Lecture 16: Vector RVs, Functions of several RVs Prof ...ece-research.unm.edu/.../Lecture16_VectorRVFunctionofSeveralRV.pdf · pdf of Linear Transformations We consider first the

ECE 340Probabilistic Methods in Engineering

M/W 3-4:15

Prof. Vince CalhounProf. Vince Calhoun

Lecture 16: Vector RVs, Functions of Lecture 16: Vector RVs, Functions of several RVsseveral RVs

• Section 5.8-5.9• Function of two RV’s

• Section 6.1-6.2• Vector RV’s• Functions of several RV’s

Section 6.1-6.2Vector RV’s

Functions of several RV’s

remember

Section 6.3-6.4Expected value of vector RV’s

Jointly Gaussian RV’s

Section 6.3-6.41.Expected value of vector RV’s2.Jointly Gaussian RV’s

Functions of Several Random Variables

Types of Transformations– A Single Function of n RV’s

– Functions of n RV’s

)X(gZ =

nk);X(gZ kk ≤≤= 1

Functions of Several Random Variables

Functions of n RV’s

)X,...,X,X,X(gnk);X(gZ

nk

kk

321

1=

≤≤=

Assume Invertability

)Z,...,Z,Z,Z(hnk);Z(hX

nk

kk

321

1=

≤≤=

pdf of Linear Transformations

We consider first the linear transformation of two random variables :

or

Denote the above matrix by A. We will assume A has an inverse, so each point (v, w) has a unique corresponding point (x, y) obtained from

In Fig. 4.15, the infinitesimal rectangle and the parallelogram are equivalent events, so their probabilities must be equal. Thus

eYcXWbYaXV

+=+= .⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡YX

ecba

WV

(4.56) .1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡ −

wv

Ayx

dPwvfdxdyyxf WVYX ),(),( ,, ≅

where dP is the area of the parallelogram. The joint pdf of V and W is thus given by

where x an y are related to (v, w) by Eq. (4.56)It can be shown that so the “stretch factor” is

where |A| is the determinant of A. Let the n-dimensional vector Z be

where A is an invertible matrix. The joint of Z is then

(4.57) ,),(

),( ,,

dxdydP

yxfwvf YX

WV =

( ) ,dxdybcaedP −=

( )( ) ,Abcaedxdy

dxdybcaedxdydP

=−=−

=

,XZ A=

nn×

EXAMPLE 4.36 Linear Transformation of Jointly Gaussian Random Variables

Let X and Y be the jointly Gaussian random variables. Let V and Wbe obtained from (X, Y) by

Find the joint pdf of V and W.

|A| = 1,

( ) ( ) ( ) (4.58) z z x

Z AAf

Axxf

zzffzAx

nXXnZZ

n

n

11,,

1,,1

1

1

,,,,)(

−

=

==≡−

…… …

…

.1111

21

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡YX

AYX

WV

,1111

21

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡WV

YX

so

where

By substituting for x and y, the argument of the exponent becomes

Thus

( ) ( ) .22 WVYWVX +=−= and

,2

,2

),( ,, ⎟⎠

⎞⎜⎝

⎛ +−=

wvwvfwvf YXWV

( ) ( ) ( ) .121,

222 12/2

2,ρρ

ρπ−+−−

−= yxyx

YX eyxf

( ) ( )( ) ( )( ) ( ) ( ) .

1212122/2/22/ 22

2

22

ρρρπρ

−+

+=

−+++−−− wvwvwvwvwv

( )( )[ ] ( )[ ]{ }.

121),( 12/12/

2/12,22 ρρ

ρπ−++−

−= wv

WV ewvf

pdf of General Transformations

Let the r.v’s V and W be defined by two nonlinear functions of X and Y :

Assume that the functions v(x, y) and w(x, y) are invertible, then

In Fig. 4.17(b) , make the approximation

and similarly for the y variable. The probabilities of the infinitesimal rectangle and the parallelogram are approximately equal. therefore

(4.59) and .),(),( 21 YXgWYXgV ==

.),(),( 21 wvhywvhx == and

,2 1),(),(),( =∂∂

+≅+ kdxyxgx

yxgydxxg kkk

dPwvfdxdyyxf WVYX ),(),( ,, =

and

where dP is the area of the parallelogram. The “stretch factor” at the point (v, w) is given by the determinant of a matrix of partial derivatives :

The determinant J(x, y) is called the Jacobian of the transformation.

