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C
2007 Biometrika Trust Printed in Great Britain
Supplementary Material for “On the Asymptotic Efficiency
of Approximate Bayesian Computation Estimators”
BY WENTAOLI
School of Mathematics, Statistics and Physics, Newcastle University, Newcastle upon Tyne NE1 7RU, U.K.
AND PAULFEARNHEAD
Department of Mathematics and Statistics, Lancaster University, Lancaster LA1 4YF, U.K.
Technical lemmas and proofs of the main results are presented. In the following, the data are considered to be random, andO(·)andΘ(·)denote the limiting behaviour whenngoes to∞. For two setsAandB, the sum of integrals´Af(x)dx+´Bf(x)dxis written as(´A+´B)f(x)dx. Recall thatTobs=anA(θ0)−1/2{sobs−s(θ0)} and by Condition4,Tobs→N(0, Id) in
distri-bution, whereIdis the identity matrix with dimensiond. For a constantd×pmatrixA, let the
minimum and maximum eigenvalues ofATA beλ2min(A) andλ2max(A). Obviously for anyp -dimension vectorx,λmin(A)kxk ≤ kAxk ≤λmax(A)kxk. For two matricesAandB, we sayA
is bounded byBifλmax(A)≤λmin(B).
1. PROOF OFRESULTS FROMSECTION3 1·1. Overview and Notation
We first give an overview of the proof to Theorem1. The convergence of the maximum like-lihood estimator based on the summary follows almost immediately fromCreel & Kristensen
(2013). The minor extensions we used are summarized in Lemmas1and2below.
The main challenge with Theorem1are the results about the posterior mean of approximate Bayesian computation. For the convergence of posterior means of approximate Bayesian com-putation we need to consider convergence of integrals over the parameter space,Rp. We will divideRp into Bδ ={θ:kθ−θ0k< δ} andBcδ for some δ < δ0, and introduce the notation
π(h) =´h(θ)π(θ)fABC(sobs |θ)dθ. The posterior mean of approximate Bayesian computa-tion ishABC=π(h)/π(1). We can writeπ(h), say, asπ(h) =πBδ(h) +πBδc(h), where
πBδ(h) =
ˆ
Bδ
h(θ)π(θ)fABC(sobs|θ)dθ, πBc δ(h) =
ˆ
Bc δ
h(θ)π(θ)fABC(sobs |θ)dθ.
As n→ ∞ the posterior distribution of approximate Bayesian computation concentrates around θ0. The first step of our proof is to show that, as a result, the contribution that comes from integrating overBcδcan be ignored. Hence we need consider onlyπBδ(h)/πBδ(1).
Second, we perform a Taylor expansion ofh(θ)aroundθ0. LetDh(θ)andHh(θ)denote the vector of first derivatives and the matrix of second derivatives ofh(θ)respectively. Then
h(θ) =h(θ0) +Dh(θ0)T(θ−θ0) + 1
2(θ−θ0)
THh(θ∗)(θ−θ
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for some θ∗, that depends on θ and that satisfies ||θ∗−θ0||<||θ−θ0||. We plug this into
πBδ(h), but re-express the integrals in term of the rescaled random vector
t(θ) =an,ε(θ−θ0),
and lett(Bδ)be the set{φ:φ=t(θ)for someθ∈Bδ}. This gives
πBδ(h)
πBδ(1)
=h(θ0) +a−n,ε1Dh(θ0)T πBδ(t)
πBδ(1)
+1 2a
−2
n,ε
πBδ{t
THh(θ t)t}
πBδ(1)
, (1)
where we writetfort(θ), andθtis the valueθ∗from remainder term in the Taylor expansion for
h(θ). We use the notationθtto emphasize its dependence ont, and note thatθtbelongs toBδ.
