We first test the three cases simulated in [4] in which the numerical analysis approach was dealt with. Namely on the interval [0,4], we consider the cases where g≡1 and first B =B1 ≡1,second B(x) =B2(x) = 1 for x≤1.5, then linear toB2(x) = 5 for x ≥1.7.
This particular form is interesting because due to this fast increase on B, the solution N is not that regular and exhibits a 2-peaks distribution (see Figure 1). Finally, we test B(x) =B3(x) = exp(−8(x−2)2) + 1.
In Figures 1 and 2, we show the simulation results with n = 5.104 (a realistic value forin vitroexperiments on E. Colifor instance) for the reconstruction ofN, ∂x∂ (gN), BN and B.
One notes that the solution can well capture the global behavior of the division rateB, but, as expected, has more difficulties in recovering fine details (for instance, the difference betweenB1 and B3) and also gives much more error when B is less regular (case ofB2).
One also notes that even if the reconstruction ofN is very satisfactory, the critical point is the reconstruction of its derivative. Moreover, for large values of x, even if N and its derivative are correctly reconstructed, the method fails in finding a proper division rateB.
This is due to two facts: first,N vanishes, so the division byN leads to error amplification.
0 1 2 3 4 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1 true N with B=1
reconstructed N with B=1 true N with B=B
2
reconstructed N with B=B2 true N with B=B
3 reconstructed N with B=B
3
0 1 2 3 4
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
3 true (gN)' with B=1
reconstructed (gN)' with B=1 true (gN)' with B=B
2
reconstructed (gN)' with B=B2 true (gN)' with B=B
3 reconstructed (gN)' with B=B
3
Figure 1: Reconstruction of N (left) and of ∂x∂ (gN) (right) obtained with a sample of n= 5.104 data, for three different cases of division rates B.
Second, the values taken byB(x) for largexhave little influence on the solutionsN of the direct problem: whatever the values of B, the solutions N will not vary much, as shown by Figure 1 (left). A similar phenomenon occurred indeed when solving the deterministic problem in [4] (for instance, we refer to Fig. 10 of this article for a comparison of the results).
We also test a case closer to biological true data, namely the case B(x) = x2 and g(x) = x. The results are shown on Figures 3 and 4 forn-samples of size 103, 5.103,104 and 5.104.
One notes that reconstruction is already very good for N when n = 103, unlike the reconstruction of ∂x∂ (gN) that requires much more data.
Finally, in Table 3.2 we give average error results on 50 simulations, for n = 1000, g ≡ B ≡ 1. We display the relative errors in L2 norms, (defined by ||φ−φ||ˆ L2/||φ||L2), and their empirical variances. In Table 3.2, for the case g(x) = x and B(x) = x2, we give some results on standard errors for various values of n, and compare them to n−1/5, which is the order of magnitude of the expected final error on BN, since with a Gaussian kernel we have s= 1 in Proposition 1. We see that our numerical results are in line with the theoretical estimates: indeed, the error onHis roughly twice as large asn−1/5.
0 1 2 3 4 -0.2
0 0.2 0.4 0.6 0.8 1
1.2 true BN with B=1
reconstructed BN with B=1 true BN with B=B
2
reconstructed BN with B=B2 true BN with B=B
3 reconstructed BN with B=B
3
0.5 1 1.5 2
0 5 10
15 true B=1
reconstructed B=1 true B2
reconstructed B2 true B3
reconstructed B3
Figure 2: Reconstruction ofBN (left) and ofB(right) obtained with a sample ofn= 5.104 data, for three different cases of division rates B.
Error onN: average Variance Error on ∂x∂ (gN) : average Variance
0.088 0.15 0.51 0.28
Error on BN: average Variance Average ˆh Average ˜h
0.39 0.29 0.12 0.40
n n−15 ˆh ˜h error on N error onD error on H
103 0.25 0.1 0.5 0.06 0.68 0.42
5.103 0.18 0.07 0.3 0.03 0.45 0.28
104 0.16 0.08 0.3 0.035 0.46 0.29
5.104 0.11 0.04 0.2 0.014 0.31 0.19
4 Proofs
In Section 4.1, we first give the proofs of the main results of Section 2. This allows us, in Section 4.2, to prove the results of Section 2.5, which require the collection of all the results of Section 2, i.e. the oracle-type inequalities on the one hand and a numerical analysis result on the other hand. This illustrates the subject of our paper that lies at the frontier between these fields. Finally, we state and prove the technical lemmas used in Section 4.1.
