which gives the desired result.
Remark 5.1.11. Theorem 5.1.9 and theorem 5.1.10 generalise [BW11, theorem 1, p. 318].
Finally, note that consistency also carries over to noncontinuous piecewise polynomials, with only minor changes of the proof.
5.2 Asymptotic distribution
5.2.1. We now turn to the asymptotic distribution of the least squares estimator for continuous, piecewise polynomials. At first, note that the partial derivatives ∂tµ(t, τ) and ∂τiµ(t, τ), τ ∈ Rν ×Sl, i = ν + 1, . . . , ν +l may not exist in the classical sense but only in the sense of absolutely continuous functions. Hence, the operators ∂t and
∂τi are always understood in the sense of the absolutely continuous functions. In con-trast to that, the corresponding left and right derivatives of ∂t±µ(t, τ) and ∂τi±µ(t, τ)
τ ∈Rν ×Sl, i=ν+ 1, . . . , ν+l
always exist.
For brevity, left and right partial derivatives with respect to theτi will be denoted by
∂i+ and ∂i+. Partial derivatives in the sense of an absolutely continuous function will be denoted by ∂i. Note that ∂i+µ(t, θ) =∂i−µ(t, θ) =∂iµ(t, θ) for almost all t∈[0,1].
Recall that the left and right derivative share some properties with the classical deriva-tive. In particular, they are linear operators and the product rule applies. More precisely, let J ⊂ R be an open interval and let f, g : J → R be continuous functions such that
∂t±f(t) and∂t±g(t) exists for allt ∈R. Then
∂t+(f +g)(t) = lim
h↓0
f(t+h)+g(t+h)−f(t)−g(t)
h =∂t+f(t) +∂t+g(t) and
∂t+(f ·g)(t) = lim
h↑0
f(t+h)·g(t+h)−f(t)·g(t) h
= lim
h↑0
f(t+h)·g(t+h)−f(t+h)·g(t)+f(t+h)·g(t)−f(t)·g(t) h
= lim
h↑0 f(t+h)g(t+h)−g(t) h + lim
h↑0g(t)f(t+h)−fh (t)
=f(t)·∂t+g(t) +g(t)·∂t+f(t).
Analogous results for ∂t− is obtained in a similar manner.
As an equivalent to stationary points, we have the following relation: lett ∈J0 such that f has a local minimum int0. Then
∂t−f
t=t0 = lim
s↑t0
f(s)−f(t0) s−t0 ≤0 and
∂t+f
t=t0 = lim
s↓t0
f(s)−f(t0) s−t0 ≥0.
Hence, there exists c∈[0,1] such that c·∂t−f
t=t0 + (1−c)·∂t−f
t=t0 = 0.
One should note that there is a vast amount of literature on generalised directional derivatives in context of optimisation and implicit functions, see e.g. [Cla76, Pou77].
5.2.2. Defining the n×(ν+l) matrices
Qn±= [∂i±µ(t/n)]t=1,...,n;i=1,...,ν+l∈Rn×(ν+l), (5.3) we find by (4.5), that
n→∞lim n−1(Q0n+WnQn+)ik = Z 1
0
µ(i)(s, θ)w(s)µ(k)(s, θ) ds. (5.4) In analogy to M0nWnMn, the matrix Q0n+WnQn+ has full rank if n is large enough so that
Λw = lim
n n Q0n+WnQn+−1
(5.5) is well defined. The essential step to derive the asymptotic distribution is to show that
n
D(n) θˆ−θ
is asymptotically equivalent to D(n)n (Q0n+WnQn+)−1Q0n+Wnen, see theorem 5.2.3 and 5.2.4. Once this asymptotic equivalence has been established, the asymptotic distribution of the (weighted) least squares estimator can easily be derived by virtue of section 3.2, see theorem 5.2.8 and theorem 5.2.9
Theorem 5.2.3. Let µbe an identified, continuous piecewise polynomial and let x(t) = µ nt
+ξ(t) with ξ(·) as in theorem 5.1.9 and let w be a weight function satisfying the conditions of definition 4.2.1. Denote r(n) = Dn
ξ(n). Assume that either s7→µ(s, θ)∈ C(1)([0,1]) ∀θ∈Θ
or
s7→µ(s, θ)∈ C([0,1]) ∀θ ∈Θ andξ(·) is strictly stationary and ergodic.
Then, for any ∆ >0, P
r(n)
θˆwn −θ−(Q0n+WnQn+)−1Q0n+Wnen >∆
=o(1) (5.6)
as n→ ∞.
Theorem 5.2.4. Let µbe an identified, continuous piecewise polynomial and let x(t) = µ nt
+ξ(t) with ξ(·) as in theorem 5.1.10. Denote r(n) = Dn
ξ(n). Assume that s 7→
µ(s, θ)∈ C([0,1]) for all θ∈Θ. Then, for any ∆>0, P
r(n)
θˆn−θ−(Q0n+Qn+)−1Q0n+en >∆
=o(1) (5.7)
as n→ ∞.