Let feABC(sobs |θ) =
´
e
fn(sobs+εnv |θ)K(v)dv, which is the likelihood approximation
that we get if we replace the true likelihood by its Gaussian limit, and define eπBδ(h) =
´
Bδh(θ)π(θ)feABC(sobs |θ)dθ. Our third step is to re-write (1) as
πBδ(h)
πBδ(1)
=h(θ0) +a−n,ε1Dh(θ0)T eπBδ(t) e πBδ(1)
+a−n,ε1Dh(θ0)T
e πBδ(t)
e πBδ(1)
− πBδ(t)
πBδ(1)
+1 2a
−2
n,ε
πBδ{t
THh(θ t)t}
πBδ(1)
.
We bound the size of the last two terms, so that asymptoticallyhABCbehaves as
h(θ0) +a−n,ε1Dh(θ0)T e
πBδ(t)
e πBδ(1)
.
If we introduce the densitygn(t, v), defined asgn(t, v, τ)in Section4·3of the main text but with
τ = 0, so
gn(t, v)∝
N
n
Ds(θ0)t;anεnv+A(θ0)1/2Tobs, A(θ0)
o
K(v), anεn→c <∞,
NnDs(θ0)t;v+an1εnA(θ0)1/2Tobs,a21
nε2nA(θ0) o
K(v), anεn→ ∞,
then we can show that
e πBδ(t)
e πBδ(1)
≈ ´
t(Bδ)
´
Rdtgn(t, v)dtdv
´
t(Bδ)
´
Rdgn(t, v)dtdv ,
with a remainder that can be ignored. Putting this together, we get that asymptoticallyhABCis
h(θ0) +a−n,ε1Dh(θ0)T ´
t(Bδ)
´
Rdtgn(t, v)dtdv
´
t(Bδ)
´
Rdgn(t, v)dtdv ,
and the proof finishes by calculating the form of this.
A recurring theme in the proofs for the bounds on the various remainders is the need to bound expectations of polynomials of either the rescaled parameter t, or a rescaled difference in the summary statistic fromsobs, or both. Later we will present a lemma, stated in terms of a general polynomial, that is used repeatedly to obtain the bounds we need.
To define this we need to introduce a set of suitable polynomials. For any integerland vector
x, if a scalar function ofxhas the expressionPl
i=0αi(x, n)Txi, where for eachi,xidenotes the
vector with all monomials ofxwith degreeias elements andαi(x, n)is a vector of functions of
xandn, we denote it byPl(x). LetPl,xbe the set
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To simplify the notations, for two vectorsx1andx2,Pl{(xT1, xT2)T}andPl,(xT
1,xT2)T are written asPl(x1, x2)andPl,(x1,x2). Where the specific form of the polynomial does not matter, and we only use the fact that it lies inPl,x, we will often simplify expressions by writing it asPl(x).
1·2. Proof of Theorem1
For the maximum likelihood estimator based on the summary, Creel & Kristensen (2013) gives the central limit theorem forθˆMLES when an=n1/2 andPis compact. According to the
proof inCreel & Kristensen (2013), extending the result to the generalan is straightforward.
Additionally, we give the extension for generalP.
LEMMA1. Assume Conditions1,4-6. Thenan(ˆθMLES−θ0)→N{0, I−1(θ0)}in distribution
asn→ ∞.
Given Condition3, by Lemma1and the delta method (Lehmann,2004), the convergence of the maximum likelihood estimator for generalh(θ)holds as follows.
LEMMA2. Assume the conditions of Lemma 1 and Condition 3. Then an{h(ˆθMLES)−
h(θ0)} →N{0, Dh(θ0)TI−1(θ0)Dh(θ0)}in distribution asn→ ∞.
The following lemmas are used for the result about the posterior mean of approximate Bayesian computation, proofs of these are given in Section1·3. Our first lemma is used to justify ignoring integrals overBδc.
LEMMA3. Assume Conditions2,3–6. Then for anyδ < δ0, πBc
δ(h) =Op(e
−aαδn,εcδ)for some positive constantscδandαδdepending onδ.