These technical tools are concerned with probabilistic results, namely concentration and Rosenthal-type inequalities that are often the main bricks to establish oracle inequalities,
0 1 2 3 4 0
0.2 0.4 0.6 0.8 1 1.2 1.4
true N n=1 000 n=5 000 n=10 000 n=50 000
0 1 2 3 4
-1 -0.5 0 0.5 1 1.5 2
true (gN)' n=1 000 n=5 000 n=10 000 n=50 000
Figure 3: Reconstruction of N (left) and of ∂x∂ (gN) (right) obtained for g(x) = x and B(x) =x2,for various sample sizes.
and also the boundedness of L−1k . In the sequel, the notation θ1,θ2,... denotes a generic positive constant depending onθ1, θ2, . . .(the notationsimply denotes a generic positive absolute constant). It means that the values ofθ1,θ2,... may change from line to line.
4.1 Proofs of the main results of Section 2 Proof of Proposition 2
For anyh∗∈ H, we have:
kNˆ −Nk2 ≤ kNˆˆh−Nˆh,h∗k2+kNˆh,h∗−Nˆh∗k2+kNˆh∗−Nk2
≤ A1+A2+A3, with
A1 :=kNˆˆh−Nˆh,h∗k2 ≤A(h∗) + χ p
nˆh kKk2, A2 :=kNˆh,h∗−Nˆh∗k2 ≤A(ˆh) + χ
√
nh∗kKk2 and
A3 :=kNˆh∗−Nk2.
0 1 2 3 4
Figure 4: Reconstruction ofBN (left) and ofB (right) obtained forg(x) =x andB(x) = x2,for various sample sizes.
We obtain
we derive
kE( ˆNh∗,h0)−E( ˆNh0)k2 ≤ kKk1kEh∗k2, (4.3) where
Eh∗(x) := (Kh∗? N)(x)−N(x), x∈R+
represents the approximation term. Combining (4.1), (4.2) and (4.3) entails kNˆ −Nk2 ≤ kNˆh∗−Nk2+ 2kKk1kEh∗k2+ 2χ
Since the first term can be bounded as follows
E combined with the Cauchy-Schwarz inequality:
E It remains to deal with the term E(ξn2q). By Lemma 1 below, we obtain
E[ξn2q]≤q,η,δ and the conclusion follows.
Proof of Proposition 3
The proof is similar to the previous one and we avoid most of the computations for simplicity. For anyh0∈ H,˜
kDˆ˜h−Dk2 ≤ kDˆ˜h−Dˆ˜h,h
0k2+kDˆ˜h,h
0−Dˆh0k2+kDˆh0 −Dk2
≤ A˜1+ ˜A2+ ˜A3,
with
A˜1 :=kDˆ˜h−Dˆ˜h,h
0k2 ≤A(h˜ 0) + χ˜ p
nh˜3
kgk∞kK0k2, A˜2:=kDˆ˜h,h
0 −Dˆh0k2 ≤A(˜˜ h) + χ˜
pnh30kgk∞kK0k2 and
A˜3 :=kDˆh0−Dk2. Then,
kDˆ˜h−Dk2 ≤2 ˜A(h0) + 2 ˜χ
pnh30kgk∞kK0k2+kDˆh0 −Dk2. To study ˜A(h0), we first evaluate
E[ ˆDh1,h2(x)]−E[ ˆDh2(x)]. = (Kh1? Kh2 ?(gN)0)(x)−(Kh2 ?(gN)0)(x)
= Z
D(t)(Kh1 ? Kh2)(x−t)dt− Z
D(t)Kh2(x−t)dt
= Z
D(t) Z
Kh1(x−t−u)Kh2(u)dudt− Z
D(t)Kh2(x−t)dt
= Z
D(t) Z
Kh1(v−t)Kh2(x−v)dvdt− Z
D(v)Kh2(x−v)dv
= Z
Kh2(x−v)Z
D(t)Kh1(v−t)dt−D(v) dv
= (Kh2?E˜h1)(x), where we set, for any real number x
E˜h1(x) :=
Z
D(t)Kh1(x−t)dt−D(x)
= (Kh1 ? D)(x)−D(x). (4.4) It follows that
