Y) (x1,y1) (x1,y2) … (xi,yj) …
pij p11 p12 pij
Z=g(X,Y) g(x1,y1) g(x1,y2
)g(xi,y
j)
or
EX Suppose that X and Y are independent a
nd both are uniformly distributed on 0-1 with law X 0 1
P q p Try to determine the distribution law of (1)W = X + Y ; (2) V = max(X, Y) ;(3) U = min(X, Y);(4)The joint distribution law of w and V .
(X,Y) (0,0) (0,1) (1,0) (1,1)
pij
W = X +YV =max(X, Y)
U = min(X, Y)
2q pq pq 2p
0 1 1 2
0 1 1 1
0 0 0 1
VW
0 1
0 1 2
2q 0 0
0 pq22p
Example 3.17 Suppose and are independent of
then each other, then
1 2~ ( ), ~ ( )X P Y P
1 2~ ( ).Z X Y P
Example 3.18 Suppose and are independent of , then
~ ( , ), ~ ( , )X b n p Y b m p
~ ( , ).Z X Y b m n p
Let be the joint density function of , then the density function of can be given by the following way.
and
( , )p x y ( , )X Y
Z X Y
( ) ( ) ( ) ( , )
( , ) [ ( , ) ]
z
D
z x
x y z
F z P Z z P X Y z p x y dxdy
p x y dxy p x y dy dx
( , ) ( , )z z
u y x dx p x u x du p x u x dx du
令
( ) ( , )Zp z p x z x dx
Suppose is the density function of and
has continuous partial derivatives, the inverse transformation
exists, the Jacobian determinant J is defined as follows:
Define then the joint density function of can
be determined as
( , )p x y ( , )X Y1
2
( , ),
( , ).
u g x y
v g x y
( , ),
( , )
x x u v
y y u v
1
1( , ) ( , )
0.( , ) ( , )
u ux yx yx y u vu uJ
x y v vu v x y
v v x y
1
2
( , )
( , ),
U g X Y
V g X Y
( , )U V
( , ) ( ( , ), ( , )) .p u v p x u v y u v J
Example 3.23 Suppose that is independent of with marginal density and respectively, then the density function of is specified by
Remark Under the conditions of Example 3.23, one can easily find the density function of is
( )Xp xX Y
( )Yp y
U XY
1( ) ( ) ( ) .U X Yp u p u v p v dv
v
U X Y
( ) ( ) ( ) .U X Yp u p uv p v v dv
1.Definition Suppose that the variance of r.v. X and Y exist,
define the expectation E{[XE(X)][YE(Y)]} is the covariance
of X and Y, Cov(X, Y)=E(XY-E(X)E(Y).
If Cov(X,Y)=0 , It is said that X and Y are uncorrelated
What is the different between the concept of “X and Y are in
dependent”and the one of “X and Y are uncorrelated”?
Numerical characteristics of random vectorsNumerical characteristics of random vectors
Example 2 Suppose that (X, Y) is uniformly
distributed on D={(X, Y) : x2+y21} .Prove that X
and Y are uncorrelated but not independent.
Proof
others
yxyxf0
11
),(22
11
1
1
01
)(
2
2
1
1
1
1
dyxdxXEx
x
0)(
2
2
1
1
1
1
dyxy
dxXYEx
x
0)()()(),( YEXEXYEYXCov
Thus X and Y are uncorrelated. Since
others
xxdyxf
x
xX
0
11121
)(
2
2
1
1
2
others
yydyyf
y
yY
0
11121
)(
2
2
1
1
2
)()(),( yfxfyxf YX
Thus, X is not independent of Y.
2. Properties of covariance:
(1) Cov(X, Y)=Cov(Y, X);
(2) Cov(X,X)=D(X);Cov(X,c)=0
(3) 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)
(4) 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)
(5) D(X+Y)=D(X)+D(Y)+2Cov(X, Y).
Proof
)()(2)(2)()()( YEXEXYEYDXDYXD
),(2 YXCovRemark D(X-Y)=D[X+(-Y)]
=D(X)+D(Y)-2Cov(X,Y)
Coefficients
Definition Suppose that r.v. X , Y has finite variance, dentoed by DX>0,DY>0 , respectively, then,
DYDX
)Y,Xcov(XY
is name the coefficient of r.v. X and Y .
Obviously, EX*=0 , DX*=1 and
).(),cov( **** YXEYXXY
DX
XEXX
)(* Introduce which is the standardized of X
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.
D
1
x=y
others
Dyxyxf
0
),(2),(Answer
3
22)(
0
1
0
x
dyxdxXE
3
12)(
0
1
0
x
ydydxYE
4
12)(
0
1
0
x
ydyxdxXYE
1( , ) ( ) ( ) ( )
36COV X Y E XY E X E Y
18
1
9
42)(
0
1
0
2 x
dydxxXD
18
1
9
12)(
0
21
0
x
dyydxYD
)()(
),(
YDXD
YXCOVXY
2
1D
1
2
2
1) ~ (0,1), , determine
2 ~ ( 1,1), , determine
XY
XY
X U Y X
X U Y X
)
What does Example 2 indicate ?
Answer 1)
45
4)(,
12
1)(,
4
1)(,
3
1)(,
2
1)( YDXDXYEYEXE
968.0
454
121121
XY 2) 0)(,0)( XYEXE
0XY