The following lemma is used to calculate the form of ´
t(Bδ)
´
Rdtgn(t, v)dtdv
´
t(Bδ)
´
Rdgn(t, v)dtdv ,
which is the leading term for{hABC−h(θ0)}.
LEMMA4. Assume Condition2. Let cbe a constant vector,{kn} be a series converging to
k∞∈(0,∞]and{b0n}be a series converging to a non-negative constant. Letbn=1{k∞=∞}+
b0n1{k∞<∞}. Then for anyd×pconstant matrixAand anyd×dconstant matrixB,
ˆ
Rp
ˆ
Rd
t N(At;Bnv+
1
knc,
1
k2
nId)K(v)
´
Rp
´
RdN(At;Bnv+
1
knc,
1
k2
nId)K(v)dtdv
dtdv= 1
kn
(ATA)−1ATc+R(A, Bn, kn, c) ,
where Bn=bnB, the expression of R(c;A, Bn, kn) is stated in the proof. Specifically,
R(A, Bn, kn, c) =o(1)whenBn=o(1)andO(1)otherwise.
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LEMMA5. Assume Conditions1,2and4hold. Ifεn=o(an−1/2), there exists aδ < δ0such
that
e
πBδ(1) =a
d−p n,ε
n π(θ0)
ˆ
t(Bδ)
ˆ
Rd
gn(t, v)dvdt+Op(an,ε−1) +Op(a2nε4n)
o ,
ˆ
t(Bδ)
ˆ
Rd
gn(t, v)dtdv= Θp(1),
e πBδ(t)
e πBδ(1)
= ´
t(Bδ)
´
Rdtgn(t, v)dtdv
´
t(Bδ)
´
Rdgn(t, v)dtdv
+Op(a−n,ε1) +Op(a2nε4n), (2)
andeπBδ{P2(t)}/eπBδ(1) =Op(1)for anyP2(t)∈P2,t.
LEMMA6. Assume the conditions of Lemma 5 and Conditions 3 and 5. Then if εn=
o(an−1/2), there exists aδ < δ0such that
πBδ(h)
πBδ(1)
=h(θ0) +a−n,ε1Dh(θ0)T
e πBδ(t)
e πBδ(1)
+Op(α−n1)
+1 2a
−2
n,ε
e πBδ{t
THh(θ t)t}
e πBδ(1)
+Op(α−n1)
,
(3) Now we are ready to prove Theorem1.
Proof of Theorem1. The convergence of the maximum likelihood estimator based on the sum-mary is given by Lemma1and Lemma2.
We now focus on the convergence for the posterior mean of approximate Bayesian compu-tation. The convergence of the posterior mean given the summaries follows from a similar, but simpler, argument and is omitted.
We can boundtTH(θt)tforθinBδby the quadratictTHmaxt, whereHmaxis an upper bound
onH(θt)forθtinBδ. This means that
e πBδ{t
THh(θ
t)t}=O(1).
Together with Lemmas3,5and6, we then have the expansion
hABC=h(θ0) +a−n,ε1Dh(θ0)T
(´
t(Bδ)×Rdtgn(t, v)dtdv
´
t(Bδ)×Rdgn(t, v)dtdv
+Op(a−n,ε1) +Op(a2nε4n) +Op(α−n1)
)
.
The analytical form of the integral in the above expansion, which we will denote byEgn(t),
can be obtained by applying Lemma4withA=A(θ0)−1/2DS(θ0),c=Tobs,
Bn=
(
anεnA(θ0)−1/2, cε<∞,
A(θ0)−1/2, cε=∞,
kn=
(
1, cε<∞,
anεn, cε=∞.
It can be seen that Egn(t) is Θp(k
−1
n ), and the remainder term, Op(a−n,ε1) +Op(a2nε4n) +
Op(α−n1), isop(1)asεn=o(a−n3/5)andα−n1 =o(a
−2/5
n ). Then sincea−n,ε1k−n1 =a−n1, we have
an{hABC−h(θ0)}
=Dh(θ0)T
hn
Ds(θ0)TA(θ0)−1Ds(θ0)
o−1
Ds(θ0)TA(θ0)−1/2Tobs+Rn(anεn, Tobs)
i
+op(1),
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where Rn(anεn, Tobs) is Dh(θ0)TR(A, Bn, kn, c) with R(A, Bn, kn, c) defined in Lemma4.