A(h˜ 0) = sup
h∈H˜
kDˆh0,h−Dˆhk2− χ˜
√
nh3kgk∞kK0k2 +
≤ sup
h∈H˜
kDˆh0,h−E[ ˆDh0,h]− Dˆh−E[ ˆDh]
k2− χ˜
√
nh3kgk∞kK0k2
+
+kE[ ˆDh0,h]−E[ ˆDh]k2
≤ sup
h∈H˜
kDˆh0,h−E[ ˆDh0,h]−( ˆDh−E[ ˆDh])k2− χ˜
√
nh3kgk∞kK0k2 ++kKk1kE˜h0k2
≤ (1 +kKk1) sup
h∈H˜
kDˆh−E[ ˆDh]k2−(1 + ˜ε)
√
nh3 kgk∞kK0k2
++kKk1kE˜h0k2, (4.5)
In order to obtain the last line, we use the following chain of arguments:
Dˆh0,h(x) = 1 n
n
X
i=1
g(Xi) Z
Kh0(x−Xi−t)Kh0(t)dt
= Z
Kh0(t) 1
n
n
X
i=1
g(Xi)Kh0(x−Xi−t)
dt and
EDˆh0,h(x)
= Z
Kh0(t)Z
g(u)Kh0(x−u−t)N(u)du dt, therefore
Dˆh0,h(x)−EDˆh0,h(x)
= Z
Kh0(t)G(x−t)dt=Kh0 ? G(x), with
G(x) = 1 n
n
X
i=1
g(Xi)Kh0(x−Xi)− Z
g(u)Kh0(x−u)N(u)du
= Dˆh(x)−EDˆh(x) . Therefore
kDˆh0,h−E[ ˆDh0,h]k2 ≤ kKh0k1kGk2
≤ kKk1kDˆh−E[ ˆDh]k2,
which justifies (4.5). In the same way as in the proof of Proposition 2, we can establish the following:
E
kE˜h0k2q2
=E
kDˆh0 −Dk2q2
≤q
kE˜h0k2q2 + g
∞kK0k2 pnh30
2q .
Finally, we successively apply (4.4), (4.5) and Lemma 1 in order to conclude the proof.
Proof of Proposition 4
We use the notation and definitions of Section 2.4. We have kL−1k (ϕ)− L−1(ϕ)k22,T =
k−1
X
i=0 xi+1,k
Z
xi,k
Hi,k − L−1(ϕ)(x)2
dx:=
k−1
X
i=0
Li,k.
We prove by induction that for all i, one has Li,k ≤ C2Tk22kϕk2W1. The result follows by summation over i.We first prove the two following estimates:
xi+1,k size Tk. Thus (4.7) is simply Wirtinger inequality applied to ϕ ∈ W1 on the interval [xi,k, xi+1,k].F or(4.7), we use the Cauchy-Schwarz inequality:
|ϕi+1,k−ϕi,k|2 = k2 We are ready to prove by induction the two following inequalities:
Li,k ≤C12Tk22kϕk2W1, (4.8)
|Hi+1,k−Hi,k|2≤C22Tkkϕk2W1. (4.9) for two constantsC1 andC2 specified later on. First, fori= 0,we have
L0,k =
This proves the first induction assumption fori= 0,and
proves the second one. Let us now suppose that the two induction assumptions are true for all j≤i−1,and takei≥1.Let us first evaluate
We distinguish the case wheniis even and wheniis odd. Letibe even: then, by definition Li,k ≤ 1
Hence, re-organizing terms, we can write
Li,k ≤ 1
Putting together the four inequalities above (the estimates for ϕ and the induction as-sumptions), we obtain
and (4.8) is proved. It remains to establishe (4.9). Let us write it for ieven (the case of an oddiis similar):
|Hi+1,k−Hi,k|2 = 1
Hence, as previously, we obtain
|Hi+1,k−Hi,k|2 ≤ 1 16
T
kkϕk2W1(C22+ 1).