Proof of theorem 5.2.3 and theorem 5.2.4. We only prove (5.6) explicitly, as we can derive (5.7) by virtually the same argument. For what follows, set ˆu= ˆθwn−θ. In analogy to the left and right derivative, we can define the right and left gradient by
∇+ = ∂1+, . . . , ∂(ν+l)+0
and
∇− = ∂1−, . . . , ∂(ν+l)−
0
. Since θ+ ˆu minimises Lwn (τ), there exist ci ∈[0,1] such that
ci·∂i+Lwn(θ+ ˆu) + (1−ci)·∂i−Lwn(θ+ ˆu) = 0 (i= 1, . . . , ν+l). (5.8) To state matters more concisely, we introduce the following notation: Forc= (c1, . . . , cν+l) and u∈Rν+l definecu and (1−c)u by
cu= (c1·u1, . . . , cν+l·uν+l)0 and
(1−c)u= ((1−c1)·u1, . . . ,(1−cν+l)·uν+l)0. Likewise, for U∈R(ν+l)×(ν+l) define
cU=
c1u11 c1u12 · · · c1u1(ν+l) c2u21 c2u22 · · · c2u2(ν+l)
... ... ...
cν+lu(ν+l)1 cν+lu(ν+l)2 · · · c2u(ν+l)(ν+l)
and
(1−c)U=
(1−c1)u11 (1−c1)u12 · · · (1−c1)u1(ν+l) (1−c2)u21 (1−c2)u22 · · · (1−c2)u2(ν+l)
... ... ...
(1−cν+l)u(ν+l)1 (1−cν+l)u(ν+l)2 · · · (1−c2)u(ν+l)(ν+l)
.
Note that is compatible with the standard matrix-vector operations, i.e.
c[U+V] =cU+cV, c(Uu) = (cU)v for U,V ∈R(ν+l)×(ν+l) and v ∈Rν+l. Moreover, sinceci ∈[0,1],
kcUvkp ≤ kUvkp
for any p-norm on Rν+l. In particular, kcUk ≤ kUkfor any induced matrix norm k·k.
In terms of this notation, (5.8) may be written as
c ∇+Lwn(θ+ ˆu) + (1−c) ∇−Lwn(θ+ ˆu) = 0. (5.9) Observe that the function z 7→ µ nt, θ+zuˆ
is differentiable w.r.t. z in the classical sense for almost all z ∈[0,1]. For suchz, the chain rule implies
∂z+µ nt, θ+zuˆ
=
ν+l
X
i=1
∂i+µ nt, θ+zu)ˆ ˆ
ui =∇+µ nt, θ+zuˆ0
ˆ u.
In particular, µ nt, θ+ ˆu
−µ nt, θ
= Z 1
0
∂sµ nt, θ+suˆ ds=
Z 1 0
∇+µ nt, θ+sˆu ds
0u.ˆ Thus, by linearity and product rule,
∂i+L(θ+ ˆu) =−2
n
X
t=1
w n+1t x(t)−µ(nt, θ+ ˆu)
∂i+µ(nt, θ+ ˆu)
=−2
n
X
t=1
w n+1t
ξ(t)∂i+µ nt, θ+ ˆu
−
2
n
X
t=1
w n+1t µ nt, θ
−µ nt, θ+ ˆu
∂i+µ nt, θ+ ˆu
=−2
n
X
t=1
w n+1t
ξ(t)∂i+µ nt, θ
−
2
n
X
t=1
w n+1t ξ(t)
∂i+µ nt, θ+ ˆu
−∂i+µ nt, θ + 2
n
X
t=1
w n+1t
∂i+µ nt, θ+ ˆu Z 1
0
∇+µ nt, θ+sˆu ds
0u.ˆ Denoting
rn+ =
n
X
t=1
w n+1t ξ(t)
∂i+µ nt, θ+ ˆu
−∂i+µ nt, θ
!ν+l
i=1
,
Qn = Z 1
0
∂i+µ nt, θ+suˆ0
ds
t=1,...,n;i=1,...,ν+l
∈Rn×(ν+l), and
Qbn+ =
∂i+µ nt, θ+ ˆu
t=1,...,n;i=1,...,ν+l ∈Rn×(ν+l), we obtain
−12∇+Lwn(θ+ ˆu) = Q0n+Wnen+rn+−Qb0n+WnQnu.ˆ (5.10) Likewise,
−12∇−Lwn(θ+ ˆu) = Q0n−Wnen+rn−−Qb0n−WnQnu,ˆ (5.11) whereQbn− andrn− are defined in analogy to their counterpartsQbn+and rn+. Combining (5.10) and (5.11) with (5.9), we find
cQ0n+Wnen+crn+−cQb0n+WnQnuˆ
+(1−c)Q0n−Wnen+ (1−c)rn−−(1−c)Qb0n−WnQnuˆ= 0, in other words
−Q0n+Wnen =yn−Q0n+WnQn+uˆ+Cnuˆ+crn++ (1−c)rn− (5.12) where we denote
yn= (1−c)
Q0n−−Q0n+
Wnen and
Cn = h
Q0n+−Qb0n−
i
WnQn++Qb0n−Wn
Qn+−Qn
+ch
Qb0n−−Qb0n+
i
WnQn. The following lemma clarifies the structure of the remaindersrn+andrn−, respectively.
Lemma 5.2.5. There exist sequences of random matrices (Rn+)n∈N ⊂ R(ν+l)×(ν+l) and (Rn−)n∈N⊂R(ν+l)×(ν+l) such that
(i) rn±=Rn±uˆ+vn±.
(ii) P(n−1kRn±k>∆) =o(1) for all ∆>0.
The vectors vn± are equal to zero if and only if µ has a continuous derivative. More precisely, defineJ ={k = 1, . . . , l: b1,k = 1} andJc={k= 1, . . . , l : b1,k >1}with b1,k
as in 4.1.1. Then
(iii) vn+ = (v1;n+, . . . , vν+l;n+)∈Rν+l with
vi;n+ = 0 (i= 1, . . . , ν), vν+k;n+ = 0 (k ∈ Jc), and
|vν+k;n+|2 =
a1,k
dn(ηˆwk;n∨ηk)−1e X
t=dn(ηˆk;nw ∧ηk)e
ξ(t)w n+1t
2
(k∈ J).
