410
Homework STAT #3
(due Friday, February 8, by 3:00 p.m.)
Spring 2008
1.
Suppose that the random variables X and Y have joint p.d.f. f ( x, y ) given by
f ( x, y ) = C x 2 y,
a) Sketch the support of ( X , Y ).
0 < x < y, x + y < 2.
b)
What must the value of C be so that f ( x, y ) is a valid joint
p.d.f.?
- -
Must have
f ( x, y ) dx dy = 1.
1 2- x 0
C x 2 y dy dx
1 0
x
=
C
2
x2 y2
y = 2-x dx y=x
=
1 0
C
2
x 2 ( 2 - x )2 - x 2
[
]
dx
=
1 0
( 2 C x 2 - 2 C x 3 )dx
2C 3 C 4 x - x 3 2
1 = 0
=
C
6
= 1.
C = 6.
c)
Find P ( X + Y < 1 ).
0.5 1- x 0
6 x 2 y dy dx
0.5 0
x
=
(3 x 2 y 2 ) y y= =1 -x x dx
=
0.5 0
(3 x 2 [ ( 1 - x ) 2 - x 2 ] )dx
( 3 x 2 - 6 x 3 )dx
3 4 x 2
- 3 1 2 2
0.5 0
4
=
0.5 0
=
x3 -
1 2
3
=
=
1 3 1 - = = 0.03125. 8 32 32
2.
Suppose that the random variables X and Y have joint p.d.f. f ( x, y ) given by
f ( x, y ) = 6 x 2 y,
a)
0 < x < y, x + y < 2.
Find the marginal probability density function for X. First, X can only take values in ( 0 , 1 ).
fX( x) =
f ( x, y ) dy = -
2- x
6 x 2 y dy =
x
(3 x 2 y 2 ) y y= =2 -x x
0 < x < 1.
= 3x2
{( 2 - x ) 2 - x 2 } = 12 x 2 12 x 3 = 12 x 2 ( 1 x ),
b)
Find the marginal probability density function for Y. First, Y can only take values in ( 0 , 2 ).
y
6 x 2 y dx
0 < y <1
fY( y) =
f (x, y ) dx = -
0 2- y 0
6 x 2 y dx
1< y < 2
( 2 x 3 y ) xx == 0y
=
2 ( 2 x 3 y ) x = = -0 y x
0 < y <1
1< y < 2
2y4
=
0 < y <1 1< y < 2
2 y ( 2 - y )3
From the textbook:
2.1.6
FZ( z ) = P( Z z ) = P( Y z X ) = z
0
z-x
0
e- x - y d y d x
z-x
0
=
z
0
e- x
e- y d y d x
=
z
0
e- x 1 - e- z + x d x e
-x
(
)
=
z
0
dx-
z
0
e- z d x = 1 - e-z - z e-z ,
z > 0.
In particular, P ( Z 0 ) = 0, P ( Z 6 ) = 1 7 e 6 0.98265.
f Z ( z ) = F Z( z ) = e z e z + z e z = z e z, ' 2.1.7
FZ( z ) = P( Z z ) = P( Y z X
z > 0.
)
=
z 1
0 0
1d y d x +
1
z
x 1d y d x
z
1
0
=
z
0
1d x +
z
z dx x
0 < z < 1. 0 < z < 1.
= z z ln z,
f Z ( z ) = F Z( z ) = ln z, '
0
FZ( z ) =
z - z ln z
1
z0 0 < z <1 z 1
fZ( z ) =
- ln z 0
0 < z <1 otherwise
2.1.9
a)
x2 x1
0 1 0 1 2
p1( x1 )
7 12 5 12
2 12 2 12 4 12
3 12 2 12 5 12
2 12 1 12 3 12
p2( x2 )
b)
P ( X 1 + X 2 = 1 ) = p ( 0, 1 ) + p ( 1, 0 ) =
3 2 5 + = . 12 12 12
2.1.10 f X ( x1 ) =
1
1
x1
x2
0
2 15 x1 x2 dx2 =
15 2 x1 2 x2 dx2 2 x1
x2
0 2 3 x1 dx1
1
=
15 2 2 x1 1 - x1 , 2
(
)
0 < x1 < 1.
f X ( x2 ) =
2
2 15 x1
x2 dx1 = 5 x2
1 2 1- 0
4 = 5 x2 ,
0 < x2 < 1.
P( X 1 + X 2 1 ) = 15 2 = x1 2 0
12
x1
x1
2 15 x1 x2 dx2 dx1
12
1-
x1
x1 2 x2 dx2 dx1
=
15 2 x1 2 0
( (1 - x1 ) 2 - x12 )dx1
12
=
15 2 x1 ( 1 - 2 x1 ) dx1 2 0 15 2 3 x1 - 15 x1 dy = 2 5 3 15 4 x - x1 2 1 4 12 0 =
12
=
0
5 15 5 - = . 16 64 64
2.1.12 x2 x1
1 2 1 2
p1( x1 )
5 12 7 12
2 12 3 12 5 12
3 12 4 12 7 12
p2( x2 )
E( X1 ) = 1 E( X2 ) = 1
5 7 19 +2 = . 12 12 12 5 7 19 +2 = . 12 12 12
2 E( X1 ) = 1 2 E( X2 ) = 1
5 7 33 +4 = . 12 12 12 5 7 33 +4 = . 12 12 12
E( X1 X2 ) = 1 No,
2 6 4 30 +2 +4 = . 12 12 12 12
30 19 19 = E( X1 X2 ) E( X1 ) E( X2 ) = . 12 12 12
2 2 E( X1 2 6 X2 + 7 X1 X2 ) = 2 E( X1 ) 6 E( X2 ) + 7 E( X1 X2 )
= 2
19 33 30 50 6 +7 = . 12 12 12 12
2.1.14
M ( t 1, t 2 ) =
1 e t1 x1 + t 2 x 2 2 x1 =1 x 2 =1
x1 + x 2
=
1 e t1 x1 2 x1 =1
x1 =1
x1
1 e t 2 x2 2 x 2 =1
x2
=
e t1
2
x1
e t2
2
x2
x 2 =1
Must have
e t1
2
< 1,
e t2
2
< 1 for the series to converge
t1 < ln 2, t2 < ln 2.
e t1
M ( t 1, t 2 ) =
e t2
1- 2
2 1-
e t1
2
e t2
2
=
(
e t1 + t 2 , 2 - e t1 2 - e t 2
)(
)
t1 < ln 2, t2 < ln 2.
