Online Supplementary Material to
“Exploratory Analysis and Modeling of Stock Returns”
Kimihiro Noguchi
Alexander Aue
Prabir Burman∗
Abstract
In this document, a detailed proof of Theorem 5.1 in Noguchi, Aue and
Burman (2014) is given. It concerns the rate of convergence of the spline-based
estimators as developed in Section 5 of that paper.
1
Proof of Theorem 5.1
The proof of Theorem 5.1 requires the use of the following auxiliary results. The first
lemma can be established by repeatedly applying integration by parts.
Lemma 1. Let r be an integrable function on [0, 1]2 and let
Z x1 Z x2
R(x1 , x2 ) =
r(u1 , u2 )du1 du2 .
0
0
Let h1 and h2 be functions on [0, 1] whose derivatives ḣ1 and ḣ2 are integrable. If
R(1, 1) = 0, it follows that
Z 1Z 1
h1 (u1 )h2 (u2 )r(u1 , u2 )du1 du2
0
∗
0
K. Noguchi (E-mail:
Kimihiro.Noguchi@wwu.edu) is Assistant Professor, Department of
Mathematics, Western Washington University, Bellingham, WA 98225. Alexander Aue (E-mail:
aaue@ucdavis.edu) is Associate Professor and Prabir Burman (E-mail: pburman@ucdavis.edu) is
Professor, Department of Statistics, University of California, One Shields Avenue, Davis, CA 95616.
This research was partially supported by NSF grants DMS 09-05400, DMS 09-07622, DMS 12-09226,
DMS 13-05858, and DMS 14-07530.
1
Z
1
Z
1
= − h1 (1)
ḣ2 (u2 )R(1, u2 )du2 − h2 (1)
ḣ1 (u1 )R(u1 , 1)du1
0
0
Z 1Z 1
ḣ1 (u1 )ḣ2 (u2 )R(u1 , u2 )du1 du2 .
+
0
0
Lemma 2. If Assumptions 5.1 and 5.2 are satisfied, then with probability approaching
P
one all the eigenvalues of kD, where D = T −1 Tt=1 Bt Bt0 , are between two positive
constants.
Proof. The proof is given in two parts. In step (a), it will be shown that the matrix
k D̄, with D̄ = E[D], has all eigenvalues between two positive constants. This requires
the use of part (a) of Assumption 5.2. In step (b), it will then be verified that
kkD−D̄k → 0 in probability. Under the α-mixing condition of part (b) in Assumption
5.2, this requires that k 3 /T → 0 as assumed in Theorem 5.1. The relationship between
k and T can be improved further: Under the simplified ρ-mixing condition
sup sup Corr h1 (Xt ), h2 (Xt+h ) ≤ ρ(h),
t
only k 2 /T → 0 and
h1 ,h2
P∞
h=0
ρ(h) < ∞ are needed. Under the simplified Ψ-mixing
condition
sup sup ft,t+h (x1 , x2 ) − ft (x1 )ft+h (x2 ) ≤ Ψ(h),
t x1 ,x2
P
a further simplification to k/T → 0 and ∞
h=0 Ψ(h) < ∞ is possible. A good summary
of properties of various strong mixing conditions may be found in Bradley (2005).
(a) Note that for any vector β = (β00 , β10 )0 ,
T Z
2
1X 0
0
β D̄β =
β0 B0 (t/T ) + β10 Φ1 (x) ft (x)dx.
T t=1
Part (a) of Assumption 5.2 implies that
T Z
0
2
f X
0
β0 B0 (t/T ) + β10 Φ1 (x) dx,
β D̄β ≥
T t=1
T Z
2
f X 0
0
β D̄β ≤
β0 B0 (t/T ) + β10 Φ1 (x) dx.
T t=1
R
Since β 0 Φ(x)dx = 0, it follows that
T Z
2
1X 0
β0 B0 (t/T ) + β10 Φ1 (x) dx
T t=1
2
T
T Z
2 1 X
2
0
1 X 0
=
β0 B0 (t/T ) +
β1 Φ1 (x) dx
T t=1
T t=1
X
Z
T
0
0 1
0
0
B1 (x)B1 (x) dx Ψ01 β1 .