(iv) vn− = (v1;n−, . . . , vν+l;n−)∈Rν+l with
vi;n− = 0 (i= 1, . . . , ν), vν+k;n− = 0 (k ∈ Jc), and
|vν+k;n−|2 =X
k∈I
a1,k
bn(ηˆwk;n∨ηk)c X
t=bn(ηˆwk;n∧ηk)+1c
ξ(t)w n+1t
2
(k ∈ J).
We postpone the proof of the lemma and proceed with the proof of the main result:
Define Rn and vn by
Rn=cRn++ (1−c)Rn−, and
vn=cv++ (1−c)vn−. Then, (5.12) can be written as
−Q0n+Wnen =yn+vn+
Cn+Rn−Q0n+WnQn+
ˆ
u. (5.13)
By (5.5), there exists K ∈R+ and N ∈N, such that for alln > N
n Q0n+WnQn+−1
≤K (n > N). We thus obtain for n > N
P
r(n)
uˆ− Q0n+WnQn+
−1
Qn+Wnen
>∆
=P
r(n)
Q0n+WnQn+−1
[yn+vn+ [Cn+Rn] ˆu]
>∆
≤P r(n)
n K|yn+vn+Cnuˆ+Rnu|ˆ >∆ .
By conditioning on the distance
ˆηwk;n−ηk
(k ∈ J), we will show that the last probability indeed vanishes asn → ∞. More specifically, letI ⊂ J andκ∈(0,1) such that r(n)nκ →0.
Denote Ic=J \ I and byEnI the event EnI =
ˆηwk;n−ηk
< n−κ :k∈ I ∩
ηˆk;nw −ηk
≥n−κ :k ∈ Ic . Note that EnI1 ∩ EnI2 =∅ if I1 6=I2 and S
I⊂J EnI = Ω. Hence P
nr(n)
n K|yn+vn+Cnuˆ+Rnu|ˆ >∆o
=P nr(n)
n K|yn+vn+Cnuˆ+Rnu|ˆ >∆ o
∩ [
I⊂J
EnI
=X
I⊂J
P
nr(n)
n K|yn+vn+Cnuˆ+Rnu|ˆ >∆o
∩ EnI . It is thus sufficient to check that for all ∆>0 and all I ⊂ J
P
nr(n)
n |yn+vn+Cnuˆ+Rnu|ˆ >∆o
∩ EnI
→0. (5.14)
So, fix I ⊂ J and define vnI = vI1;n, . . . , vIν+l;n by
vi;nI =
vi;n if i=ν+k for some k ∈ I, 0 else.
Observe thatvn =vInc+vIn. Define further the diagonal matrixOIn = oIij
i=1,...,ν+l;j=1,...,ν+l
by
oIij =
vIi;nc ˆ
ui if i=j =ν+k with k ∈ Ic, 0 else.
Note that OIn is well-defined if the event EnI occurs, as |ˆuν+k| ≥n−κ for all k ∈ Ic in this case. By the very definition of OIn, we have
vnIc =OInu.ˆ Hence, (5.14) follows, if we can show that
P
nr(n) n
yn+vnI +
OIn+Cn+Rn ˆ u
>∆o
∩ EnI
→0. (5.15)
But lemma 5.2.5 implies
P n−1kRn|>∆
→0 for all ∆>0.
From lemma 5.2.6 and lemma 5.2.7 below, we conclude that, P
r(n)
n |yn|>∆
→0 for all ∆ >0
and
P n−1kCnk>∆
→0 for all ∆>0.
Consequently, (5.15) follows, if we can show that P
n−1 OIn
>∆ ∩ EnI
→0 for all ∆ >0, (5.16) P
nr(n) n
vnI >∆o
∩ EnI
→0 for all ∆>0, (5.17) and that r(n)ˆu is tight in the following sense: for all ε >0 exists ˜∆>0 such that
P n
r(n)|ˆu|>∆˜o
∩ EnI
≤ε. (5.18)
Proof of (5.16): We need to ascertain that P
|vν+k;n|
|nˆuν+k| >∆
∩ EnI
→0 for all ∆ >0, k ∈ Ic. (5.19) So, fix k ∈ Ic and ∆>0. By definition of vn,
P
|vν+k;n|
|nˆuν+k| >∆
∩ EnI
≤P
|vν+k;n+|
|nˆuν+k| >∆/2
∩ EnI
+P
|vν+k;n−|
|nˆuν+k| >∆/2
∩ EnI
. To show that
P
|vν+k;n+|
|nˆuν+k| >∆/2
∩ EnI
→0, (5.20)
choose δ >0 such that [ηk−δ, ηk+δ]⊂(0,1). Since ˆθwn is consistent, we find P
|vν+k;n+|
|nˆuν+k| >∆/2
∩ EnI
=P
|vν+k;n+|
|nˆuν+k| >∆/2
∩ EnI∩ {|ˆu| ≤δ}
+o(1).
By definition of vn+ in lemma 5.2.5, P
|vν+k;n+|
|nˆuν+k| >∆/2
∩ EnI ∩ {|ˆu| ≤δ}
=P
1 nˆuν+k
Pdn(ηk∨ˆηwk;n)−1e
t=dn(ηk∧ηˆk;nw )eξ(t)w n+1t
>∆/2
∩ {|ˆu| ≤δ} ∩ EnI
≤P
sup
0≤u≤δ,|u|>n−κ
1 nu
Pdn(ηk+u)−1e
t=dnηke ξ(t)w n+1t
>∆/4
+ P
sup
−δ≤u≤0,|u|>n−κ
1 nu
Pdnηke−1
t=dn(ηk+u)eξ(t)w n+1t
>∆/4
=Π1+ Π2,
say. Since [η−δ, η+δ]⊂(0,1), we may assume that t+dηne−1n+1 ∈K for some compact set K ⊂ (0,1). Denote Ydnηke(t) =Pt
s=dnηkeξ(s), Y(t) = Pt
s=1ξ(s) and by VK(w) the total variation of w on K. Note that VK(w) is finite, since w is continuously differentiable on K. Then
Pdn(ηk+u)−1e
t=dnηke ξ(t)w n+1t =
Pdn(ηk+u)−1e t=dnηke
Ydnηke(t)−Ydnηke(t−1)
w n+1t
=
Pdn(ηk+u)−2e
t=dnηke Ydnηke(t)
w n+1t
−w n+1t+1 +Ydnηke(dn(ηk+u)−1e)wdn(η
k+u)−1e n+1
≤2dn(ηksup+u)−1e
t=dnηke
Ydnηke(t)
VK(w).