= M ( t 1 , t 2 ).
M ( t 1, 0 ) M ( 0, t 2 ) =
e t1 e t2 2 - e t1 2 - e t2
2.1.15
M ( t 1, t 2 ) =
0
0
x1
e t1 x1 + t 2 x 2 x1 e - x2 d x2 d x1
=
x1 e t1 x1 x1 e t1 x1
x1
e t 2 x 2 e - x 2 d x2 d x1 e x2 ( t 2 -1 ) d x2 d x1
=
0
x1
Must have t2 < 1 for the integral to converge. M ( t 1, t 2 ) =
0
0
e x1 e t1 x1
x 2 ( t 2 -1 )
x1
t2 - 1
d x1
=
x1 e t1 x1
e x1 ( t 2 -1 ) d x1 1 - t2
=
1 x1 e x1 ( t1 + t 2 -1 ) d x1 1 - t2 0
Must have t1 + t2 < 1 for the integral to converge.
x1 e x1 ( t1 + t 2 -1 ) e x1 ( t1 + t 2 -1 ) 1 M ( t 1, t 2 ) = - 1 - t2 t1 + t 2 - 1 ( t1 + t 2 - 1 ) 2
=
0
1
( 1 - t 2 ) ( 1 - t1 - t 2 ) 2
,
t2 < 1, t1 + t2 < 1.
1 1
No, M ( t 1 , t 2 ) M ( t 1 , 0 ) M ( 0 , t 2 ) =
( 1 - t1 ) 2
(1 - t2 ) 3
.
2.3.2
2 f 1 | 2 ( x 1 | x 2 ) = c1 x1 x 2 ,
0 < x 1 < x 2 , 0 < x 2 < 1.
4 f2( x2) = c2 x2 ,
0 < x 2 < 1.
1 0 4 c 2 x 2 dx 2 =
(a)
1 =
-
f 2 ( x 2 ) dx 2 =
c2
5
.
c 2 = 5.
1 =
-
f 1| 2 ( x1 | x 2 ) dx1 =
x2
0
2 c1 x1 x 2 dx1 =
c1
2
.
c 1 = 2.
(b)
2 f ( x 1 , x 2 ) = f 1 | 2 ( x 1 | x 2 ) f 2 ( x 2 ) = 10 x1 x 2 ,
0 < x 1 < x 2 < 1.
(c)
1 1 5 P < X1 < X2 = 4 2 8
|
12
=
14
2 x1 (5 8)
2
64 dx1 = 25
1 2
2
1 - 4
2
=
64 3 12 = . 25 16 25
12 12
(d)
1 1 P < X1 < 4 2
=
14
1
x1
2 10 x1 x 2 5
dx 2 dx1 =
10 3 x1 1 - x1 dx1 3 14
(
)
=
2 10 x1 x1 - 3 2 5
12 1 2 5 12 = 5 x1 - 2 x1 14 14 3
=
(
)
=
1 5 1 5 1 - - + 3 4 16 16 512
1 449 449 = . 3 512 1536
________________________________________________________________________ ________________________________________________________________________
If you are registered for 4 credit hours: 2.1.16
P( 2 X + 3 Y < 1 )
12
( to be handed in separately )
=
0 12
(1 - 2 x )
0
6 ( 1 - x - y ) dy dx
2
3
=
0
(1 - 2 x ) 2 ( 1 - x )( 1 - 2 x ) -
3 5 14 8 - x + x 2 dx 3 3 3
dx
12
= =
0
=
5 7 1 13 - + = . 6 12 9 36
OR
12 0
P( 2 X + 3 Y < 1 ) =
(1 - 2 x )
0
6 ( 1 - x - y ) dy dx
3
let u = 1 x y, then d u = d y, 0 1 x, ( 1 2 x ) / 3 ( 2 x ) / 3
12
=
0 12
-
(2 - x)
6 u du dx = 1- x 0
3
12
3 (1 - x ) 2 -
( 2 - x ) 2 dx
3
=
0
5 14 8 5 7 1 13 - x + x 2 dx = - + = . 3 3 3 6 12 9 36
E( X 2 ) =
1 0
1- x 0
x 2 6 ( 1 - x - y ) dy dx =
1 0
1 0
( 6 x 2 (1 - x ) 2 - 3 x 2 (1 - x ) 2 )dx
= 1-
=
1 0
3 x ( 1 - x ) 2 dx
2
=
( 3 x 2 - 6 x 3 + 3 x 4 )dx
=
1 0
6 3 + = 0.10. 4 5
E( X Y ) =
1 0
1- x 0
x y 6 ( 1 - x - y ) dy dx
1 0
( 3 x (1 - x ) 3 - 2 x (1 - x ) 3 )dx
=
=
1 0
x ( 1 - x ) 3 dx =
( x - 3 x 2 + 3 x 3 - x 4 )dx
1 3 1 - 1 + - = 0.05. 2 4 5
E ( X Y + 2 X 2 ) = E ( X Y ) + 2 E ( X 2 ) = 0.25.