WINSEM2015 16 CP1481 04 Jan 2016 RM01 Two Dim Random Variables

8/18/2019 WINSEM2015 16 CP1481 04 Jan 2016 RM01 Two Dim Random Variables

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Two Dimensional Random Variable / Random Vectors / Bivariate distribution:

Let S be the sample space. Let )( s X X and )( sY Y be two functions each

assigning a real number to each outcome S s . Then ),( Y X is called a two dimensionalrandom variable.

Joint Probability mass function (or) Probability function ),( Y X :

If ),( Y X is a two-dimensional discrete random variable such that

ij ji ji p y x P yY x X P ),(),( is called the joint probability function or joint

probability mass function of ),( Y X satisfying the following conditions:

i) ,0ij p for all i and j

ii) 1 j i

ij p

Marginal Probability Function of X andY :

i)

The marginal probability function of X is given by

......),(),()( 21 yY x X P yY x X P x X P iii

= ....21 ii p p

= j

ij p

= i p ii) The marginal probability function of Y is given by

......),(),()( 21 j j j yY x X P yY x X P yY P

= ....21 j j p p

= i

ij p

= j p

Conditional Probability Function of X andY :

i) The conditional probability function of X given j

yY is given by

j

ij

j

ji ji

p

p

yY P

yY x X P yY x X P

}{

},{}{

ii) The conditional probability function of Y given i x X is given by

i

ij

i

jii j

p

p

x X P

yY x X P x X yY P

}{

},{}{


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Note:

The two R.V’s of X andY are said to be independent if

)()(}{ ji ji yY P x X P yY x X P

That is, jiij p p p

Joint Probability Density Function:

If ),( Y X is a two dimensional continuous R.V such that

dxdy y x f dy

yY dy

yand dx

x X dx

x P ),(}2222

{ then ),( y x f is

called the joint probability density function and it satisfies the following conditions:

i) 0),( y x f , x

ii)

1),( dxdy y x f

Marginal probability density function of X andY :

i) Marginal probability density function of X is given by

dy y x f x f X ),()(

ii) Marginal probability density function of Y is given by

dx y x f y f Y ),()(

Conditional density function of X andY :

i) The conditional density of X given Y is given by

)(

),()(

y f

y x f y x f

Y

ii) The conditional probability function of Y given X is given by

)(

),()(

x f

y x f x y f

X

Cumulative distribution function:

If ),( Y X is a two dimensional R.V (discrete or continuous), then

)(),( yY and x X P y x F is called the cumulative distribution function of ),( Y X .

i) In discrete case, i

ij j

p y x F ),(

ii) In continuous case, dydx y x f y x F x y

),(),(


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1. The joint distribution of X and Y is given by

,21

),( y x

y x f

3,2,1 x , 2,1 y . Find the marginal distributions.

Solution:

Given: ,21

),( y x

y x f

3,2,1 x , 2,1 y

X )( y P Y

Y

1 2 3

1212

213

214

219

2213

214

215

2112

)( x P X 215

217

219 1

Marginal distribution of X

)1( X P = )1,1( P + )2,1( P =212 +

213 =

215

)2( X P = )1,2( P + )2,2( P =213 +

214 =

217

)3( X P = )1,3( P + )2,3( P =214 +

215 =

219

Marginal distribution of Y

)1( Y P = )1,1( P + )1,2( P + )1,3( P =212 +

213 +

214 =

219

)2( Y P = )2,1( P + )2,2( P + )2,3( P = 213

+ 214

+ 215

= 2112

2. For the bivariate distribution of ),( Y X given below find )1( X P ,

)3( Y P , )3,1( Y X P , )31( Y X P , )13( X Y P .