We can interpret Rn(anεn, Tobs) as the extra variation brought by εn: an[hABC−E{h(θ)|
sobs}].
By the delta method, the first term in the right hand side of (4) converges toI(θ0)−1/2Z. For the second term, sinceA(ATA)−1AT is a projection matrix, by eigen decompositition
I−A(ATA)−1AT =U
0 0 0Id−p
UT, (ATA)−1/2AT =
Ip0
0 0
UT,
whereUis an orthogonal matrix. For a vectorx, letxk1:k2be the(k2−k1+ 1)-dimension vector containing thek1th–k2th coordinates ofx. Letv0 =UTA(θ0)−1/2v, andTobs0 =UTTobs. Then
Rn(anεn, Tobs)can be written as
Rn(anεn, Tobs)
=Dh(θ0)T(ATA)−1/2anεn
´
v01:pN{v(0p+1):d;− 1
anεnT
0
obs,(p+1):d,
1
a2
nε2nId−p}K{A(θ0)
1/2U v0}dv0
´
N{v(0p+1):d;−a1
nεnT
0
obs,(p+1):d,
1
a2
nε2nId−p}K{A(θ0)
1/2U v0}dv0 .
(5) Denote the weak limit ofRn(anεn, Tobs)asR(cε, Z). Whend=p, obviouslyRn(anεn, Tobs) = 0 and therefore R(cε, Z) = 0. When d > p, if εn=o(1/an), Rn(anεn, Tobs) =op(1) by
Lemma4and thereforeR(cε, Z) = 0. When the covariance matrix ofK(·)isc2A(θ0), for
con-stant c >0, K(v)∝K{ckA(θ0)−1/2vk2}. Then K{A(θ0)1/2U v0} in (5) can be replaced by
K(ckv0k2)and for fixedv0
(p+1):d, the integrand in the numerator, as a function ofv
0
1:p, is
sym-metric around zero. ThereforeRn(anεn, Tobs) = 0andR(cε, Z) = 0.
Otherwise,Rn(anεn, z)is not necessarily zero. Since for anyn,Rn(anεn, z)as a function of
zis symmetric around0,R(cε, z)is also symmetric andR(cε, Z)has mean zero. SinceI−1(θ0)
is the Cramer-Rao lower bound, var{I(θ0)−1/2Z+R(cε, Z)} ≥I−1(θ0).
For (i), the asymptotic normality holds forh(ˆθ)by Lemma2.
1·3. Proof of Lemmas
Here we give the proofs of lemmas from Section1·2.
Proof of Lemma3. It is sufficient to show that for any δ, supθ∈Bc
δfABC(sobs |θ) =
Op(e−a
αδ
n,εcδ). By dividingRdinto{v:kε
nvk ≤δ0/3}and its complement, we have
sup
θ∈Bc δ
fABC(sobs|θ) = sup
θ∈Bc δ
ˆ
Rd
fn(sobs+εnv|θ)K(v)dv
≤ sup
θ∈Bcδ\Pc
0
(
sup
ks−sobsk≤δ0/3
fn(s|θ)
)
+ sup
θ∈Pc
0
(
sup
ks−sobsk≤δ0/3
fn(s|θ)
)
+ ¯K(λmin(Λ)ε−n1δ0/3)ε−nd,
whereλmin(Λ)is positive. In the above, asn→ ∞, the third term is exponentially decreasing by Conditions2(iv). For the second term, by Condition4, with probability1,
ks−s(θ)k=k{s(θ0)−s(θ)}+{sobs−s(θ0)}+εnvk
≥δ0−δ0/3−δ0/3 =δ0/3.
