Graphics Programs Reference
In-Depth Information
3
∂
EX
1
X
2
X
3
[
]
=
C
X
ω
1
ω
2
ω
3
(
,
,
)
at
ω
=
0
(13.89)
∂
ω
1
ω∂
3
∂
Example:
The vector
X
is a 4-variate Gaussian with
t
µ
x
=
2
110
6321
3432
2343
123
3
C
x
=
Define
X
1
X
2
X
3
X
4
X
1
=
X
2
=
Find the distribution of
X
1
and the distribution of
2
X
1
Y
=
X
1
+
2
X
2
X
3
+
X
4
Solution:
X
1
has a bivariate Gaussian distribution with
2
1
63
3
4
µ
x
1
=
C
x
1
=
The vector
Y
can be expressed as
X
1
X
2
X
3
X
4
2000
1200
001
1
Y
=
=
AX
It follows that
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