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Hence,
+
∞
exp
(
x
(
x
µ
1
)
2
2
σ
1
µ
1
)
2
2
σ
2
dx
=
σ
1
2
σ
2
1
σ
1
√
2
π
−
−
−∞
+
∞
exp
(
x
(
µ
1
−
µ
1
)
2
2
σ
1
µ
2
)
2
2
σ
2
µ
2
)
2
2
σ
2
1
σ
1
√
2
π
−
dx
=
(
µ
1
−
−∞
exp
(
x
(
x
+
∞
1
σ
1
√
2
π
−
µ
1
)
2
2
σ
1
−
µ
1
)(
µ
1
−
µ
2
)
dx
=0
.
2
σ
2
−∞
Finally, one gets
D
(
p
1
,p
2
)=ln
σ
2
σ
1
µ
2
)
2
2
σ
2
2
+
σ
1
1
+
(
µ
1
−
−
.
2
σ
2
Then
∆
can be computed as
µ
2
)
2
4
σ
2
µ
2
)
2
4
σ
1
2
+
σ
1
+
σ
2
4
σ
1
1
+
(
µ
1
−
+
(
µ
1
−
∆=
−
4
σ
2
µ
2
)
2
(
σ
1
+
σ
2
)
2
σ
1
σ
2
+
σ
1
+
σ
2
+(
µ
1
−
=
−
4
σ
1
σ
2
=
σ
1
−
σ
2
+(
µ
1
−
µ
2
)
2
σ
1
+
σ
2
4
σ
1
σ
2
.
2.10.8 Computation of the Leverages
Many discussions of the computation of leverages can be found in the litera-
ture. The present one is from [Monari 1999].
Z
is an (
N
,
q
) matrix, with
N
,z
N
]
T
. The leverages
to be computed are the diagonal terms of the orthogonal projection matrix
H
=
Z
(
Z
T
Z
)
−
1
Z
,
q
:
Z
=[
z
1
,
≥
···
h
kk
=
z
k
T
(
Z
T
Z
)
−
1
z
k
.
As diagonal elements of an orthogonal projection matrix, the terms
{
h
kk
}
k
=1
,...,N
are defined only if
Z
has full rank, i.e., if
Z
T
Z
is invertible. If it is, the following
relations are valid:
1
,
trace(
H
)=
N
∀
k
∈
[1
,...,N
]
,
0
≤
h
kk
≤
h
kk
=rank(
Z
)
.
k
=1
A first leverage computation technique consists in computing matrix
Z
T
Z
,
inverting it with a classical method (Cholesky, LU decomposition, etc.), then
in left and right multiplying by
z
k
and
z
kT
. That method is satisfactory only
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