Recall that Wn(s) =anA(θ)−1/2{s−s(θ)}. Then by Condition 6, the second term is
expo-nentially decreasing. For the first term, when θ∈Bδc\Pc
0 and ks−sobsk ≤δ0/3, kWn(s)k ≥
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sum of a normal density and α−n1rmax(w), which are both exponentially decreasing, so supθ∈Bc
δ\P c
0supks−sobsk≤δ0/3fn(s|θ) is also exponentially decreasing. Finally, the sum of all the above isO(e−aαδn,εcδ)by noting thata
n,ε≤min(ε−n1, an).
The following additional lemma will be used repeatedly to bound error terms that appear in Lemmas5and6.
LEMMA7. Assume Condition 2. For t∈Rp and v∈
Rd, let {An(t)} be a series of d×p
matrix functions,{Cn(t)}be a series ofd×dmatrix functions,Qbe a positive definite matrix
andg1(v) andg2(v)be probability densities inRd. Letcbe a random vector,{kn}be a series
converging to k∞∈(0,∞] and {b0n} be a series converging to a non-negative constant. Let
bn=1{k∞=∞}+b
0
n1{k∞<∞}. If
(i)g1(v)andg2(v)are bounded inRd;
(ii)g1(v)andg2(v)depend onvonly throughkvkand are decreasing functions ofkvk;
(iii) there exists an integerlsuch that´ Ql+p
k=1vikgj(v)dv <∞,j= 1,2, for any coordinates
(vi1,· · · , vil)ofv;
(iv) there exists a positive constant m such that for any t∈Rp and n, λ
min{An(t)} and
λmin{Cn(t)}are greater thanm;
then for anyPl(t, v)∈Pl,(t,v),
ˆ
Rp
ˆ
Rd
Pl(t, v)kndg1[knCn(t){An(t)t−bnv−k−n1c}]g2(Qv)dvdt=Op(1),
ˆ
Rp
ˆ
Rd
kndg1[knCn(t){An(t)t−bnv−k−n1c}]g2(Qv)dvdt= Θp(1).
Proof. For simplicity, here´ denotes the integration over the whole Euclidean space. Accord-ing to (ii),g1(v)can be written as¯g1(kvk). Whenk∞<∞, assumekn= 1without loss of
gen-erality. For anyPl(t, v)∈Pl,(t,v), by Cauchy–Schwarz inequality, there exists aPl(ktk,kvk)∈
Pl,(ktk,kvk) with coefficient functions taking positive values such that |Pl(t, v)| is bounded by
Pl(ktk,kvk)almost surely. Therefore for the first equality, it is sufficient to consider the
equal-ity where Pl(t, v) is replaced byPl(ktk,kvk)and the coefficient functions of Pl(ktk,kvk)are
positive almost surely. For eachn, divideRp intoV ={t:kAn(t)tk/2≥ kb0nv+ck}andVc.
InV,kCn(t){An(t)t−b0nv−c}k ≥m2ktk/2; inVc,ktk ≤2m−1kb0nv+ck. With probability
tending to1, ˆ
Pl(ktk,kvk)g1[Cn(t){An(t)t−b0nv−c}]g2(Qv)dvdt≤ ˆ
Pl(ktk,kvk)¯g1(m2ktk/2)g2(Qv)dvdt+ sup
v∈Rd
g1(v) ˆ ˆ
Vc
dt Pl(2m−1kb0nv+ck,kvk)g2(Qv)dv.