Y )( y P Y

X

1 2 3 4 5 6

1 0 0321

322

322

323

328

2161

161

81

81

81

81

1610

3321

321

641

641 0

642

648

)( x P X

323

323 1


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Solution:

i) )1( X P = P(X=0) + P(X=1) =328 +

1610 =

87

Now,

)0( X P = )1,0( P + )2,0( P + P(0,3) + P(0,4) + P(0,5) + P(0,6)

= 0 + 0 +321 +

322 +

322 +

323

=328

)1( X P = )1,1( P + )2,1( P + P(1,3) + P(1,4) + P(1,5) + P(1,6)

=161 +

161 +

81 +

81 +

81

=16

10

ii) )3( Y P = P(Y=1) + P(Y=2) + P(Y=3)

Now,

P(Y=1) = P(0,1) + P(1,1) +P(2,1) = 0 + 161 + 321 = 323

P(Y=2) = P(0,2) + P(1,2) + P(2,2) = 0 +161 +

321 =

323

P(Y=3) = P(0,3) + P(1,3) + P(2,3) =321 +

81 +

6411 =

6423

iii) )3,1( Y X P =

3

1),0(

j jY X P +

3

1),1(

j jY X P

= (0 + 0 +321 ) + (

161 +

161 +

81 )

=329

iv) )31( Y X P =)3()3,1(

Y P Y X P =

64/2332/9 =

2318

v) )13( X Y P =)1(

)3,1(

X P

Y X P =

8/7

32/9=

28

9

3. The joint probability mass function of (X,Y) is given by

)32(),( y xk y x p ; ;2,1,0 x 3,2,1 y . Find all the marginal and

conditional probability distributions.

Solution: Given: )32(),( y xk y x p

Y )( y P Y

X

1 2 3

0 3k 6k 9k 18k

1 5k 8k 11k 24k

2 7k 10k 13k 30k

)( x P X 15k 24k 33k


5/21

Wkt,

3

1

2

01),(

j i ji y x p

ie) 15k + 24k + 33k = 1

ie) 72k = 1

Therefore,72

1k

Marginal Probability Distribution of X:

P(X = 0) = P(0,1) + P(0,2) + P(0,3) =72

18

P(X =1) = P(1,1) + P(1,2) + P(1,3) =72

24

P(X =2) = P(2,1) + P(2,2) + P(2,3) =72

30

Marginal Probability Distribution of Y:

P(Y = 1) = P(0,1) + P(1,1) + P(2,1) =72

15

P(Y =2) = P(0,2) + P(1,2) + P(2,2) =72

24

P(Y = 3) = P(0,3) + P(1,3) + P(2,3) =72

33

Wkt, the conditional probability function of X given j yY is given by

j

ij

j

ji ji

p

p

yY P

yY x X P yY x X P

}{

},{}{

a) The conditional probability function of X given 1Y is given by

}1{

}1,{}1{

Y P

Y x X P Y x X P ii

i)5

1

15

3

}1{

}1,0{}10{

k

k

Y P

Y X P Y X P

ii)31

155

}1{}1,1{}11{

k k

Y P Y X P Y X P

iii)15

7

15

7

}1{

}1,2{}12{

k

k

Y P

Y X P Y X P


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b) The conditional probability function of X given 2Y is given by