Consequently, by strict stationarity of ξ(·), Π1 ≤P
sup
u>n−κ 1 nu
dn(ηk+u)−1e
sup
t=dnηke
Ydnηke(t)
VK(w)>∆/8
=P
sup
u>n−κ 1 nu
dn(ηk+u)e−dnηke
sup
t=1
|Y(t)|VK(w)>∆/8
≤P
sup
u>n−κ 1 nu
dnuesup
t=1
|Y(t)|VK(w)>∆/8
. Likewise, we find
Π2 ≤P
sup
u>n−κ 1 nu
dnuesup
t=1
Ydnηke(t)
VK(w)>∆/8
. Hence, (5.20) follows, if we can show that
P
sup
u>n−κ 1 nu¸dunesup
t=1
|Yt|>∆∗
=o(1) (5.21)
for all ∆∗ >0. But by assumption, limnn1|Yn|= 0 almost surely and thus lim sup
k
1n1
k|Yk|>∆∗o = 0 a.s. for all ∆∗ >0 from which we conclude by contraposition that
P
T∞ k=1
S
t≥k
n|Y
t|
t >∆∗o
= 0 for all ∆∗ >0.
Thus, for allε >0, there exists N(ε)∈Nsuch that P
S
t≥N(ε)
n|Yt| t >∆∗
o
≤ε.
Since un > n1−κ, we may assume that dnue> N(ε). In this case P
sup
u>n−κ dunesup
t=1
|Yt| nu >∆∗
≤P
sup
u>n−κ
Sdnue t=1
n|Y
t|
nu ≥∆∗o
≤P
SN(ε)
t=1 {nκ−1|Yt|>∆∗}
+P
S∞ t=N(ε)+1
n|Y
t|
t >∆∗o . By construction,
P
S∞ t=N(ε)+1
n|Y
t|
t >∆∗o
≤ε,P
SN(ε)
t=1 {nκ−1|Yt|>∆∗}
=O(nκ−1).
Thus, (5.21) and in turn (5.20) follow.
By virtually the same argument, we can show that P
|vν+k;n−|
|nˆuν+k| >∆/2
∩ EnI
→0, and so (5.16) follows.
Proof of equation (5.17): We must show that P
nr(n)
n |vν+k;n|>∆ o
∩ EnI
→0 for all ∆>0 and all k ∈ I.
To that end, fixk ∈ Iand ∆ >0. We may w.l.o.g. assume that [ηk−n−κ, ηk+n−κ]⊂ [a, b] for 0< a < b <1. By definition of vn and lemma 5.2.5,
P
nr(n)
n |vν+k;n|>∆o
∩ EnI
=P
nr(n) n
cν+k·v(ν+k)+;n+ (1−cnu+k)·v(ν+k)−;n >∆o
∩ EnI
=P
r(n)
n |a1,k|Pdn(ˆηwk;n∨ηk)e t=bn(ˆηk;nw ∧ηk)c
ξ(t)w n+1t >∆
∩ EnI
≤P
r(n)
n |a1,k|Pdn(ηk+n−κ)e t=bn(ηk−n−κ)c
ξ(t)w n+1t >∆
≤P
r(n) n
Pdn(ηk+n−κ)e
t=bn(ηk−n−κ)c|ξ(t)|>∆∗
where ∆∗ = ∆
|a1,k|supx∈[a,b]|w(x)|. SinceE(|ξ(t)|) does not depend ont, we obtain by Markov’s inequality
P
r(n) n
dn(ηk+n−κ)e
X
t=bn(ηk−n−κ)c
|ξ(t)|>∆∗
≤ n∆r(n)∗
dn(ηk+n−κ)e
X
t=bn(ηk−n−κ)c
E|ξ(t)| →0 which proves the claim.
Proof of equation (5.18): From (5.13), we obtain Q0n+Wnen+yn+vnI =
Q0n+WnQn+−OIn−Cn−Rn ˆ
u. (5.22)
Since all eigenvalues 1nQ0n+WnQn+ are bounded away from zero as n → ∞, there exist K ∈R∗+ such that
|ˆu| ≤ Kn
Q0n+Wnen+yn+vnI whenever 1n
OIn+Cn+Rn
is sufficiently small, say 1n
OIn+Cn+Rn
≤ ∆0. Define therefore the event
Fn =1
n
OIn+Cn+Rn
≤∆0 . Then
P {|r(n)ˆu|>∆∗} ∩ EnI
≤P {|r(n)ˆu|>∆∗} ∩ EnI ∩ Fn
+P EnI ∩ Fnc . But by lemma 5.2.5, lemma 5.2.7 and (5.16)
P EnI ∩ Fnc
→0.
On the other hand
P {|r(n)ˆu|>∆∗} ∩ EnI∩ Fn
≤P
nr(n)K n
Q0n+Wnen+yn+vnI
>∆∗o
∩ EnI . So, for fixed ε >0, we find by 3.2.4, lemma 5.2.6 and (5.17), a ∆∗ >0 such that
P
nr(n)K n
Q0n+Wnen+yn+vnI
>∆∗o
∩ EnI
≤ε, which is indeed the desired result.
