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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
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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 }
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=
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
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(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
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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
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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
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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
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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 )()()(
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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 .
8/18/2019 WINSEM2015 16 CP1481 04 Jan 2016 RM01 Two Dim Random Variables
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