Download Solution-Homework 2

Solution-Homework 2
4.3 Let X be N3 (µ, Σ) with µ0 = [−3, 1, 4] and


1 −2 0


Σ =  −2 5 0 
0
0 2
Which of the following random variables are independent? Explain.
(a) X1 and X2
(b) X2 and X3
(c) (X1 , X2 ) and X3
(d)
X1 +X2
2
and X3
(e) X2 and X2 − 52 X1 − X3
Solution:
(a) No, because the covariance between X1 and X2 is σ12 = −2 6= 0.
(b) Yes, because the covariance between X2 and X3 is σ23 = 0.
(c) Yes, because the covariance between (X1 , X2 ) and X3 is (σ13 , σ23 ) = (0, 0).
Ã
X1 +X2
2
X3
(d) Yes, since
!
Ã
=
Ã
so the covariance matrix of
Ã
Σ=
1
2
0
1
2
0
0 1
X1 +X2
2
X3
1
2
0
0 1
0
1
2
!
!
X
!
is



1
Ã
!
0
1 −2 0
1
0
  12


2
 −2 5 0   2 0  =
0 2
0
0 2
0 1
2
and cov( X1 +X
, X3 ) = 0.
2
Ã
(e) No, since
X2
X2 − 52 X1 − X3
Ã
so the covariance matrix of
Ã
Σ=
0
− 52
!
Ã
=
0
− 52
1 0
1 −1
X2
5
X2 − 2 X1 − X3
1 0
1 −1
!
!
!
is


X

Ã
!
0 − 25
1 −2 0
5 10



 −2 5 0   1 1  =
10 93
4
0
0 2
0 −1
and cov(X2 , X2 − 52 X1 − X3 ) = 10 6= 0.
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4.4 Let X be N3 (µ, Σ) with µ0 = [2, −3, 1] and


1 1 1


Σ= 1 3 2 
1 2 2
(a) Find the distribution of 3X1 − 2X2 + X3 .
"
(b) Relabel the variables if necessary, and find a 2×1 vector a such that X2 and X2
−a0
X1
X3
#
are independent.
Solution:
(a) Since 3X1 − 2X2 + X3 =
³
´
3 −2 1
X
so 3X1 − 2X2 + X3 also
 follows
 normal distribution with mean
2
³
´


µ = 3 −2 1  −3  = 13
1
and variance



1 1 1
3
³
´



σ = 3 −2 1  1 3 2   −2  = 9.
1 2 2
1


X"2
#


X1  can be presented by
(b) Let a0 = (a1 a2 ) then the covariance matrix of 
0
X2 − a
X3
Ã
3
−a1 + 3 − 2a2
−a1 + 3 − 2a2 a21 − 2a2 + 2a1 a2 + 3 − 4a2 + 2a22
"
!
.
#
X1
Independency require cov(X2 , X2 − a0
) = 3 − a1 − 2a2 = 0. Thus any a0 = (a1 , a2 )
X3
of the form a0 = (3 − 2a2 , a2 ) will meet the requirement. As an example, a0 = (1, 1).
4.5 Specify each of the following.
(b) The conditional distribution of X2 , given that X1 = x1 and X3 = x3 for the joint
distribution in Exercise 4.3.
(c) The conditional distribution of X3 , given that X1 = x1 and X2 = x2 for the joint
distribution in Exercise 4.4.
Solution:
(b) X2 |X1 = x1 , X3 = x3 is N (−2x1 − 5, 1).
(c) X3 |X1 = x1 , X2 = x2 is N ( 12 (x1 + x2 + 3), 12 ).
Add Question. Assume x ∼ Np (µ, Σ). Partition x, µ and Σ into
2
Ã
x=
x1
x2
!
Ã
, µ=
µ1
µ2
!
Ã
, Σ=
Σ11 Σ12
Σ21 Σ22
!
Prove x1 and x2 − Σ21 Σ−1
11 x1 are independent.
Proof: Since
−1 0
−1
cov(x1 , x2 − Σ21 Σ−1
11 x1 ) = cov(x1 , x2 ) − var(x1 )(Σ21 Σ11 ) = Σ12 − Σ11 Σ11 Σ12 = 0,
then we can conclude that x1 and x2 − Σ21 Σ−1
11 x1 are independent.
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