Proof of lemma 5.2.5. We consider only the case of the right derivative, as the arguments for left derivative are virtually identical.
For what follows, denote byri;n+andvi;n+thei-th component ofrn+andvn+(i= 1, . . . , ν+l) and by rim;n+ the im-th component of Rn+ (i, m= 1, . . . , ν+l).
(i) If i= 1, . . . , p0, then ∂i+µ(s, θ) does not depend onθ, i.e.
∂i+µ nt, θ+ ˆu
=∂i+µ nt, θ . In particular, ri;n+ = 0 (i= 1, . . . , p0). Hence
vi;n+ = 0, rim;n+= 0 (i= 1, . . . , p0, m= 1. . . , ν +l).
(ii) Let i=p0+ 1, . . . , ν. Thenθi =aj,k for some k = 1, . . . , l and j = 1, . . . , pk. Hence
∂i+µ(s, θ+ ˆu)−∂i+µ(s, θ) = (s−uˆν+k−ηk)b+j,k −(s−ηk)b+j,k
= Z 1
0
∂z(s−uˆν+kz−η)b+j,k dz
=−uˆν+k
Z 1 0
bj,k(s−uˆν+kz−η)b+j,k−1dz, thus
ri;n+ =
n
X
t=1
w n+1t ξ(t)
∂i+µ nt, θ+ ˆu
−∂i+µ nt, θ
=−
n
X
t=1
ξ(t)w n+1t ˆ uν+k
Z 1 0
bj,k(nt −uˆν+kz−η)b+j,k−1dz.
Therefore
rim;n+ = 0, vi;n+ = 0 (i=p0+ 1, . . . , ν, m 6=ν+k) and
ri(ν+k);n+=−
n
X
t=1
ξ(t)w n+1t Z 1
0
bj,k nt −uˆν+kz−ηbj,k−1
+ dz.
(iii) Let i=ν+k (k = 1. . . , l), in other words θi =ηk. In this case θj+p =aj,k
j = 1, . . . , pk; p=
k−1
X
k0=0
pk0
.
Assume b1,k = 1. Then
∂i+µ(s, θ+ ˆu)−∂i+µ(s, θ)
=
pk
X
j=1
bj,k[aj,k+ ˆuj+p]·(s−ηk−uˆν+k)b+j,k−1−
pk
X
j=1
bj,kaj,k·(s−ηk)b+j,k−1
=
pk
X
j=1
ˆ
uj+pbj,k·(s−ηk−uˆν+k)b+j,k−1+
pk
X
j=1
bj,kaj,k·(s−ηk−uˆν+k)b+j,k−1−
pk
X
j=1
bj,kaj,k·(s−ηk)b+j,k−1
=
pk
X
j=1
ˆ
uj+pbj,k·(s−ηk−uˆν+k)b+j,k−1+a1,k ·
(s−ηk−uˆν+k)0+−(s−ηk)0+ +
pk
X
j=2
aj,kbj,k·h
(s−ηk−uˆν+k)b+j,k−1−(s−ηk)b+j,k−1i
=
pk
X
j=1
ˆ
uj+pbj,k·(s−ηk−uˆν+k)b+j,k−1+a1,k ·
(s−ηk−uˆν+k)0+−(s−ηk)0+
−
pk
X
j=2
aj,kbj,k(bj,k −1) ˆuν+k
Z 1 0
(s−ηk−uˆν+kz)b+j,k−2 dz.
Hence, ri;n+=−
pk
X
j=1
ˆ uj+pbj,k
n
X
t=1
w n+1t
ξ(t) nt −ηk−uˆν+kbj,k−1
+ −
n
X
t=1
w n+1t
ξ(t)a1,k·h
t
n−ηk−uˆν+k0
+− nt −ηk0 +
i + ˆ
uν+k n
X
t=1
w n+1t ξ(t)
pk
X
j=2
aj,kbj,k(bj,k −1) Z 1
0 t
n −ηk−uˆν+kzbj,k−2
+ dz
and thus
r(ν+k)m;n+ =−
n
X
t=1
ξ(t)w n+1t
bj,k nt −ηk−uˆν+kbj,k−1 +
if m =p+j, j = 1, . . . , pk, r(ν+k)m;n+ =
n
X
t=1
ξ(t)w n+1t
pk
X
j=2
aj,k b2j,k−bj,k Z 1
0 t
n −ηk−zuˆν+kbj,k−2
+ dz
if m =ν+k, and
r(ν+k)m;n+ = 0 else. By 4.2.5,
|vν+k;n+|=
a1,k
dn(ˆηwk;n∨ηk)−1e
X
t=dn(ˆηk;nw ∧ηk)e
ξ(t)w(n+1t ) .
If bj,k >1, then all quantities remain the same, except for vν+k;n+ which is equal to zero.
In order to complete the proof, we need to show that 1nkRn+k →0 in probability, i.e. for all ∆>0
P n−1|rim;n|>∆
=o(1) (i= 1, . . . , ν+l;m = 1, . . . ν+l). (5.23) To that end, it suffices to check that for all k= 1, . . . , l,m ∈N∪ {0} and ∆>0
P
n−1
n
X
t=1
ξ(t)w n+1t t
n −ηk−uˆν+km +
>∆
=o(1), (5.24)
and since each rjk;n can be represented as a linear combination of expressions of this kind.
So, fix m ∈ N ∪ {0}, k ∈ {1, . . . , ν} and ∆ > 0. At first, note that there exists Therefore, define the events
Am =
and thus P
By 5.1.2 and 5.1.4, P(V0c) = o(1) and, by 3.2.4, P(Acm) = o(1). Theorem 5.1.9 implies P(U0c) = o(1).