To complete the proof, we remark thatC22 = 151 andC12 = 4π12+18(1 +151)< 16 are suitable.
It is consequently sufficient to takeC =C1.
4.2 Proof of Theorem 1 and Proposition 1 Proof of Theorem 1
It is easy to see that
kHb −Hk2,T = kL−1k (ˆκDˆ + ˆλnNˆ)− L−1(L(BN))k2,T
≤ kL−1k (ˆκDˆ + ˆλnNˆ)− L−1k (L(BN))k2,T
+kL−1k (L(BN))− L−1(L(BN))k2,T
≤ kL−1k (ˆκDˆ + ˆλnNˆ −(κD+λN))k2,T +1
3
√T
kkL(BN)kW1,
thanks to Proposition 4. Note that L(BN) =κ(gN)0+λN so that we can write kL(BN)kW1 ≤C(kNkW1 +kgNkW2).
We obtain, thanks to Lemma 4 that gives the boundedness of the operator L−1k : kHb −Hk2,T ≤
ˆκnDˆ −κD 2,T +
λˆnNˆ −λN
2,T + (kNkW1+kgNkW2) T
√k
≤ |ˆκn|kDˆ−Dk2+|ˆλn|kNˆ −Nk2+|ˆκn−κ|kDk2+|λˆn−λ|kNk2 + (kNkW1+kgNkW2) T
√ k
≤ |λˆn||ρˆn|kDˆ −Dk2+|ˆλn| kNˆ −Nk2+|ˆρn−ρg(N)|kDk2 + kNk2+|ρg(N)|kDk2
|ˆλn−λ|+ (kNkW1 +kgNkW2) T
√ k
. Taking expectation and using Cauchy-Schwarz inequality, we obtain for anyq ≥1,
E[kHb −Hkq2,T] ≤ qh
E[ˆλ2qn]1/2n
E[ ˆρ4qn ]1/4
E[kDˆ −Dk4q2 ]1/4
+ E[kNˆ−Nk2q2 ]1/2
+kDkq2 E[|ˆρn−ρg(N)|2q]1/2 + kNk2+ρg(N)
D 2
q
E[|λˆn−λ|q] + (kNkW1+kgNkW2) T
√ k
q .
Now, Lemma 3 gives the behaviour of E[|ˆρn−ρg(N)|2q]. In particular, we obtain E[ ˆρ4qn ]≤q,g,N,c.
We finally apply successively Propositions 2 and 3 to obtain the proof of Theorem 1.
Proof of Proposition 1
We have already proved (1.9). It remains to prove (1.10). We introduce the event Ωn={2 ˆN(x)≥mfor any x∈[a, b]}.
The first term of the right hand side is handled by (1.9) and Proposition 2. The second term is handled by Lemma 2 that establishes thatP(Ωcn) =O(n−q).
4.3 Technical lemmas Concentration inequalities
We first state the following concentration result. Note that a more general version of this result can be found in [13]. We nevertheless give a proof for the sake of completeness.
Lemma 1. We have the following estimates
• Assume that
∞ are finite. For every q > 0, introduce the grid H ⊂ {1,1/2, ...,1/Dmax} and Dmax = δn for some δ > 0. Then, for every
• Assume that kK0k2,kK0k1,kgk∞ and kNk∞ are finite. For every q > 0, introduce the grid H ⊂ {1,˜ 1/2, ...,1/D˜max} andD˜max=p
δn˜ for some δ >˜ 0. Then for every η >0
E sup
h∈H˜
kDˆh−E[ ˆDh]k2− (1 +η)
√
nh3 kgk∞kK0k2 2q+
≤q,η,δ,kK˜ 0k2,kK0k1,kgk∞,kNk∞n−q. Proof.Let X be a real random variable and let us consider the random process
∀t∈R, w(t, X) =ϕ(X)Ψ(t−X),
where ϕ and Ψ are measurable real-valued functions. Let X1, ..., Xn be n independent and identically distributed random variables with the same distribution as X and let us consider the process
∀t∈R, ξϕ,Ψ(t) =
n
X
i=1
w(t, Xi)−E
w(t, X) . First, let us study the behavior of
ξϕ,Ψ
2 =Z
ξϕ,Ψ2 (t)dt1/2
.