In the above,´Vc dt is the volume ofVcinRp and is proportional tokb0nv+ckp. By (iii), the
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Whenk∞=∞, letv∗ =kn{A(t)t−v−kn−1c}. Then for anyPl(t, v)∈Pl,(t,v), with prob-ability1,
ˆ
Pl(t, v)kndg1[knCn(t){A(t)t−v−kn−1c}]g2(Qv)dvdt
=
ˆ
Pl(t, v∗)g2[Q{A(t)t−k−n1v∗−k−n1c}]g1(Cn(t)v∗)dv∗dt
,
≤ ˆ
Pl(ktk,kv∗k)g2[Q{A(t)t−kn−1v∗−k−n1c}]g1(mkv∗k)dv∗dt
for somePl(t, v∗)∈Pl,(t,v∗)andPl(ktk,kv∗k)∈Pl,(ktk,kv∗k). The right hand side of the above
inequality is similar to the integral whenk∞<∞ with g1(·)and g2(·) replaced by g2(·) and
g1(·)respectively. Therefore it isOp(1)by the same reasoning.
For Pl(t, v) = 1, by considering only the integral in a compact region, it is easy to see the
target integral is larger than0. Therefore the lemma holds.
Proof of Lemma4. LetP =ATA. By matrix algebra,
NAt;Bnv+
1
kn
c, 1 k2
n
Id
K(v) =Nnt;P−1AT
Bnv+
1
kn
c
, 1 k2
n
P−1or(v;A, Bn,kn, c),
where
r(v;A, Bn,kn, c) =
kdn−p
(2π)(d−p)/2 exp
n
−k 2
n
2
Bnv+
c kn
T
(I−AP−1AT)
Bnv+
c kn
o
K(v).
Then the target integral can be expanded as ˆ
t N(At;Bnv+
1
knc,
1
k2
nId)K(v)
´
N(At;Bnv+k1nc,k12
nId)K(v)dtdv
dtdv= ˆ
P−1AT
1
kn
c+Bnv
r(v;A, Bn,kn, c)
´
r(v;A, Bn,kn, c)dv
dv
= 1
kn
(ATA)−1ATc+R(A, Bn, kn, c) ,
where
R(A, Bn, kn, c) = (ATA)−1ATBn
ˆ
knv
r(v;A, Bn,kn, c)
´
r(v;A, Bn,kn, c)dv
dv.
The remainder term R(A, Bn, kn, c) depends on the mean of the probability density
pro-portional to r(v;A, Bn,kn, c) in the directions of (ATA)−1ATB. If Bn does not
degener-ate to 0 as n→ ∞, then in the directions orthogonal to those of (I−A(ATA)−1AT)1/2B,
r(v;A, Bn,kn, c) is symmetric around 0; in the directions of (I−A(ATA)−1AT)1/2B,
r(v;A, Bn,kn, c)is a product of a normal density whose mean isO(1/kn)and a rescaledK(v),
which is symmetric around0, so its mean value isO(1/kn). Therefore when the spaces expanded
by(ATA)−1ATB and{I−A(ATA)−1AT}B are orthogonal,R(A, B
n, kn, c) = 0; when it is
not the case,R(A, Bn, kn, c) =O(1).
If Bn=o(1) as n→ ∞, which implies kn→c∈(0,∞), it is easy to see that
´
knvr(v;A, Bn,kn, c)dv/
´
r(v;A, Bn,kn, c)dv is upper bounded as n→ ∞ and hence
R(A, Bn, kn, c)iso(1).
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Proof of Lemma5. First considerπeBδ(1). With the transformationt=t(θ),
e
πBδ(1) =a
−p n,ε
ˆ
t(Bδ)
ˆ
Rd
π(θ0+a−n,ε1t)fen(sobs+εnv|θ0+an,ε−1t)K(v)dvdt. (6)
We can obtain an expansion of πeBδ(1)by expanding fen(sobs+εnv|θ0+a
−1
n,εt)K(v) as
fol-lows. The expansion needs to be discussed separately for two cases, depending on whether the limit ofanεnis finite or infinite.