(4.60) ,)),(),,((

),( 21,,

dxdydP

wvhwvhfwvf YX

WV =

.det),(

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

yw

xw

yv

xv

yxJ

The Jacobian of the inverse transformation is given by

It can be shown that

We therefore conclude that the joint pdf of V and W can be found using either of the following expressions :

.det),(⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

wy

vy

wx

vx

wvJ

.),(

1),(yxJ

wvJ =

( )( ) (4.61b)

(4.61a)

wvJwvhwvhf

yxJwvhwvhf

wvf

YX

YXWV

,)),(),,((

,)),(),,((

),(

21,

21,,

=

=

Functions of Several Random Variables(Gaussian Example)

X and Y are two i.i.d. Zero mean Gaussian RV’s:

⎟⎠⎞

⎜⎝⎛=

+=

XYarctanW

YXV 22

x

y

vw

)Wsin(VY)Wcos(VX

==

Inversions:

Functions of Several Random Variables(Gaussian Example)

Jacobian

T

wvwvww

T

wy

wx

vy

vx

wv ⎥⎦

⎤⎢⎣

⎡−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=ℑ)cos()sin(

)sin()cos(),(

Then:

v|v||)w,v(| ==ℑ

Functions of Several Random Variables(Gaussian Example)Jacobian

( ))wsin(v),wcos(vvf)w,v(f XYVW =

2

22

2)(

22

1),( σπσ

yxeyxXYf

+−

=

Where

ππσ

πσ

σ

σ

20 2

2

2

2

2

22

2

2

<<=

=

−

+−

w;)v(Uev

ev)w,v(f

v

])wsinv()wcosv[(

VW

Thus

Functions of Several Random Variables(Gaussian Example)

Marginals:

2

2

22

2

0

)v(Uev

dw)w,v(f)v(f

v

VWw

V

σ

π

σ

−

=

=

∫=

Rayleigh RV

ππ

2w0 ; 21

),(

0

)(

<<=

∞

=

= ∫ dvwvVWf

v

wWf

Uniform RV

Consider the problem of finding the joint pdf for n functions of n random variables X = (X1,…, Xn):

We assume as before that the set of equations

has a unique solution given by

The joint pdf of Z is then given by

,)(,)(,)( 2211 X ,X X nn gZgZgZ === …

,)(,)(,)( 2211 x ,x x nn gzgzgz === … )62.4(

,)(,)(,)( 2211 x ,x x nn hxhxhx === …

( )( )

( ) ( ) (4.63b)xxx

(4.63a) xxx

,,,,)(,),(),(

,,,)(,),(),(

),,(

2121,,

21

21,,1,,

1

1

1

nnXX

n

nXXnZZ

zzzJhhhf

xxxJhhhf

zzf

n

n

n

……

……

…

…

……

=

=

where are the determinants of the transformation and the inverse transformation, respectively,

and

( ) ( )nn zzJxxJ ,,,, 11 …… and

( )

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

n

nn

n

n

xg

xg

xg

xg

xxJ …

1

1

1

1

1 det,,

( )

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

n

nn

n

n

zh

zh

zh

zh

zzJ …

1

1

1

1

1 det,,

4.5 MULTIPLE RANDOM VARIABLES

Joint Distributions

The joint cumulative distribution function of X1, X2,…., Xn is defined as the probability of an n-dimensional semi-infinite rectangle associate with the point (x1,…, xn):

The joint cdf is defined for discrete, continuous, and random variables of mixed type.

[ ] (4.38) .,,,),,( 221121,, 21 nnnXXX xXxXxXPxxxFn

≤≤≤= ………

EXAMPLE 4.27

Let the event A be defined as follows :

Find the probability of A .

The maximum of three numbers is less than 5 if and only if each of the three numbers is less than 5 ; therefore

( ){ }.5,,max 321 ≤= XXXA

[ ] { } { } { }[ ].)5,5,5(

555

321 ,,

321

XXXFXXXPAP

=≤∩≤∩≤=

The joint probability mass function of n discrete random variables is defined by

The probability of any n-dimensional event A is found by summing the pmf over the points in the event

One-dimensional pmf of Xj is found by adding the joint pmf over all variables other than xj:

The marginal pmf for X1,…,Xn-1 is given by

[ ] (4.39) .,,),,( 221121,, 21 nnnXXX xXxXxXPxxxpn

==== ………

( )[ ] ∑ ∑= (4.40) in Ain

.),,(,, 21,,21 21 nXXXxn xxxpAXXXPn

…… …

[ ] (4.41) .),,()( 21,,,1 1

21

1

∑∑∑∑− +

===n

n

j j

jx

nXXXx xx

jjjX xxxpxXPxp ……

(4.42) .),,(),,( 21,,,121,,, 21121 ∑=−−

n

nnx

nXXXnXXX xxxpxxxp …… ……

A family of conditional pmf’s is obtained from the joint pmf by conditioning on different subsets of the random variables.

