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3. If
J
(
w
(
i
))
<J
(
w
(
i
−
1)), then retrieve
w
(
i
−
1), divide
µ
i
by
r
and go to
step 2.
4. Otherwise multiply
µ
i
by
r
. Iterate until a suitable value of
µ
i
is found.
2.10.7 Kullback-Leibler Divergence Between two Gaussians
The expression of the Kullback-Leiibbler divergence between two Gaussians
with mean and standard deviation (
µ
1
,
σ
1
)and(
µ
2
,
σ
2
) respectively is derived.
The following relations are useful:
exp
(
x
d
x
=1
+
∞
1
σ
√
2
π
µ
)
2
2
α
2
−
−∞
x
exp
(
x
d
x
=
µ
+
∞
µ
)
2
2
α
2
1
σ
√
2
π
−
−∞
+
∞
µ
)exp
(
x
d
x
=
σ
2
.
µ
)
2
2
α
2
1
σ
√
2
π
−
(
x
−
−∞
The Kullback-Leibler divergence is defined as
D
(
p
1
,p
2
)=
+
∞
−∞
p
1
(
x
)ln
p
1
(
x
)
p
2
(
x
)
d
x.
Because that definition is not symmetrical with respect to the two distri-
butions, the following quantity is preferred:
∆=[
D
(
p
1
,p
2
)+
D
(
p
2
,p
1
)]
/
2
+
∞
exp
(
x
−
µ
1
)
2
2
σ
1
1
σ
1
√
2
π
D
(
p
1
,p
2
)=
−∞
ln
σ
2
d
x
µ
1
)
2
2
σ
1
µ
2
)
2
2
σ
2
(
x
−
+
(
x
−
×
σ
1
−
+
∞
exp
(
x
ln
σ
2
µ
1
)
2
2
σ
1
1
σ
1
√
2
π
−
=
σ
1
d
x
−∞
+
∞
exp
(
x
(
x
µ
1
)
2
2
σ
1
µ
1
)
2
2
σ
1
−
−
−
d
x
−∞
exp
(
x
(
x
d
x
+
+
∞
−∞
µ
1
)
2
2
σ
1
µ
2
)
2
2
σ
2
−
−
The first two terms are equal to ln(
σ
2
/σ
1
)
−
(1
/
2).
For the third term, one writes
µ
2
)
2
=(
x
µ
2
)
2
(
x
−
−
µ
1
+
µ
1
−
µ
1
)
2
+(
µ
1
−
µ
2
)
2
+2(
x
=(
x
−
−
µ
1
)(
µ
1
−
µ
2
)
.
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