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(
h
decreases less than 1
/n
)the
f
n
estimate
2. If
h
→
n→∞
0 with
nh
n→∞
∞
→
verifies:
nhV
[
f
n
(
x
)] =
f
(
x
)
K
2
(
y
)
dy .
lim
n
(E.6)
→∞
For the Gaussian kernel: lim
n→∞
V
[
f
n
(
x
)] =
f
(
x
)
2
nh
√
π
.
f
n
is then
3. If
K
satisfies the conditions stated in 1 and 2, the estimate
MSE-consistent:
MSE
(
f
n
(
x
))
n→∞
→
0 (with
MSE
(
f
n
(
x
)) =
[(
f
n
(
x
)
E
−
f
(
x
))
2
]).
f
n
, for a density having
r
derivatives, verifies:
4. The MSE-consistent
∞
f
(
x
)
nh
MSE
(
f
n
(
x
)) =
K
2
(
y
)
dy
+
h
2
r
k
r
|
f
(
r
)
(
x
)
2
,
|
(E.7)
−∞
where
f
(
r
)
(
x
) is the derivative of order
r
and
k
r
is the
characteristic ex-
ponent
of the Fourier transform of
K
(
x
) (denoted
k
(
u
)) defined as:
1
−
k
(
u
)
k
r
= lim
u→
0
.
(E.8)
|
u
|
r
Any symmetric kernel such that
x
2
K
(
x
)
1
, has a nonzero finite
k
r
for
r
=2. In particular, for the Gaussian kernel
k
r
=1
/
2 for
r
=2.
∈L
f
n
, has the following
5. The optimal (smallest) MSE of the MSE-consistent
value at a given
x
∈
X
:
MSE
opt
(
f
n
(
x
)) =
(2
r
+1)
f
(
x
)
2
nr
∞
K
2
(
y
)
dy
2
r/
(2
r
+1)
k
r
f
(
r
)
(
x
)
2
r/
(2
r
+1)
.
−∞
(E.9)
Thus, the decrease of the optimal MSE with the sample size
n
is of order
n
−
2
r/
(2
r
+1)
.
6. The optimal IMSE (the integrated MSE for the whole
X
support) of the
MSE-consistent
f
n
, for a symmetric kernel such that
x
2
K
(
x
)
1
,is
∈ L
obtained with a bandwidth given by:
h
(
n
)=
n
−
1
/
5
α
(
K
)
β
(
f
)
,
(E.10)
with
α
(
K
)=
K
2
(
y
)
dy
1
/
5
and
β
(
f
)=
dy
−
1
/
5
f
(2)
(
y
)
2
(
y
2
K
(
y
)
d
y
)
2
.
(E.11)