IfB denotes the unit ball in L2 and A is a countable subset ofB, we have
ξϕ,Ψ
2 = sup
a∈B
Z
a(t)ξϕ,Ψ(t)dt
= sup
a∈A
Z
a(t)ξϕ,Ψ(t)dt
= sup
a∈A n
X
i=1
Z
a(t) w(t, Xi)−E
w(t, X) dt.
Hence one can apply Talagrand’s inequality (see the version of Bousquet in the independent and identically distributed case [16, p 170]). For allε, x >0, one has:
P ξϕ,Ψ
2 ≥(1 +ε)E[ ξϕ,Ψ
2] +√
2vx+c(ε)bx
≤e−x, wherec(ε) = 1/3 +ε−1,
v=nsup
a∈AE Z
a(t)
w(t, X)−E w(t, X) dt2
, and
b= sup
y∈R,a∈A
Z a(t)
w(t, y)−E w(t, X) dt.
We study each term of the right hand term within the expectation.
• Obviously, one has:
E
kξϕ,Ψk2
≤ E
Z Xn
i=1
w(t, Xi)−E
w(t, X)2
dt1/2
=
Z n X
i=1
E
w(t, Xi)−E
w(t, X)2 dt
1/2
≤ n
Z E
w(t, X)2 dt
1/2
. But we easily see that for ally,
Z
w2(t, y)dt≤ kϕk2∞kΨk22, (4.10) hence
E[ ξϕ,Ψ
2]≤√ n
ϕ ∞
Ψ 2.
• Since v is a supremum of variance terms, v ≤ nsup
a∈AE Z
a(t)w(t, X)dt2
≤ nsup
a∈AE Z
|w(t, X)|dt Z
a2(t)|w(t, X)|dt
≤ nsup
y∈R
Z
|w(t, y)|dt×sup
t∈R
E[|w(t, X)|]
≤ nkϕk2∞kNk∞kΨk21.
• The Cauchy-Schwarz inequality and (4.10) give b = sup
y∈R
kw(•, y)−E[w(., X)]k2
≤ sup
y∈R
kw(•, y)k2+ E Z
w(t, X)2dt1/2
≤ 2kϕk∞kΨk2. The main point here is that√
vmay be much smaller thanE[kξϕ,Ψk2]. So, for allε, x >0, P kξϕ,Ψk2 ≥(1 +ε)√
nkϕk∞kΨk2+kϕk∞kNk1/2∞ kΨk1√
2nx+ 2c(ε)kϕk∞kΨk2x
≤e−x.
Now we consider a familyMof possible functionsϕand Ψ. Let us introduce some strictly positive weights Lϕ,Ψ and let us apply the previous inequality to x=Lϕ,Ψ+u foru >0.
Hence with probability larger than 1−P
(ϕ,Ψ)∈Me−Lϕ,Ψe−u, for all (ϕ,Ψ)∈ M, one has
Then, let us takeu such that
x=f(u)2q := kϕk∞kNk1/2∞ kΨk1√
Finally, we have proved that Rq≤q,ε
Now let us evaluate what this inequality means for each set-up.
• First, when ϕ= 1/nand Ψ = 1/hK(•/h), the family Mcorresponds to the family H. In that case Mϕ,Ψ and Lϕ,Ψ will respectively be denoted by Mh and Lh. The upper bound given in (4.12) becomes
Rq≤q,ε
X
h∈H
e−Lh
n−qkNkq∞kKk2q1 +n−2qh−qkKk2q2
. (4.13)
Now it remains to choose Lh. But Mh = (1 +ε) 1
√
nhkKk2+kNk1/2∞ kKk1 r2Lh
n + 2c(ε) 1 n√
hkKk2Lh. Letθ >0 and let
Lh= θ2kKk22 2kNk∞kKk21√
h.