Whenanεn→cε <∞,an,ε=an. We apply a Taylor expansion tos(θ0+a−n1t)andA(θ0+
a−n1t)−1/2and have
e
fn(sobs+εnv|θ0+a−n1t) =
adn
|A(θ0+a−n1t)|1/2
×NnA(θ0)−1/2+a−n1rA(t, 2)
oh
A(θ0)1/2Tobs+anεnv− {Ds(θ0) +a−n1rs(t, 1)}t
i
; 0, Id
,
(7) wherers(t, 1)is thed×pmatrix whoseith row istTHsi{θ0+1(t)}, rA(t, 2) is thed×d matrix Pp
k=1
d
dθkA{θ0+2(t)}
−1/2t
k, and 1(t) and 2(t) are from the remainder terms of
the Taylor expansions and satisfy k1(t)k ≤δ and k2(t)k ≤δ. For a d×d matrix τ2, let
gn(t, v;τ1, τ2)be the function gn(t, v;τ1), defined in Section4·3of the main text, withA(θ0)
replaced by{A(θ0)−1/2+τ2}−2. Applying a Taylor expansion to the normal density in (7), we have
e
fn(sobs+εnv|θ0+a−n1t)K(v)
= a
d
n|A(θ0)|1/2 |A(θ0+a−n1t)|1/2
h
gn(t, v) +a−n1P3(t, v)gn{t, v;en1rs(t, 1), en1rA(t, 2)}
i
, (8)
whereP3(t, v)is the function 1
2|A(θ0)−1/2+r2(a−n1t)|
× d
dx n
A(θ0)−1/2+xrA(t, 2)
o h
A(θ0)1/2Tobs+anεnv− {Ds(θ0) +xrs(t, 1)}t
i
2
x=en1
,
and en1 is from the remainder term of Taylor expansion and satisfies |en1| ≤a−n1 . Since
ken1tk ≤δandrs(t, 1)andrA(t, 2)belongP1,t, thisP3(t, v)belongs toP3,(t,v). Furthermore, sincers(t, 1)andrA(t, 2)have no constant term, for any smallσ,en1rs(t, 1)anden1rA(t, 2)
can be bounded byσIdandσIpuniformly innandt, ifδis small enough.
When anεn→ ∞, an,ε=εn−1. Let v∗(v) =A(θ0)1/2Tobs+anεnv−anεnDs(θ0)t. Under the transformationv∗ =v∗(v), the expansion offen(sobs+εnv|θ0+εnt)obtained by applying
a Taylor expansion tos(θ0+εnt)andA(θ0+εnt)−1/2is
e
fn(sobs+εnv|θ0+εnt)
= a
d n
|A(θ0+a−n1t)|1/2
N
n
A(θ0)−1/2+anε2n
rA(t, 4)
anεn
on
v∗−anε2nrs(t, 3)t
o
; 0, Id
385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432
where3(t)and4(t)are from the remainder terms of the Taylor expansion and satisfyk3(t)k ≤
δandk4(t)k ≤δ. Letg∗n(t, v∗;τ1, τ2)be the function
gn∗(t, v∗;τ1, τ2)
=Nhv∗;anεnτ1t,{A(θ0)−1/2+τ2}−2
i K
Ds(θ0)t+ 1
anεn
v∗− 1
anεn
A(θ0)1/2Tobs
,
so that (anεn)dgn∗(t, v∗;τ1, τ2) is gn(t, v;τ1, τ2) with transformed variable v∗=v∗(v), and
g∗n(t, v∗) =gn∗(t, v∗; 0,0). Denote a k1×k2 matrix with element being Pl(t) by P(k1
×k2)
l (t).
Then by applying a Taylor expansion to the normal density in the expansion above,
e
fn(sobs+εnv|θ0+εnt)K(v)
= ε
−d
n |A(θ0)|1/2 |A(θ0+εnt)|1/2
h
gn∗(t, v∗) +anε2n
P2(d×1)(t)v∗+ 1
anεn
v∗TP1(d×d)(t)v∗
g∗n(t, v∗)
+ (anε2n)2P4(t, v∗)gn∗{t, v∗;en2rs(t, 3), en2rA(t, 4)}
i
(anεn)d, (9)
where P2(d×1)(t) is the function tTrs(t, 3)TA(θ0)−1/2/2, P1(d×d)(t) is the function −A(θ0)−1/2rA(t, 4), en2 =e0n2/(anεn), e0n2 is from the remainder term of the Taylor expan-sion and satisfies |e0n2| ≤anε2n, and P4(t, v∗) is a linear combination of {dρ(w)/dw}2 and
d2ρ(w)/dw2atw=e0n2withρ(w)being the function
n
A(θ0)−1/2+wrA(t, 4) anεn
on
v∗−wrs(t, 3)t
o
2
.
