Hw 8 (due Apr. 15, Thursday)
5.20
[Sol]
(a)
y2
p(y2 )
(b) P (Y1 = 3 | Y2 = 1) =
P (Y1 =3,Y2 =1)
P (Y2 =1)
-1
1/8
=
1/8
4/8
1
4/8
2
2/8
3
1/8
= 1/4.
5.26
!∞
!1
(a) By definition, f1 (y1 ) = −∞ f (y1 , y2 )dy2 = 0 4y1 y2 dy2 = 2y1 for 0 ≤ y1 ≤ 1 and
!∞
!1
f2 (y2 ) = −∞ f (y1 , y2 )dy1 = 0 4y1 y2 dy1 = 2y2 for 0 ≤ y2 ≤ 1
(b) P (Y1 ≤ 1/2 | Y2 > 3/4) =
7/64
7/16
P (Y1 ≤1/2,Y2 >3/4)
P (Y2 >3/4)
=
= 1/4.
R 1/2 R 1
0
3/4 f (y1 ,y2 )dy1 dy2
R1
3/4 f2 (y2 )dy2
=
R 1/2 R 1
0
3/4
4y1 y2 dy1 dy2
3/4
2y2 dy2
R1
=
1 ,y2 )
(c) By definition of conditional density function, f (y1 | y2 ) = ff(y2 (y
= 4y2y1 y2 2 = 2y1 for
2)
0 ≤ y1 ≤ 1.
1 ,y2 )
(d) By definition of conditional density function, f (y2 | y1 ) = ff(y1 (y
= 4y2y1 y1 2 = 2y2 for
1)
0 ≤ y2 ≤ 1.
! 3/4
! 3/4
(e) P (Y1 ≤ 3/4 | Y2 = 1/2) = 0 f (y1 | y2 = 1/2)dy1 = 0 2y1 dy1 = 9/16.
5.46
[Sol] No. Y1 and Y2 are not independent. Consider P (Y1 = 3, Y2 = 1) and P (Y1 = 3)P (Y2 =
1) : p(3, 1) = 1/8 "= (1/8)(4/8) = p1 (3)p2 (0).
5.53
[Sol] No. First, the range of Y1 and! Y2 are related to each other.
! y2 Second, f (y1 , y2 ) =
1
6(1 − y2 ) "= f (y1 )f (y2 ) where f (y1 ) = y1 6(1 − y2 )dy2 and f (y2 ) = 0 6(1 − y2 )dy1 .
5.72
[Sol]
(a) E(Y1 ) = np = 2/3
(b) V (Y1 ) = np(1 − p) = 4/9
(c) E(Y1 − Y2 ) = E(Y1 ) − E(Y2 ) = 2/3 − 2/3 = 0.
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5.77
!1
!y
[Sol] f (y1 ) = y1 6(1 − y2 )dy2 = 3(1 − y1 )2 for 0 ≤ y1 ≤ 1 and f (y2 ) = 0 2 6(1 − y2 )dy1 =
6y2 (1 − y2 ) for!0 ≤ y2 ≤ 1.
!1
!1
1
(a) E(Y1 ) = 0 y1 f1 (y1 )dy1 = 0 y1 3(1 − y1 )2 dy1 = 1/4 and E(Y2 ) = 0 y2 f2 (y2 )dy2 =
!1
y 6y2 (1 − y2 )dy2 = 1/2 .
0 2
!1
!1
(b) V (Y1 ) = 0 y12 f1 (y1 )dy1 − E(Y1 )2 = 0 y12 3(1 − y1 )2 dy1 − (1/4)2 = 1/10 − 1/16 = 3/80
!1
!1
and V (Y2 ) = 0 y22 f2 (y2 )dy2 − E(Y2 )2 = 0 y22 6y2 (1 − y2 )dy2 − (1/2)2 = 3/10 − 1/4 = 1/20.
(c) E(Y1 − 3Y2 ) = E(Y1 ) − 3E(Y2 ) = 1/4 − 3/2 = −5/4.
5.92
!1!y
!1!y
[Sol] Cov(Y1 , Y2 ) = E(Y1 Y2 ) − E(Y1 )E(Y2 ) = 0 0 2 6y1 y2 (1 − y2 )dy1 dy2 − 0 0 2 6y1 (1 −
!1!y
y2 )dy1 dy2 × 0 0 2 6y2 (1 − y2 )dy1 dy2 = 3/20 − (1/4) ∗ (1/2) = 1/40. Since Cov(Y1 , Y2 ) "= 0,
Y1 and Y2 are not independent.
5.95
[Sol] The marginal distribution for Y1 and Y2 are shown in the following tables : Since
y1
p(y1 )
-1
1/3
0
1/3
1
1/3
y2
p(y2 )
0
2/3
1
1/3
p(−1, 0) "= p(−1)p(0), Y1 and Y2 are not independent. However, E(Y1 ) = −1(1/3) + 0(1/3) +
1(1/3) = 0, E(Y2 ) = 0(2/3) + 1(1/3) = 1/3, E(Y1 Y2 ) = (−1)(0)(1/3) + (0)(1)(1/3) +
(1)(0)(1/3) = 0 so that Cov(Y1 , Y2 ) = 0 − 0 ∗ 1/3 = 0.
5.103
[Sol] Use Theorem 5.12. E(3Y1 +4Y2 −6Y3 ) = 3(2)+4(−1)−6(4) = −22, V (3Y1 +4Y2 −6Y3 ) =
9(4) + 16(6) + 36(8) + (2)(3)(4)(1) + (2)(3)(−6)(1) + (2)(4)(−6)(0) = 480.
5.106
[Sol] V (Y1 − 3Y2 ) = V (Y1 ) + V (Y2 ) − 3Cov(Y1 , Y2 ) = 3/80 + 9(1/20) − 6(1/40) = 27/80.
Example 5.27
[Sol] see text book for Example 5.27
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