}2{

}2,{}2{

Y P

Y x X P Y x X P ii

i) 4

1

24

6

}2{

}2,0{

}20{

k

k

Y P

Y X P

Y X P

ii)3

1

24

8

}2{

}2,1{}21{

k

k

Y P

Y X P Y X P

iii)12

5

24

10

}2{

}2,2{}22{

k

k

Y P

Y X P Y X P

c) The conditional probability function of X given 3Y is given by

}3{

}3,{}3{

Y P

Y x X P Y x X P ii

i)339

339

}3{}3,0{}30{

k k

Y P Y X P Y X P

ii)3

1

33

11

}3{

}3,1{}31{

k

k

Y P

Y X P Y X P

iii)33

13

33

13

}3{

}3,2{}32{

k

k

Y P

Y X P Y X P

The conditional probability function of Y given i x X is given by

i

ij

i

jii j

p

p

x X P

yY x X P x X yY P

}{

},{}{

a) The conditional probability function of Y given 0 X is given by

}0{

},0{}0{

X P

yY X P X yY P

j j

i)6

1

18

3

}0{

}1,0{}01{

k

k

X P

Y X P X Y P

ii)3

1

18

6

}0{

}2,0{}02{

k

k

X P

Y X P X Y P

iii)2

1

18

9

}0{

}3,0{}03{

k

k

X P

Y X P X Y P

b) The conditional probability function of Y given 1 X is given by

}1{

},1{}1{

X P

yY X P X yY P

j j

i)24

5

24

5

}1{

}1,1{}11{

k

k

X P

Y X P X Y P


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ii)3

1

24

8

}1{

}2,1{}12{

k

k

X P

Y X P X Y P

iii)24

11

24

11

}1{

}3,1{}13{

k

k

X P

Y X P X Y P

c) The conditional probability function of Y given 2 X is given by

}2{

},2{}2{

X P

yY X P X yY P

j j

i)30

7

30

7

}2{

}1,2{}21{

k

k

X P

Y X P X Y P

ii)3

1

30

10

}2{

}2,2{}22{

k

k

X P

Y X P X Y P

iii)30

13

30

13

}2{

}3,2{}23{

k

k

X P

Y X P X Y P

4. The joint probability density function of the 2-dimensional R.V is

otherwise

y x xy y x f

,0

21,9

8

),(

i) Find the marginal density functions of X andY

ii) Find the conditional density function of Y given x X .

Solution:

Marginal density function of X is given by

dy y x f x f x f X ),()()(

= 2

9

8

x

xydy {since 2 y x }

=

22

29

8

x

y x

= ]4[9

4 2 x

x

,

Marginal density function of Y is given by

dx y x f x f y f Y ),()()(

= y

xydx19

8 {since y x 1 }


8/21

=

y x

y

1

2

29

8

= ]4[9

4 2 y y

,

The conditional density function of Y given x X is

)(

),()(

y f

y x f y x f

Y

=

)4(9

49

8

2 x x

xy

=

)4(

2

2 x

y

5. The bivariate R.V X and Y has the p.d.f

20

2,)8(),(

2

x

x y x y Kx y x f .

Find K.

Solution:

We know that,

1),( dydx y x f

x

x

dydx y Kx2

22

0

1)8(

12

8

22

0

22

dx

y y x K

x

x

12

82

416

2

0

222

dx

x x

x x x K

12

38

2

0

43

dx

x x K

152

3

4

82

0

54

x x K

112

5 K

6. The joint density function of the R.V’s X and Y is given by

otherwise

x y x xy y x f

0

0,10,8),( .

Find (i) )( x f X (ii) )( y f Y (iii) )( x y f


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Solution:

dy y x f x f x f X ),()()( =

x xydy

0

8 = 34 x

dx y x f x f y f Y ),()()( = 1

0

8 xydx = y4

)(

),()(

x f

y x f x y f

Y

=23

2

4

8

x

y

x

xy

7. The joint pdf of R.V. X and Y is given by

0,0,),( )22( y xkxye y x f y x . Find the value of k and also

prove that X and Y are independent.

Solution:

We know that,

1),( dydx y x f

0

)22(

0

1dydxkxye y x

0

2dy y yek .

0

2dx x xe =1

2

1.

2

1k = 1 { since

0

2dx x xe =2

1}

4k

To prove that: X and Y are independent

ie, to prove that : ),()().( y x f y f x f

Now,

dy y x f x f x f X ),()()(

=

0

)22

(dykxye

y x

=

0

22

dy yekxe

y x

=

2

14

2 x

xe

= 0,22

x xe x


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dx y x f x f y f Y ),()()( = 0,22

y ye y

Now, )().( y f x f 2

2 x xe .2

2 y ye = )22(4 y x xye = ),( y x f

Therefore, X and Y are independent .

8. Given

otherwise

x y x x y xCx y x f

0

,20,)(),(

(i) Find C (ii) Find )( x f (iii) )( x y f

Solution:

(i) We know that,

1),( dydx y x f

2

0

1)( dxdy y xCx x

x

8

1C

(ii)

dy y x f x f x f X ),()()( =

x

x

dy y x x )(8

1 =

4

3 x

(iii))(

),()(

x f

y x f x y f

Y

= )(2

1

4

)(8

1

23 y x x x

y x x

9. The joint pdf of a two-dimensional RV ),( Y X is given by

,8

),(2

2 x xy y x f 20 x , 10 y . Compute ),1( X P

),2

1( Y P )

211( Y X P .