Obviously the series in (4.13) is finite and for anyh∈ H, sinceh≤1, we have:
Mh≤(1 +ε+θ)kKk2
√
nh + c(ε)θ2kKk32 kNk∞kKk21
1 nh. Since Dmax=δn, one obtains that
Mh ≤ 1 +ε+θ+ c(ε)θ2kKk32√ δ kNk∞kKk21
kKk2
√ nh.
It remains to choose ε=η/2 andθ small enough such that θ+ c(ε)θ2kKk32
√ δ
kNk∞kKk21 =η/2 to obtain the desired inequality.
• Secondly, ifϕ=g/n and Ψ = 1/h2K0(•/h)the family Mcorresponds to the family H. So,˜ Mϕ,Ψ andLϕ,Ψwill be denoted byMh0 andL0h respectively. The upper bound given by (4.12) now becomes
Rq≤q,ε
X
h∈H˜
e−L0h
n−qh−2qkNkq∞kgk2q∞kK0k2q1 +n−2qh−3qkgk2q∞kK0k2q2
. (4.14)
But
Mh0 = (1+ε) 1
√nh3/2kgk∞kK0k2+kgk∞kNk1/2∞ kK0k1 r2L0h
nh2+2c(ε)kgk∞kK0k2 L0h nh3/2. Letθ >0 and let
L0h = θ2kK0k22 2kNk∞kK0k21h.
Obviously the series in (4.14) is finite and we have:
Mh0 ≤(1 +ε+θ)kgk∞kK0k2
√
nh3 +c(ε)θ2kK0k32kgk∞ kNk∞kK0k21
1 nh5/2. Buth2≥(˜δn)−1. Hence
Mh0 ≤ 1 +ε+θ+c(ε)θ2kK0k32kgk∞
pδ˜ kNk∞kK0k21
kK0k2
√ nh3. As previously, it remains to chooseεand θaccordingly to conclude.
The second result is based on probabilistic arguments as well.
Lemma 2. Under Assumptions and notations of Proposition 1, if there exists an interval [a, b] in (0, T) such that
[m, M] := [ inf
x∈[a,b]N(x), sup
x∈[a,b]
N(x)]⊂(0,∞), Q:= sup
x∈[a,b]
|H(x)|<∞,
and if ln(n) ≤ Dmin ≤ n1/(2m0+1) and n1/5 ≤ Dmax ≤ (n/ln(n))1/(4+η) for some η > 0 fixed, then there exists Cη, a constant depending on η, such that for n large enough,
P(Ωcn)≤Cηn−q.
Proof.Let x, ybe two fixed point of [a, b] and for h∈ H0, first let us look at Zh(x, y) = ˆNh(x)−Nˆh(y)−[Kh? N(x)−Kh? N(y)].
One can apply Bernstein inequality toZh(x, y) (seee.g.(2.10) and (2.21) of [16]) to obtain:
E
eλZh(x,y)
≤exp λ2v2(x, y) 2(1−λc(x, y))
, ∀λ∈(0,1/c(x, y)), with
c(x, y) = 1 3nsup
u∈R
|Kh(x−u)−Kh(y−u)|
and
v2(x, y) = 1 n
Z
R
|Kh(x−u)−Kh(y−u)|2N(u)du.
But, withK the Lipschitz constant of K, c(x, y)≤ K
3nh2|x−y|
and
v2(x, y)≤ K2
nh4|x−y|2.
Letx0 be a fixed point of [a, b]. One can apply Theorem 2.1 of [2] which gives that, if Zh = sup
x∈[a,b]
|Zh(x, x0)|
then for all positive x,
P(Zh ≥18(p
v2(x+ 1) +c(x+ 1))≤2e−x, with
v2= K2
nh4|b−a|2 and
c= K
3nh2|b−a|.