Proof of (5.25): We apply similar arguments. Assume w.l.o.g. that ˆuν+k ≥ 0.
At first, we need to calculate the integral R1 0
t
n −ηk−zuˆν+km
+ dz: If nt > ηk+δ and 0<|ˆuν+k| ≤δ, then
Z 1 0
t
n −ηk−zuˆν+k m
+
dz = 1 m+ 1
1 ˆ uν+k
h t
n−ηkm+1
− nt −ηk−uˆν+km+1i
= 1
m+ 1 1 ˆ uν+k
"
t
n−ηkm+1
−
m+1
X
i=0 m+1
i
t
n −ηki
(−ˆuν+k)m+1−i
#
= 1
m+ 1
" m X
i=0 m+1
i
t
n−ηki
(−ˆuν+k)m−i
# . If nt > ηk+δ and ˆuν+k = 0, this simplifies to
Z 1 0
t
n−ηk−zuˆν+k
m
+ dz = nt −ηk
m +. If |nt −ηk| ≤δ, we have the (rather trivial) upper bound
Z 1 0
t
n−ηk−zuˆν+km
+ dz ≤1.
Hence, for ω ∈Am∩U0∩V0, 1
n
n
X
t=1
ξ(t)w n+1t Z 1
0 t
n−ηk−zuˆν+km + dz
≤
X
t n>ηk+δ
ξ(t)w n+1t 1
m+1
m X
i=0 m+1
i
t
n −ηk
i
(−ˆuν+k)m−i
+ 1
n X
|t n−ηk|≤δ
ξ(t)w n+1t
≤
m
X
i=0
(−ˆuν+k)m−i m+11 m+1i X
t n>ηk+δ
ξ(t)w n+1t t
n −ηki
+ 1
n X
|t n−ηk|≤δ
ξ(t)w n+1t
≤∆ 2 +
v u u t
1 n
X
|t n−ηk|≤δ
ξt2 v u u t
1 n
X
|t n−ηk|≤δ
wd2 n+1t
<∆
and thus again P n−1
n
X
t=1
ξ(t)w n+1t Z 1
0 t
n −ηk−zuˆν+k
m + dz
>∆
!
≤P(Acm∪U0c∪V0c).
Lemma 5.2.6. For all ∆ >0 P
r(n)
n |yn|>∆
=O
r(n) n
(5.26) Proof. Since ∂j+µ(t, θ) = ∂j−µ(t, θ) for allt ∈[0,1] and i= 1, . . . ν, we have
|yi;n|= 0 (i= 1, . . . , ν).
If i=ν+k (k = 1, . . . , l), then ∂j+µ(t, θ) and ∂j−µ(t, θ) differ at most at t=ηk. Hence
r(n)
n |yν+k;n| ≤ r(n)n
[∂j+µ(ηk, θ)−∂j−µ(ηk, θ)]w n+1ηkn ξdnηke
But, by Markov’s inequality,
P r(n)
n |yν+k;n|>∆
≤ r(n)n E
[∂j+µ(ηk, θ)−∂j−µ(ηk, θ)]w n+1ηkn ξdnηke
→0, which proves the assertion.
Lemma 5.2.7. For all ∆ >0
P 1nkCnk>∆
=o(1).
Proof. From the triangle inequality, we obtain the upper bound kCnk ≤
h
Q0n+−Qb0n−i
WnQn+
+
Qb0n−Wn
Qn+−Qn +
h
Qb0n−−Qb0n+i
WnQn
Applying case-by-case arguments, we will show that each term converges to zero as n → ∞. To that end, we assume (as we may by theorem 5.1.9) that |ˆu| ≤ δ for some δ >0. Note that there exists a constant C ∈R+ such that
sup
i=1,...,ν+l,t∈[0,1],|u|≤δ
|∂i±µ(t, θ+u)| ≤C.
As for the first term, we obtain by definition 1
n hh
Q0n+−Qb0n−i
WnQn+i
ij
=1 n
n
X
t=1
∂i−µ nt, θ+ ˆu
−∂i+µ nt, θ w nt
∂j+µj nt, θ .
The functions (s, u)7→∂i+µ(s, θ+u) and (s, u)7→∂i−µ(s, θ+u) coincide and are contin-uous in all (s, u) whenever |s−ηk| >|u| (k = 1, . . . , l). In particular, they coincide and are uniformly continuous on the set
D(δ1, δ2) = {(s, u) :s ∈[0,1] such that |s−ηk| ≥δ1,|u| ≤δ2} if δ1 ≥2δ2. By choosingδ1 and δ2 sufficiently small, we thus obtain for any ε >0
∂i+µ nt, θ
−∂i−µ nt, θ+u ≤ε, whenever nt, u
∈D(δ1, δ2). Thus, fixing ∆>0 and assuming |ˆu| ≤δ2,
1 n
hh
Q0n+−Qb0n−i
WnQn+i
ij
≤2Cn2 X
˛
˛
˛ t n−ηk
˛
˛
˛≤2δ
w nt
+εC (i, j = 1, . . . , ν+l)
<∆ if δ1, δ2 sufficiently small (i, j = 1, . . . , ν+l). Hence, for any sufficiently small δ >0,
P
1 n
hh
Q0n+−Qb0n−i
WnQn+i
ij
>∆
≤P(|ˆu|> δ), and thus
P
h
Q0n+−Qb0n−i
WnQn+
>∆
→0.