= β0
B0 (t/T )B0 (t/T ) β0 + β1 Ψ1
T t=1
P
Following the arguments of Burman (1991), all eigenvalues of kT −1 t B0 (t/T )B0 (t/T )0
R
and B1 B10 are bounded between two positive constants. Since Ψ1 Ψ01 = I, it follows
that all eigenvalues of
T Z
2
kX 0
β0 B0 (t/T ) + β10 Φ1 (x) dx
T t=1
and hence those of k D̄ are bounded between two positive constants as well.
(b) Note that it is enough to show that kkV01 k → 0 in probability and kkV11 k → 0
in probability, where
V01
T
0
1X
=
B0 (t/T ) B1 (Xt ) − E[B1 (Xt )] ,
T t=1
V11
T
1 X
=
B1 (Xt )B1 (Xt )0 − E[B1 (Xt )B1 (Xt )0 ] .
T t=1
The result is proved if it can be shown that k 2 E[tr(V010 V01 )] → 0 and k 2 E[tr(V110 V11 )] →
0. Observe that
k
2
E[tr(V010 V01 )]
2 X X
T
k
=
B0i (s/T )B0i (t/T )Cov(B1j (Xs ), B1j (Xt ))
T
i,j s,t=1
2 X X
k
=
B0i (s/T )B0i (t/T )Cov(B1j (Xs ), B1j (Xt )),
T
i,j s,t∈I
i
where Ii ⊂ {1, . . . , T } and B0i (t/T ) 6= 0 for t ∈ Ii . Then, Ti = card(Ii ) ∼ T /k.
Furthermore,
Cov B1j (Xs ), B1j (Xt ) =
Z
1
B1j (x1 )B1j (x2 ) fs,t (x1 , x2 ) − fs (x1 )ft (x2 ) dx1 dx2 ,
0
where fs,t is the joint probability density function of Xs and Xt . Now let r = fs,t −fs ft
and h1 = h2 = B1,j . Since |R(x1 , x2 )| = |Fs,t (x1 , x2 ) − Fs (x1 )Ft (x2 )| ≤ α(|s − t|) by
R
part (b) of Assumption 5.2 and since |Ḃ1j | ≤ c for all j and some constant c > 0,
3
Lemma 1 yields that |Cov(B1j (Xs ), B1j (Xt ))| = O(1)α(|s − t|). Hence,
k
2
E[tr(V010 V01 )]
2 X X
k
= O(1)
B0i (s/T )B0i (t/T )α(|s − t|)
T
i,j s,t∈Ii
2 X X
2 2 X
∞
∞
k
k k T
= O(1)
Ti
α(h)
α(h) = O(1)
T
T
k h=0
i,j
h=0
∞
k3 X
= O(1)
α(h) = o(1).
T h=0
In the above arguments, it was used that B0 (u) ≤ 1 for all u and that
P∞
h=0
α(h) < ∞
by 5.2. Next, consider
k
2
2 X X
T
k
Cov B1i (Xs )B1j (Xs ), B1i (Xs )B1j (Xt ) .
=
T
i,j s,t=1
E[tr(V110 V11 )]
Using Lemma 1 again, this time with r = fs,t − fs ft and h1 = h2 = B1i B1j , it follows
that |Cov B1i (Xs )B1j (Xs ), B1i (Xs )B1j (Xt ))| ≤ α(|s − t|) for all i and j for some
constant c > 0. Thus,
k
2
E[tr(V110 V11 )]
2 X X
T
k
α(|s − t|),
= O(1)
T
s,t=1
|i−j|≤2d
where the double over i and j has been reduced to a sum over |i − j| ≤ 2d, since
B1i (x)B1j (x) = 0 whenever |i − j| > 2d. Consequently,
k
2
E[tr(V110 V11 )]
∞
k3 X
α(h) = o(1).
= O(1)
T h=0
This completes the proof.
Lemma 3. If Assumptions 5.1 and 5.2 are satisfied, then
(a)
(b)
ST,1
T
k
1X
(m̂t − m̄t )2 = OP
,
=
T t=1
T
ST,2
T
1X
=
(m̄t − mt )2 = OP k −2ν ,
T t=1
as T → ∞.