Solution:

(i)

)1(

1

),()1(

X

R

dxdy y x f X P

= dxdy x

xy

1

0

2

1

22

8 =

24

19


11/21

(ii)

)2

1(

2

),()2

1(

Y

R

dxdy y x f Y P

= dxdy x

xy

2

1

0

2

0

22

8 =

4

1

(iii) dxdy x

xyY X P

2

1

0

2

1

22

8)21,1( =

24

5

(iv) )211( Y X P =

)2

1(

)2

1,1(

Y P

Y X P

=

4124

5

=6

5

Mathematical Expectation:

Mean (or) expectation is a significant number representing the behavior of a random

variable.

Mathematical expectation of ‘X’ is denoted by E(X)

i)

dx x xf X E )()( , [for a single dimensional continuous random variable]

ii)

x

x xp X E )()( , [for a single dimensional discrete random variable]

iii) x

x p x X E )()( 22 , [for a single dimensional discrete random variable]

iv)

dxdy y x xf X E ),()( [ for a two dimensional continuous random variable]

v)

dxdy y x f x X E ),()( 22 [ for a two dimensional continuous random variable]

vi)

dxdy y x xyf XY E ),()( , [ for a two dimensional continuous random variable]

vii) 22 )]([)()( X E X E X Var

Note:

If X and Y are independent random variables then )().()( Y E X E XY E


12/21

1. Two R.V’s X and Y have joint pdf

elsewhere

y x xy

y x f

,0

51,40,96),(

Find (i) E(X) (ii) E(Y) (iii) E(XY) (iv) E(2X + 3Y) (v) Var(X)

(vi) Cov(X,Y).

Solution:

i)

dxdy y x xf X E ),()(

= dxdy xy

x

5

1

4

0 96

=3

8

ii)

dxdy y x yf Y E ),()(

= dxdy xy

y

5

1

4

0 96

=9

31

iii)

dxdy y x xyf XY E ),()(

= dxdy xy

xy

5

1

4

0 96

=27

248

iv) )(3)(2]32[ Y E X E Y X E = 2.3

8+ 3.

9

31 =

3

47

v) We know that, 22 )]([)()( X E X E X Var

Now,

dxdy y x f x X E ),()( 22

= dxdy xy

x

5

1

4

02

96

= 8

22 )]([)()( X E X E X Var

= 8 -

2

3

8

=

9

8


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vi) )().()(),( Y E X E XY E Y X Cov

=27

248 -

9

31

3

8

= 0

2.

If the joint pdf of (XY) is given by 10),1(24),( x y x y y x f ,

then find E(XY).

Solution:

We know that,

dxdy y x xyf XY E ),()(

= 1

0

1

),( y

dxdy y x xyf {since x varies from y to 1,

y varies from 0 to 1}

= 1

0

12

)1(24 y

dxdy x xy

=

1

0

322

326

124 dy

y y y

=

1

0

653

18101824

y y y

= 15

4

3. If X and Y is a two dimensional R.V uniformly distributed over the

triangular region R bounded by3

43,0

x yand x y . Find

),( x f ),( y f ),( X E Var(X), ),(Y E XY .

Solution:

Given X and Y are uniformly distributed .

Therefore, )tan(),( t consak y x f We know that, 1),( dxdy y x f

That is, 4

0

3

4

3

1 y

kdxdy


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1][4

0

3

4

3 dy xk y

14

33

4

0

dy

yk

16 k 6

1 k

3

4

3

),()( y

dx y x f y f = 3

4

3 6

1

y

dx = 40),4(8

1 y y

3

4

0 6

1)(

x

dy x f = 30,9

2 x x

dx x xf X E )()( =

3

0

2

9

2dx x = 2

dy y yf Y E )()( =3

4)4(

8

4

0

dy y y

2

9)()( 22

dx x f x X E

3

8)()( 22

dy y f yY E

22

)]([)()( X E X E X Var

= 2

1

22 )]([)()( Y E Y E Y Var =9

8

4

0

3

4

36

1)(

y

xydxdy XY E = 3

Now,Y X

XY Y E X E XY E

.