Similar arguments apply to the second term: By definition, 1
n h
Qb0n−Wn
Qn+−Qni
ij
=1 n
n
X
t=1
∂i−µ nt, θ+ ˆu w nt
∂j+µ nt, θ
− Z 1
0
∂jµ nt, θ+ ˆus ds
. By above arguments, we can choose δ >0 such that for all nt, u
∈D(δ)
∂j−µ nt, θ
− Z 1
0
∂jµ nt, θ+su ds
≤ε, Hence, for fixed ∆>0 and |ˆu| ≤δ,
1 n
h
Qb0n−Wn
Qn+−Qni
ij
≤2Cn2 X
˛
˛
˛ t n−ηk˛
˛
˛≤2δ
w nt
+εC (i, j = 1, . . . , ν +l)
<∆ ifδ sufficiently small (i, j = 1, . . . , ν+l), which proves the claim.
Finally, consider the third term: By definition, 1
n hh
Qb0n−−Qb0n+i
WnQni
ij
=1 n
n
X
t=1
∂i−µ
t n,θˆwn
−∂i+µ
t
n,θˆwn w nt
Z 1 0
∂jµ nt, θ+sˆu ds.
As above, ∂i+µ t,θˆwn
6= ∂i−µ t,θˆnw
only if i = ν +k, k = 1, . . . , l and t = ˆηwk;n. Hence, for fixed ∆>0,
1 n
hh
Qb0n−−Qb0n+i
WnQni
ij = 0 (i= 1, . . . , ν) and
1 n
hh
Qb0n−−Qb0n+i
WnQni
ij
≤2C2 n
X
|nt−ηk|≤|uˆν+k|
w nt
(i=ν+k)
<∆ if |ˆu| sufficiently small (i=ν+k), which completes the proof.
Theorem 5.2.8. Let w, ξ(·) and µbe as in theorem 5.2.3.
(i) Assume that the finite dimensional distributions of Zξ,n(·) converge to a fractional Brownian motionBH(·) with Hurst parameterH = 0.5+d. Denote by Ξ the random vector
Ξ :=
Z
w(s)∂j+µ(s) dBH(s)
j=1,...,ν+l
. Then, as n → ∞,
r(n) ˆθwn −θ d
−
→ΛΞ. (5.27)
with Λ as in (5.5).
(ii) Assume that the finite dimensional distributions of Zξ,n(·) converge to a Hermite process HmH(·) with Hurst parameterH = 0.5 +d. Denote by Ξ∗ the random vector
Ξm :=
Z
w(s)∂j+µ(s) dHmH(s)
j=1,...,ν+l
. Then, as n → ∞,
r(n) ˆθwn −θ d
−
→ΛΞm. (5.28)
Theorem 5.2.9. Letξ(·) andµbe as in theorem 5.2.4. Setw= 1 in the above definitions of Λ, Ξ and Ξm.
(i) Assume that the finite dimensional distributions of Zξ,n(·) converge to a fractional Brownian motion BH(·) with Hurst parameter H = 0.5 +d. Then, as n→ ∞,
r(n) ˆθn−θ d
−
→ΛΞ. (5.29)
(ii) Assume that the finite dimensional distributions of Zξ,n(·) converge to a Hermite process HmH(·) with Hurst parameter H = 0.5 +d. Then, as n → ∞,
r(n) ˆθn−θ d
−
→ΛΞm. (5.30)
Proof of theorem 5.2.8 and theorem 5.2.9. We only prove (5.27) explicitly, as (5.28), (5.29) and (5.30) follow by similar reasoning: By theorem 5.2.3, it suffices to show that
r(n) Q0n+WnQn+−1
Q0n+Wnen→ΛΞ.
By definition
n Q0n+WnQn+−1
→Λ, so the claim follows, if we can show that
r(n)
n Q0n+Wnen−→d Ξ, i.e. if we can show that
r(n) n
α,Q0n+Wnen
→ hα,Ξi ∀a∈Rν+l. (5.31) But this a a consequence of section 3.2. By theorem 3.2.14,
r(n) n
α,Q0n+Wnen
→ Z
µα+(s) dBH(s) with µ(α+)(t) :=Pν+l
j=1αj∂j+µ(t) which proves the claim.
5.2.10. The covariance matrix Σ0 of Ξ follows immediately from the isometric property of the Wiener integral established in section 3.1. It can be expressed in terms of fractional integrals, see paragraph 3.2.4, as
Σ0 =
Γ(d+ 1)2 c21(d)
Z
R
I−d(w ∂j+µ) (s)
I−d (w ∂k+µ) (s)ds
j,k=1,...,p+1
, (5.32)
Hence, under the assumptions of theorem 5.2.8 (i) and for w = 1 under the assumptions of theorem 5.2.9 (i) we have, as n→ ∞,
r(n) ˆθwn −θ d
−
→N(0,ΛΣ0Λ). (5.33)
Using a computer algebra system it is possible to obtain closed form formulas for the entries of the covariance matrix ΛΣ0Λ in the case of w = 1. Consider, for instance, the variance of ˆηfor a cubic spline with an unknown knot η. denote byPM⊥
n(ˆη) the projection matrix on the subspace that is orthogonal to the columns ofM⊥n(ˆη). The first four columns of M⊥n(ˆη) correspond to the basis functions gj(x) = xj−1 (j = 1, . . . ,4), the last column corresponds to g5(x) = (x−η)ˆ 3+. Since minimisation of
X−PMn(˜η)(X)
2 =
PM⊥
n(˜η) en+
4
X
j=1
ajmjn(η) +a5m5n(η)
!