But one can similarly prove, using Bernstein’s inequality, that P |Nˆh(x0)−Kh? N(x0)| ≥
rN,Kx nh +K
x nh
≤2e−x. Hence
P
∃x∈[a, b],|Nˆh(x)−Kh? N(x)| ≥
rN,Kx nh4 +K
x nh2
≤4e−x. We apply this inequality withx=qlog(n) +Dη,D=D0min, . . . , Dmax0 . We obtain
P
∃h∈ H0,∃x∈[a, b],|Nˆh(x)−Kh? N(x)| ≥
rN,K,qlog(n)
nh4+η +K,qlog(n) nh2+η
≤4
Dmax
X
D=1
e−Dηn−q =ηn−q.
Consequently, since nhlog(n)4+η →0 uniformly inh∈ H0, fornlarge enough, P(∃h∈ H0,∃x∈[a, b],|Nˆh(x)−Kh? N(x)| ≥m/4)≤ηn−q. But since N ∈ Ws+1,s≥1, then N0∈ Ws soN0 is bounded onR. Then
Kh? N(x)−N(x) = Z
Kh(x−y)(N(y)−N(x))dy
= Z
K x−y h
y−x h
Z 1
0
N0(x+u(y−x))dudy
= h
Z
tK t Z 1
0
N0(x−uht)dudt.
So,
|N(x)−Kh? N(x)| ≤N,Kh.
Because of the definition of H0 this term tends to 0, uniformly in x and h. Hence for n large enough, for all x∈[a, b] and allh∈ H0
Kh? N(x)≥N(x)−m/4≥3m/4.
Consequently, for nlarge enough,
P(∃h∈ H0,∃x∈[a, b],2 ˆNh(x)< m)≤ηn−q.
Rosenthal-type inequality
The following result studies the behavior of the moments of ˆρn. Lemma 3. For any p≥2,
E
|ˆρn−ρg(N)|p
≤p,g,N,cn−p/2. Proof.We have:
E
|ˆρn−ρg(N)|p
≤p(An+Bn), with
An:=E h
Pn
i=1Xi nR
g(x)N(x)dx+c− R
R+xN(x)dx R
R+g(x)N(x)dx
pi
and
Bn:=E h
Pn
i=1Xi Pn
i=1g(Xi) +c −
Pn
i=1Xi nR
g(x)N(x)dx+c
pi .
We use the Rosenthal inequality (seee.g.the textbook [7]): if R1, . . . , Rn are independent centered variables such thatE[|R1|p]<∞,withp≥2 then
E 1 n
n
X
i=1
Ri
p
≤pn−p nE|R1|p+ nER21p/2
. (4.15)
We recall that Assumption 1 ensures that E[Xip] = R
xpN(x)dx < ∞, for any p ≥ 1.
Hence, for the first termAn, using (4.15), we have:
An = E
Since g is positive, the Bernstein inequality (see Section 2.2.3 of [16]) gives:
P(Ωn)≥1−2n−γ. (4.16)
Therefore, we bound from above the term Bn by a constantp,g,N times
where we have used (4.15) for the first term and (4.16) for the second one. This concludes
the proof of the lemma.
The boundedness of L−1k
Lemma 4. For any function ϕ, we have:
L−1k (ϕ) At the second line, we have distinguished the i’s that are even and the i’s that are odd.
At the third line, we have used the inequalities (a+b)2≤2a2+ 2b2 and (a+b+c+d)2 ≤ The Cauchy-Schwarz inequality gives:
so that
T k
k−1
X
i=0
ϕ2j,k = T k
k−1
X
i=0
k2 T2
xi+1,k
Z
xi,k
ϕ(x)dx2
≤
T
Z
0
ϕ2(x)dx and finally we obtain the desired result:
Z
|L−1k (ϕ)(x)|2dx≤ 1 3
Z
ϕ2(x)dx.
Acknowledgment:We warmly thank Oleg Lepski for fruitful discussions that eventually led to the choice of his method for the purpose of this article. The research of M. Hoffmann is supported by the Agence Nationale de la Recherche, Grant No. ANR-08-BLAN-0220-01.
The research of P. Reynaud-Bouret and V. Rivoirard is partly supported by the Agence Nationale de la Recherche, Grant No. ANR-09-BLAN-0128 PARCIMONIE. The research of M. Doumic is partly supported by the Agence Nationale de la Recherche, Grant No.
ANR-09-BLAN-0218 TOPPAZ.
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