)()()( =

2

1


15/21


16/21

3

2

3

2))(()(

2'

t t

X X ee

t M dt

d t M

Therefore,3

4

3

2

3

2)]([)originabout( 0

''1 t X t M

34

32))(()(

2

2

2

''

t t

X X eet M dt d t M

Therefore, 23

6

3

4

3

2)]([)originabout( 0

'''2 t X t M

Therefore, Mean =3

4'1 and

Variance =9

2)( 2'1

'22

2. Find the moment generating function of a random variable X whose probability

function is ;1)(n

x X P .,...,2,1 n x

Solution:

By definition, we know that )()( tx X e E t M x

tx x P e )(

]...[1 2 nt t t eeen

])(...)(1[ 2 nt t t t

eeen

e

t

nt t

e

e

n

e

1

1

.

3. Obtain the m.g.f for the distribution where

otherwise;0

0;2

1)( 2 xe x f

x

.

Solution: By definition, we know that,

dx x f ee E t M txtx X )()()(

0

2

2

1dxee

x

tx

0

)2

1(

2

1dxe

xt

0

)21(

2

12

1

t

et

12

1

t


17/21

4. If a random variable X has the m.g.ft

t M X 2

2)( , determine the variance of X .

Solution:1)2

1()2

1(2

2

2

2)(

t t t

t M X

........22

1 2

t t

........!2.2!1.2

12

t t

Now, coefficient of2

1

!1.

'1

t ;

coefficient of2

1

!2.

'2

2

t

;

Therefore, Variance of X

4

1)( 2'1

'2

5. Find the moment generating function of the random variable whose raw moments are

given by r r r 2)!1(' .

Solution: We know that, if X is a random variable then its m.g.f

)()( tx X e E t M ...!

....!2

1 ''2

2'1 r

r

r

t t t

...)2)!.1((!

....)2!.3(!2

)2!.2(1 222

1 r r

t t t

r

r

r )1(...321 2

where t 2 2)1(

2)21( t

Characteristic Function

In few cases the moment generating function does not exist. In this case, we use

characteristic functions which is more serviceable function than the m.g.f.

Definition: The characteristic function of a random variable X is denoted by )(t X and

is defined as


18/21


19/21

3. Find the characteristic of the exponential distribution

otherwise;0

0;)(

xe x f

x .

Hence find the mean and variance.

Solution: By definition, the characteristic function of X is given by

dxeedx x f ee E t xitxitxitX X

0

.)()()(

dxe xit 0

)(

0

)(

)( it

e xit

)( it

Note that, we can also express )(t X as,

it

t X

1

1)(

....1

2

it it

Hence, we find that '1 = coefficient of

1

!1

it

and '2 = coefficient of2

2

!2

it

Hence, mean of X is given by Mean

1'1

and variance of X is given by Var( X )2

2'1

'22

1)(

4. If X has the probability density function (Laplace distribution) x

e x f 2

1)( for

x . Show that the characteristic function of X is given by .1

1)(

2t t X

Solution: By definition, the characteristic function of X is given by

dx x f ee E t itxitX X )()()(


20/21

dxeedxee xitx xitx

0

0

2

1.

2

1.

dxeedxee xitx xitx

0

0.

2

1.

2

1

dxeedyee xitx yity 00

.21.

21

dxedye xit yit

0

)1(

0

)1(

2

1

2

1

0

)1(

0

)1(

)1(2

1

)1(2

1

it

e

it

e xit yit

)1(

1

2

1

)1(

1

2

1

it it

)1)(1(

1

2

1

it it

21

1

t

5. Find the density function )( x f corresponding to the characteristic function defined as

1,0

1,1)(

t

t t t X . .

Solution: The probability density function )( x f is given by,

dt t e x f X itx )(

2

1)(

1

0

0

1

)1()1(2

1dt t edt t e itxitx

1

0

2

0

1

2 )()1()1(

)()1()1(

2

1

ix

e

ix

et

ix

e

ix

et

itxitxitxitx

)11(

)(

1

2

1

2

ixix ee

ix

21

12

ixix ee

x

,cos12 x

x

x .


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Documents

WINSEM2015 16 CP1481 04 Jan 2016 RM01 Two Dim Random Variables