Supplementary material for “Subvector Inference When
the True Parameter Vector is near the Boundary”
C
Sufficient conditions
We present sufficient conditions for Assumptions 1 and 4. They are taken from Andrews
(1999) (A1) with appropriate notational adjustments to allow for drifting sequences of true
parameters. The following Assumption corresponds to Assumption 1 in A1.
Assumption 5. Under {γn } ∈ Γ(γ ∗ ), θ̂n − θn = op (1).
The following Assumption is sufficient for Assumption 5 and corresponds to Assumptions
1 (a) and 1∗ (b∗ ) in A1.
∗
Assumption 5∗ .
(i) Under {γn } ∈ Γ(γ ∗ ), supθ∈Θ |Qn (θ) − Q(θ; γ ∗ )| = op (1) for some non-stochastic realvalued function Q(θ; γ ∗ ).
(ii) Q(θ; γ ∗ ) is continuous on Θ ∀γ ∗ ∈ Γ.
(iii) Q(θ; γ ∗ ) is uniquely minimized by θ∗ ∀γ ∗ ∈ Γ.
(iv) Θ is compact.
The following Assumption corresponds to Assumption 2∗ in A1.
Assumption 6. Under {γn } ∈ Γ(γ ∗ ),
1
Qn (θ) = Qn (θn ) + DQn (θn )0 (θ − θn ) + (θ − θn )0 D2 Qn (θn )(θ − θn ) + Rn (θ),
2
where
sup
θ∈Θ:k(θ−θn )k≤n
|nRn (θ)|
√
= op (1)
(1 + k n(θ − θn )k)2
for all constants n → 0.
The following Assumption is sufficient for Assumption 6 and corresponds to (a variant
of) Assumptions 22∗ in A1.
1
Assumption 6∗ .
(i) Qn (θ) has continuous left/right (l/r) partial derivatives of order two on Θ ∀n ≥ 1 with
probability 1.
(ii) Under {γn } ∈ Γ(γ ∗ ), for all constants n → 0,
2
∂
∂2
sup
∂θ∂θ0 Qn (θ) − ∂θ∂θ0 Qn (θn ) = op (1),
θ∈Θ:kθ−θn k≤n
where (∂/∂θ)Qn (θ) and (∂ 2 /∂θ∂θ0 )Qn (θ) denote the J × 1 vector and J × J matrix of
l/r partial derivatives of Qn (θ) of orders one and two, respectively.
Note that Assumption 6∗ is also sufficient for Assumption 4. The following Lemma corresponds to Theorem 1 in A1 and shows that Assumptions 5 and 6 together with Assumptions
2 and 3 are sufficient for Assumption 1.
√
Lemma 4. Under {γn } ∈ Γ(γ ∗ ) and Assumptions 2, 3, 5, and 6, n(θ̂n − θn ) = Op (1).
For sake of completeness, we reproduce the proof.
1/2 √
Proof of Lemma 4. Let Jn = D2 Qn (θn ) and κn = Jn
n(θ̂n − θn ). We have
op (1) = n Qn (θ̂n ) − inf Qn (θ)
θ∈Θ
≥ n Qn (θ̂n ) − Qn (θn )
√
1
nDQn (θn )0 Jn−1/2 κn + kκn k2 + nRn (θ̂n )
2
1
2
= Op (kκn k) + kκn k + (1 + kJn−1/2 κn k)2 op (1)
2
1
= Op (kκn k) + kκn k2 + op (kκn k) + op (kκn k2 ) + op (1).
2
=
The first equality holds by definition of θ̂n , see equation (3). The inequality then follows by
the definition of the infimum. The next equality follows from Assumption 6. The following
equality holds by Assumption 2 and 3 and Assumption 6. Upon rearrangement and after
dropping op (kκn k) and op (kκn k2 ), we get
kκn k2 ≤ 2kκn kOp (1) + op (1).
Let ξn denote the Op (1) term. Then,
(kκn k − ξn )2 ≤ ξn2 + op (1) = Op (1).
2
Taking square roots gives kκn k ≤ Op (1). Given Assumption 2, this establishes the result.