if . Repeated applications of Eq. (4.43a) yield

( ) ( )( ) (4.43a) ,

,,|,,

,,|111,,

11,,11

1

1

−−

−− =

nnXX

nXXnnX xxxp

xxpxxxp

n

n

n ……

……

…

( ) 0,, 11,,1>−nXX xxp

n……

( ) ( )( ) ( ) ( )

(4.43b)

xpxxpxxxp

xxxpxxp

XXnnX

nnXnXX

n

nn

112211

1111,,

121

1

|,,|

,,|,,

……………

−−

−−

−×

=

EXAMPLE 4.28

A computer system receives messages over three communications lines. Let Xj be the number of messages received on line j in one hour. Suppose that the joint pmf of X1, X2, and X3 is given by

Find p(x1, x2) and p(x1) given that 0< ai < 1.

( ) ( )( )( )000

111,,

321

321321321,,321

321

≥≥≥

−−−=

,x,xx

aaaaaaxxxp xxxXXX

( ) ( )( )( )

( )( ) .11

111,

21

3

321

21

2121

032132121,

xx

x

xxxXX

aaaa

aaaaaaxxp

−−=

−−−= ∑∞

=

( ) ( )( )

( ) .1

11

1

2

21

1

11

021211

x

x

xxX

aa

aaaaxp

−=

−−= ∑∞

=

If r.v’s X1, X2,…,Xn are jointly continuous random variables, then the probability of any n-dimensional event A is

where is the joint probability density functionThe joint cdf of X is obtained from the joint pdf by integration :

The joint pdf (if the derivative exists) is given by

The marginal pdf for a subset of the random variables is obtained b integrating the other variables out. The marginal of X1 is

( )[ ] (4.44) Ain Ain

,),,(,, ''1

''1,,1 1∫∫= nnXXxn dxdxxxfXXP

n………… …

),,( ''1,,1 nXX xxf

n……

(4.45) ,),,(),,,( 1

121

''1

''1,,21,,, ∫ ∫∞− ∞−

=x x

nnXXnXXXn

nndxdxxxfxxxF ……… ……

(4.46) .),,(),,( 1,,1

''1,, 121 nXX

n

n

nXXX xxFxx

xxfnn

…… …… ∂∂∂

=

(4.47) .),,,()( ''2

''21,,1 11 ∫ ∫

∞

∞−

∞

∞−= nnXXX dxdxxxxfxf

n………

The marginal pdf for X1,…,Xn-1 is given by

The pdf of Xn given the values of X1,…,Xn-1 is given by

if

Repeated applications of Eq. (4.49a) yield

(4.48) .),,,(),,( ''11,,11,, 111 ∫

∞

∞− −− =− nnnXXnXX dxxxxfxxf

nn…… ……

(4.49a) ),,(

),,(),,|(

11,,

1,,11

11

1

−−

−

=nXX

nXXnnX xxf

xxfxxxf

n

n

n ……

……

…

0),,( 11,, 11>−− nXX xxf

n……

(4.49b)

)()|(),,|(

),,|(),,(

112211

111,,

121

1

xfxxfxxxf

xxxfxxf

XXnnX

nnXnXX

n

nn

……………

−−

−

−×

=

EXAMPLE 4.29

The r.v’s X1, X2, and X3 have the joint Gaussian pdf

Find the marginal pdf of X1 and X3 .

The above integral was carried out in Example 4.13 with

( ) .2

,,

2321

22

21

321

212

321,, ππ

⎟⎠⎞

⎜⎝⎛ +−+−

=xxxxx

XXXexxxf

( )( )

.2/22

, 2

22/

31,

2122

21

23

31dxeexxf

xxxxx

XX ∫∞

∞−

−+−−

=ππ

2/1=ρ

( ) .22

,2/2/

31,

21

23

31 ππ

xx

XXeexxf−−

=

Independence

X1,…,Xn-1 are independent if

for any one-dimensional events A1,…,An. It can be shown that X1,…,Xn are independent if and only if

for all x1,…,xn. If the random variables are discrete,

If the random variables are jointly continuous,

[ ] [ ] [ ]nnnn AXPAXPAXAXP in in in in … 1111 ,, =

(4.50) )()(),,( 11,, 11 nXXnXX xFxFxxFnn

……… =

. all for nnXXnXX ,x,xxpxpxxpnn

………… 111,, )()(),,(11

=

. all for nnXXnXX ,x,xxfxfxxfnn

………… 111,, )()(),,(11

=

4.6 FUNCTIONS OF SEVERAL RANDOM VARIABLES

One Function of Several Random Variables

Let the random variable Z be defined as a function of several random variables:

The cdf of Z is found by first finding the equivalent event of that is, the set then

( ) (4.51) .,,, 21 nXXXgZ …=

{ },zZ ≤( ) ( ){ },,,1 zgxxR nZ ≤== x that such x …

[ ]( ) (4.52)

in X

inx ∫∫==

.,,

)(''

1''

1,,1 nnXXR

zZ

dxdxxxf

RPzF

nz

………

EXAMPLE 4.31 Sum of Two Random Variables

Let Z = X + Y. Find FZ(z) and fZ(z) in terms of the joint pdf of X and Y.

The cdf of Z is

The pdf of Z is

Thus the pdf for the sum of two random variables is given by a superposition integral.

If X and Y are independent random variables, then by Eq. (4.21) the pdf is given by the convolution integral of the margial pdf’s of Xand Y :

.'')','()('

,∫ ∫∞

∞−

−

∞−=

xz

YXZ dxdyyxfzF

(4.53) .')','()()( ,∫∞

∞−−== dxxzxfzF

dzdzf YXZZ

(4.54) .')'()'()( ∫∞

∞−−= dxxzfxfzf YXZ

EXAMPLE 4.32 Sum of Nonindependent Gaussian Random Variables

Find the pdf of the sum Z = X + Y of two zero-mean, unit-variance Gaussian random variables with correlation coefficient ρ= -1 / 2.

After completing the square of the argument in the exponent we obtain

( )( ) ( )[ ]( )

( )[ ]

( ) .'4/32

''exp4/32

1

'12

'''2'exp12

1

')','()(

22

21

2

22

212

,

dxzzxx

dxxzxzxx

dxxzxfzf YXZ

⎭⎬⎫

⎩⎨⎧ +−−=

⎭⎬⎫

⎩⎨⎧

−−+−−

−−

=

−=

∫

∫

∫

∞

∞−

∞

∞−

∞

∞−

π

ρρ

ρπ

.2

)(2/2

π

z

Zezf−

=

Let Z = g(X, Y), and suppose we are given that Y = y, then Z = g(X, y) is a function of one random variable. And the pdf of Z given Y= y: fZ(z | Y = y). The pdf of Z is found from

.')'()'|()( ∫∞

∞−= dyyfyzfzf YZZ

EXAMPLE 4.34

Let Z = X / Y. Find the pdf of Z if X and Y are independent and both exponentially distributed with mean one.

Assume Y = y, then

The pdf of Z is

.)|()|( yyzfyyzf XZ =

.')','('

')'()'|'(')(

,∫∫∞

∞−

∞

∞−

=

=

dyyzyfy

dyyfyzyfyzf

YX

YXZ

Transformations of Random Vectors

Let X1,…, Xn be random variables associate with some experiment, and let the random variables Z1,…, Zn be defined by n functions of X= (X1,…, Xn) :

The joint cdf of Z1,…, Zn at the point z = (z1,…, zn) is equal to the probability of the region of x where

( ).0

11

''

')'()'(')(

2

0

''

0

>+

=

=

=

∫∫∞ −−

∞

zz

dyeey

dyyfzyfyzf

yzy

YXZ

.)()()( 2211 X X X nn gZgZgZ === …

:,...,1)( nkzg kk for x =≤

If X1,…, Xn have a joint pdf, then

EXAMPLE 4.35

Let the random variables W and Z be defined by

Find the joint cdf of W and Z in terms of the joint cdf of X and Y.

If z > w, the above probability is the probability of the semi-infinite rectangle defined by the point (z, z) minus the square region denote by A.

[ ] (4.55a) XX .)(,,)(),,( 111,,1 nnnZZ zgzgPzzFn

≤≤= ………

(4.55b) xx

.),...,(),,( ''1

''1,...,

)'(:'1,, 11 ∫∫

≤

= nnXXzg

nZZ dxdxxxfzzFn

kk

n……

.),max(),min( YXZYXW == and

{ } ( ){ }[ ].,max),min(),(, zYXwYXPzwF ZW ≤∩≤=

If z < w then

[ ]

{ }),(),(),(

),(),(),(),(),(),(),(

,,,

,,,,

,

,,

wwFwzFzwFwwFwzFzwFzzF

zzFAPzzFzwF

YXYXYX

YXYXYXYX

YX

YXZW

−+=

+−−−

=

−=

.),(),( ,, zzFzwF YXZW =