2
with respect to ˜η is equivalent to minimising
PM⊥n(˜η) en
a5 +m5n(η)
2
, ˆ
η depends on the parameters (a1, . . . , a5, , σξ2) via the ratio σaξ
5 only. If ξ is multiplied by a constant c, the asymptotic variance of ˆη is multiplied by the factor c2. Hence, the asymptotic variance is of the form
Varasym(ˆη) = σξ2
a25f(α, η) where f is a nonlinear function of η and α. Here, σ
2 ξ
a25 can be regarded as a noise to signal ratio. After some transformations in maple (see program 11.1.1, 11.1.2 and 11.1.3), the closed form of the asymptotic variance of ˆη normalised by D2(n) reads forα ∈(0,1) as
100(1−α)(2−α) a25(1−α)8
12
X
j=1
Vj(α, η) with (1−α)k = (k−α)(k−1−α)· · ·(1−α) and
V1(α, η) := α6 η5(1−η)5
V2(α, η) :=−3α5`
7η2−7η+ 3´ η6(1−η)6 V3(α, η) :=α4
„ 12
η4+α(1−η)7 + 12 η7(1−η)4+α
«
V4(α, η) :=α4
„145η4−290η3+ 292η2−147η+ 30 η7(1−η)7
«
V5(α, η) :=−3α3`
97η6−291η5+ 562η4−639η3+ 395η2−124η+ 14´ η8(1−η)8
V6(α, η) := 6α3
`12η2+ 8η+ 7´ η8(1−η)4+α +
`27 + 12η2−32η´ η4+α(1−η)8
!
V7(α, η) :=−α2`
506η6−1518η5−207η4+ 2944η3−3135η2+ 1410η−252´ η8(1−η)8
V8(α, η) :=−12α2 (2η+ 3)`
6η2−3η+ 7´
η8(1−η)4+α −(2η−5)`
6η2−9η+ 10´ η4+α(1−η)8
!
V9(α, η) :=6α`
280η6−840η5+ 665η4+ 70η3−521η2+ 346η−77´ η8(1−η)8
V10(α, η) := 6α
`24η3−72η2+ 4η+ 77´ η8(1−η)4+α −
`24η3−68η−33´ η4+α(1−η)8
!
V11(α, η) := 252 (2η−1)`
2η2+ 2η+ 1´
η8(1−η)4+α −(2η−1)`
2η2−6η+ 5´ η4+α(1−η)8
!
V12(α, η) := 252 (2η−1)2 η8(1−η)8
!
For fixed α, each of the terms Vj(α, η) is symmetric at η = 0.5. Hence, the asymptotic variance is symmetric at η= 0.5. Moreover, it has a pole at η= 0 of the form
f(α, η)∼ 1 η4+α
1200(6−2α2+ 3α)(1−α)(2−α) a242(8−α)(6−α)(1−α)4 . The asymptotic variance of ˆa5 is given by
g(α, η) := 392(1−α)(2−α) 2(1−α)8
n
X
j=1
Wj with
W1:=−α625η2−25η+ 6
η7(1−η)7 −3α5175η4−350η3+ 283η2+ 16 η8(1−η)8
W2:=−α43625k6−10875k5+ 15129k4−12133k3+ 5712k2−1458 + 156 η9(1−η)9
W3:=−30α4
„(2η−1)(5η−3)
η6+α(1−η)9 +(2η−1)(5η−2) η9(1−η)6+α
«
W4:=α3−216−15888η2−7275η8+ 29100η7−65214η6+ 93792η5+ 2880η+ 47580η3−84975η4 η10(1−η)10
W5:= 24α3
„75η4−290η3+ 430η2−260η+ 54
η(6 +α)(1−η)10 +75η4−10η3+ 10η2−30η+ 9 η10(1−η)6+α
«
W6:=−2α2−648 + 6654η+ 64632η3−28533η2−78176η4−25300η7+ 17076η6+ 6325η8+ 37322η5 η10(1−η)10
W7:= 6α2
„600η5−8115η2−933−3300η4+ 6970η3+ 4562η η(6+α)(1−η)10
−600η5+ 300η4−230η3+ 1005η2−958η+ 216 η10(1−η)6+α
«
W8:= 24α−4795η5−7000η7+ 9765η6+ 1750η8−99−2845η4−3311η2+ 921η+ 5515η3 η10(1−η)10
W9:=−24α
„−900η4+ 994η+ 1135η3−243 + 300η5−1385η2 η(6+α)(1−η)10
−300η5−600η4+ 535η3+ 380η2−471η+ 99 η10(1−η)6+α
«
W10:= 144(5η2−5η+ 1)(35η2−35η+ 9) η10(1−η)10
W11:=−144
„(5η2−5η+ 1)(30η3−105η2+ 119η−35)
η(6+α)(1−η)10 −(5η2−5η+ 1)(30η3+ 15η2−k−9) η10(1−η)6+α
«
As in the case of η, the asymptotic variance is symmetric at η= 0.5 for fixed alpha and has a pole at η = 0 of the form
g(α, η)∼ 1 η6+α
7056(8−5α2+ 12α)(1−α)(2−α) 2(8−α)(6−α)(1−α)4 .
As can be seen in figure 10, not all components of the asymptotic covariance matrix are symmetric. For a complete numerical implementation, see section 6.
true knot 0.3
0.4 0.5
0.6 0.7
d
−0.2
−0.1 0.0
0.1 1e+05 0.2
2e+05 3e+05 4e+05 5e+05
Asymptotic variance of ηη^
true knot 0.3
0.4 0.5
0.6 0.7
d
−0.2
−0.1 0.0
0.1 0.2 2e+07
4e+07 6e+07
Asymptotic variance of a^5
true knot 0.3
0.4 0.5
0.6 0.7
d
−0.2
−0.1 0.0
0.1 0.2 2e+07
4e+07 6e+07
Asymptotic variance of a^4
true knot 0.3
0.4 0.5
0.6 0.7
d
−0.2
−0.1 0.0
0.1 0.2
−6e+07
−4e+07
−2e+07 0e+00
Asymptotic covariance of a^
4 and a^