Statistics 351 Fall 2006 (Kozdron) Midterm #2 — Solutions
1. (a)
Recall that a square matrix is strictly positive definite if and only if the determinants
of all of its upper block diagonal matrices are strictly positive. Since
Λ =
2

2

2
3
we see that det(Λ
1
) = det(2) = 2
>
0 and det(Λ
2
) = det(Λ) = 6

4 = 2
>
0, and so
we conclude that Λ is, in fact, strictly positive definite.
1. (b)
Since
X
is multivariate normal, we conclude that its characteristic function is given by
ϕ
X
(
t
) = exp
it
1
+ 2
it
2

1
2
(2
t
2
1

4
t
1
t
2
+ 3
t
2
2
)
.
1. (c)
If
Y
1
=
X
1

2
X
2
and
Y
2
=
X
1
+
X
2
, then by Definition I,
Y
1
is normal with mean
E
(
Y
1
) =
E
(
X
1
)

2
E
(
X
2
) =

3 and variance var(
Y
1
) = var(
X
1
) + 4 var(
X
2
)

4 cov(
X
1
, X
2
) = 22, and
Y
2
is normal with mean
E
(
Y
2
) =
E
(
X
1
) +
E
(
X
2
) = 3 and
variance var(
Y
1
) = var(
X
1
) + var(
X
2
)

2 cov(
X
1
, X
2
) = 1. Since
cov(
Y
1
, Y
2
) = cov(
X
1

2
X
2
, X
1
+
X
2
) = var(
X
1
)

cov(
X
1
, X
2
)

2 var(
X
2
) =

2
,
we conclude
Y
= (
Y
1
, Y
2
)
∈
N

3
3
,
22

2

2
1
.
2.
By Definition I, we see that
X
1
and
X
1
+
X
2
are each normally distributed random
variables. Therefore, by Theorem V.7.1,
X
1
and
X
1
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 Fall '08
 MichaelKozdron
 Statistics, Normal Distribution, Probability, Probability theory, probability density function, X1, 01 J

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