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Proof. (a) Note that
ST,1
X
2
1 T
= (β̂ − β̄) D(β̂ − β̄) ≤ kD kkβ̂ − β̄k = OP (k)
Bt Zt ,
T t=1
0
−1
2
(1.1)
where the OP (k) rate follows from Lemma 2. Now
2 2 2
T
T
T
1 X
1 X
1 X
Bt Zt B0 (t/T )Zt Φ1 (Xt )Zt T
= T
+ T
.
t=1
t=1
t=1
Observe that for the first squared norm on the right-hand side, it holds that
2 X
T
1 T
1 XX 2
σZ2
2
E =
≤
B
Z
(t/T
)σ
,
B
t t
Z
0j
T
2
T
T
t=1
j t=1
P 2
on account of j B0j (u) ≤ 1. Next, using Φ1 (x) = Ψ1 B1 (x), it follows that
2
2 2
T
T
T
1 X
1 X
1 X
2
≤ kΨ1 k ≤
,
Φ
(X
)Z
B
(X
)Z
B
(X
)Z
1
t
t
1
t
t
1
t
t
T
T
T
t=1
t=1
t=1
(1.2)
where the last step is justified since kΨ1 k ≤ 1. Now
2 X X
X
2 T
1 T
1
E B1 (Xt )Zt =
B1j (Xt )Zt
E
T t=1
T t=1
j
=
T
X σ2 X
2
σZ2
Z
,
E
B
(X
)
≤
t
1j
2
T
T
t=1
j
(1.3)
where the inequality is implied by
E B1j (Xs )B1j (Xt )Zs Zt =


0,
s 6= t,

σ 2 E[B 2 (Xt )], s = t,
1j
Z
and
P
j
2
(x) ≤ 1 for any x. Equations (1.1)–(1.3) consequently show that ST,1 =
B1j
OP (k/T ) and part (a) is verified.
(b) There are splines s∗0 and s∗1 with k equispaced knots (of degree d) such that
kg0 − s∗0 k∞ = OP (k −ν )
and
kg1 − s∗1 k∞ = OP (k −ν ).
R
R
Let s̃1 = s∗1 − s∗1 (x)dx, then s̃1 (x)dx = 0. Since ḡ0 (t/T ) = β̄00 B0 (t/T ) and ḡ1 (Xt ) =
P
β̄10 Φ1 (Xt ) are obtained by minimizing T −1 t [mt − β̄00 B0 (t/T ) − β̄10 Φ1 (Xt )]2 , it follows
that
ST,2
T
2
1 X
mt − β̄00 B0 (t/T ) − β̄10 Φ1 (Xt )
=
T t=1
5
≤
T
2
1 X
mt − s∗0 (t/T ) − s̃∗1 (Xt )
T t=1
=
T
i2
1 X h
g0 (t/T ) − s∗0 (t/T ) + g1 (Xt ) − s̃∗1 (Xt )
T t=1
T
T
2 2 X
2
2 X
∗
≤
g0 (t/T ) − s0 (t/T ) +
g1 (Xt ) − s̃∗1 (Xt )
T t=1
T t=1
≤ 2kg0 − s∗0 k2∞ + 2kg1 − s̃∗1 k2∞ .
R
Now, kg0 − s∗0 k∞ = OP (k −ν ) by construction and kg1 − s̃∗1 k∞ ≤ kg1 − s∗1 k∞ + | s∗1 | ≤
R
R
R
2kg1 − s∗1 k∞ = OP (k −ν ) because | s∗1 | = | g1 − s∗1 | ≤ kg1 − s∗1 k∞ . This shows that
ST,2 = OP (k −2ν ), as required for part (b).
Proof of Theorem 5.1. The assertion follows from Lemma 3, noticing that the two
rates are balanced if k is of the order T 1/(2ν+1) . In this case, k/T = k −2ν = T −2ν/(2ν+1) .
This completes the proof.
References
[1] Noguchi, K., Aue, A., and Burman, P. (2014), “Exploratory Analysis and Modeling of
Stock Returns,” Journal of Computational and Graphical Statistics.
[2] Bradley, R.C. (2005), “Basic Properties of Strong Mixing Conditions: a Survey and
Some Open Questions,” Probability Surveys, 2, 107–144.
[3] Burman, P. (1991), “Regression Function Estimation from Dependent Observations,”
Journal of Multivariate Analysis, 36, 263–279.
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