D
Proposition 1
We reproduce Theorem 3(b) in A1 with Θ defined as below Theorem 1 and with notational
adjustments to allow for drifting sequences of true parameters. To that end, we first introduce
√
some additional notation. Let Jn = D2 Qn (θn ) and Zn = −Jn−1 nDQn (θn ) and note that by
Assumptions 2 and 3, we have
d
Zn → Z(γ ∗ ) ∼ N (0, J(γ ∗ )−1 V (γ ∗ )J(γ ∗ )−1 ).
(18)
Under Assumption 6, the objective function can be written as
Qn (θ) = Qn (θn ) −
√
1
1 0
Zn Jn Zn + qn ( n(θ − θn )) + Rn (θ),
2n
2n
(19)
where
qn (λ) = (λ − Zn )0 Jn (λ − Zn ).
Under Assumption 1, the remainder term, Rn (θ̂n ), is asymptotically negligible and the
√
asymptotic distribution of n(θ̂n − θn ) under {γn } ∈ Γ(γ ∗ ), as formalized in Proposition 1
below, is given by the distribution of
λ̂ = arg min q(λ),
(20)
λ∈Λ(γ ∗ )
where
q(λ) = (λ − Z(γ ∗ ))0 J(γ ∗ )(λ − Z(γ ∗ ))
√
and where Λ(θ∗ ) = limn→∞ n(Θ − θn ), see Remark 2 for the exact form.
√
d
Proposition 1. Under {γn } ∈ Γ(γ ∗ ) and Assumptions 2, 3, 5, and 6, n(θ̂n − θn ) → λ̂,
where λ̂ is defined in (20).
For sake of completeness, we reproduce the proof of Proposition 1 given in A1. To that
end, we first introduce several Lemmas, which are also taken from A1.
√
√
Lemma 5. Under {γn } ∈ Γ(γ ∗ ) and Assumptions 1-3, qT ( n(θ̂n − θn )) = inf θ∈Θ qT ( n(θ −
θn )) + op (1).
Lemma 5 corresponds to Theorem 2(e) in A1. Note that
√
inf qn ( n(θ − θn )) =
θ∈Θ
3
qn (λ),
√ inf
λ∈ n(Θ−θn )
where
√
n(Θ − θn ) = {λ ∈ RJ : λ =
√
n(θ − θn ) for some θ ∈ Θ}.
Proof of Lemma 5. Let θ̂nq satisfy θ̂nq ∈ cl(Θ) and
√
√
qn ( n(θ̂nq − θn )) = inf qn ( n(θ − θn )) + op (1).
θ∈Θ
(21)
With this definition in place, it is sufficient to show that
First, we show that
1/2 √
Let κqn = Jn
√
√
qn ( n(θ̂n − θn )) = qn ( n(θ̂nq − θn )) + op (1).
(22)
√ q
n(θ̂n − θn ) = Op (1).
(23)
n(θ̂nq − θn ). We have
q
√
κn − Jn1/2 Zn 2 = qn ( n(θ̂nq − θn )) ≤ qn (0) + op (1) = Zn0 Jn Zn + op (1) = Op (1),
where the inequality follows from equation (21) and the last equality follows from Assump1/2
tions 2 and 3. It follows that κqn = Jn Zn + Op (1) = Op (1), where the last equality follows
from Assumptions 2 and 3 again. Then, by Assumption 2, equation (23) follows.
Then, by equation (19), Assumption 1, and equation (23), we have
1
1 √
nQn (θ̂n ) = nQn (θn ) + Zn0 Jn Zn + qn ( n(θ̂n − θn )) + op (1)
2
2
(24)
and
1 √
1
nQn (θ̂nq ) = nQn (θn ) + Zn0 Jn Zn + qn ( n(θ̂nq − θn )) + op (1).
2
2
The result of the Lemma then follows, since
(25)
op (1) = n Qn (θ̂n ) − inf Qn (θ) ≥ n Qn (θ̂n ) − Qn (θ̂nq )
θ∈Θ
1 √
1 √
= qn ( n(θ̂n − θn )) − qn ( n(θ̂nq − θn )) + op (1)
2
2
√
1
1 √
≥ inf qn ( n(θ − θn )) − qn ( n(θ̂nq − θn )) + op (1) = op (1),
2 θ∈Θ
2
where the first and last equality follow from equations (3) and (21), respectively. The
inequalities follow from the definition of the infimum, while the equality in the middle follows
from equations (24) and (25).
√
Lemma 6. If n(Θ − θn ) → Λ as n → ∞, where Λ is a convex cone with possible nonzero
vertex, then inf λ∈√n(Θ−θn ) qn (λ) = inf λ∈Λ qn (λ) + op (1).
4
Lemma 6 corresponds to Lemma 2 in A1.
Proof of Lemma 6. For any set ∆ ⊂ RJ and any z ∈ RJ , let
dist(z, ∆) ≡ inf kλ − zk
λ∈∆
and
1
distn (z, ∆) ≡ inf ((λ − z)0 Jn (λ − z)) 2 .
λ∈∆
Note that
1
distn (Zn , Λ) = inf qn2 (λ)
λ∈Λ
and
√
distn (Zn , n(Θ − θn )) =
1
√ inf
qn2 (λ).
λ∈ n(Θ−θn )
√
Let Cn = distn (Zn , Λ) − distn (Zn , n(Θ − θn )). Then, it suffices to show that Cn = op (1).
√
√
Note that for any ZΘ,n ∈ n(Θ − θn ), we have dist(ZΘ,n , Λ) = o(1), since n(Θ − θn ) → Λ.
p
This, together with Jn → J(γ ∗ ), where J(γ ∗ ) is of full rank, implies that distn (ZΘ,n , Λ) =
√
√
op (1). Now, choose ZΘ,n ∈ n(Θ−θn ) such that distn (Zn , n(Θ−θn )) = distn (Zn , {ZΘ,n })+
op (1). Such an element always exists. Then, by the triangle inequality
√
Cn = distn (Zn , Λ) − distn (Zn , n(Θ − θn ))
√
≤ distn (Zn , {ZΘ,n }) + distn (ZΘ,n , Λ) − distn (Zn , n(Θ − θn )) = op (1).
Analogously, choose ZΛ,n ∈ Λ such that distn (Zn , Λ) = distn (Zn , {ZΛ,n }) + op (1). Again, we
√
√
√
have dist(ZΛ,n , n(Θ − θn )) = o(1), since n(Θ − θn ) → Λ, and distn (ZΛ,n , n(Θ − θn )) =
op (1). By the triangle inequality
√
Cn = distn (Zn , Λ) − distn (Zn , n(Θ − θ0 ))
√
≥ distn (Zn , Λ) − distn (Zn , {ZΛ,n }) − distn (ZΛ,n , n(Θ − θn )) = op (1).
Given the comment below Lemma 5, it follows from Lemma 5 and 6 that
√
qn ( n(θ̂n − θn )) = inf qn (λ) + op (1).
λ∈Λ
(26)
Proof of Proposition 1. The proof of Proposition 1 follows from the proof of Theorem 3(a)-
5
(b) in A1. Let λ̂n be such that λ̂n ∈ cl(Λ) and
qn (λ̂n ) = inf qn (λ).
λ∈Λ
Then, equation (26) reads
√
qn ( n(θ̂n − θn )) = qn (λ̂n ) + op (1).
(27)
√
The proof proceeds by showing that n(θ̂n −θn ) = λ̂n +op (1). The desired result then follows
by the continuous mapping theorem, see the proof of Theorem 3(b) in A1. By convexity of
Λ, there exists a unique λ∗n ∈ cl(Λ) such that
√
k n(θ̂n − θn ) − λ∗n k = op (1).
(28)
Therefore, it suffices to show that kλ∗n − λ̂n k = op (1). Define k · kn by kxkn = (x0 Jn x)1/2 .
Then, by Assumption 2 it suffices to show that
kλ∗n − λ̂n kn = op (1).
We have
√
kλ∗n − Zn kn = k n(θ̂n − θn ) − Zn kn + op (1) = kλ̂n − Zn kn + op (1),
where the first equality follows from the triangle inequality and the fact that equation (28)
also holds with k · k replaced by k · kn by Assumption 2. The last equality holds by equation
(27), which as mentioned above relies on Lemmas 5 and 6. The rest of the proof follows by
the arguments presented below equation (7.19) in the proof of Theorem 3(a) in A1 with T
replaced by n. The proof goes through despite Λ being a cone with possible nonzero vertex,
as convexity constitutes the crucial property of Λ.
6