Obviously elements of P2(d×1)(t) and P1(d×d)(t) belong to P2,t and P1,t respectively. Since
ken2tk ≤δ, the function P4(t, v∗) belongs to P4,(t,v∗) and, similar to before, en2rs(t, 3) and
en2rA(t, 4)can be bounded byσIdandσIp uniformly innandtfor any smallσ, ifδ is small
enough.
Forπ(θ0+a−n,ε1t)in the integral ofπeBδ(1)in (6), a Taylor expansion gives that
π(θ0+a−n,ε1t)
|A(θ0+a−n,ε1t)|1/2
= π(θ0) |A(θ0)|1/2 +a
−1
n,εDθ
π{θ0+5(t)} |A{θ0+5(t)}|1/2
t, |5(t)| ≤δ. (10)
As mentioned before, δ can be selected such that Ds(θ0) +en1rs(t, 1) and Ds(θ0) +
en2rs(t, 3) are lower bounded by m1Ip and A(θ0)−1/2+en1rA(t, 2) and A(θ0)−1/2+
en2rA(t, 4)are lowered bounded bym2Idfor some positive constantm1andm2. We chooseδ satisfying these and, sinceka−n,ε1tk ≤δ, this meansπ(θ0+a−n,ε1t)/|A(θ0+an,ε−1t)|1/2is bounded
433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480
By plugging (8)–(10) into (6), it can be seen that the leading term of eπBδ(1) is
adn,ε−pπ(θ0) ´
t(Bδ)×Rdgn(t, v)dtdv. The remainder terms are given in the following, apn,ε−dπeBδ(1)−π(θ0)
ˆ
t(Bδ)×Rd
gn(t, v)dtdv
=a−n,ε1
ˆ
t(Bδ)×Rd
|A(θ0)|1/2D π(θ0+5)
|A(θ0+5)|1/2
tgn(t, v)dvdt
+a−n1
ˆ
t(Bδ)×Rd
P3(t, v)gn{t, v;en1rs(t, 1), en1rA(t, 2)}dvdt1{limanεn<∞}
+anε2n
ˆ
t(Bδ)
P2(d×1)(t) ˆ
Rd
v∗gn∗(t, v∗)dv∗dt1{limanεn=∞}
+εn
ˆ
t(Bδ)×Rd
v∗TP1(d×d)(t)v∗gn∗(t, v∗)dv∗dt1{limanεn=∞}
+a2nε4n
ˆ
t(Bδ)×Rd
P4(t, v∗)g∗n{t, v∗;en2rs(t, 3), en2rA(t, 4)}dv∗dt1{limanεn=∞}, (11)
where P3(t, v), P2(d×1)(t), P1(d×d)(t) and P4(t, v∗) are products of π(θ0+an,ε−1t)/|A(θ0+
a−n,ε1t)|1/2 and corresponding terms in expansions (8) and (9). In the above, there are five re-mainder terms. For the integrals in the first two terms, it is easy to write them in the form of the first integral in Lemma 7 and conditions therein are satisfied, where g1(·) is the standard normal density andg2(·)isK(v)rescaled to have identity covariance. Then the first two terms areOp(a−n,ε1)andOp(a−n1). The integral in the fourth term can also be written in this form where
g1(·)is the rescaledK(v)andg2(·)is the standard normal density. The integral in the fifth term needs to use the transformationv∗∗=v∗−anεnen2rs(t, 3)t, after which it can be written in a
similar form, asP5{t, v∗∗+anεnen2rs(t, 3)t} ∈P5,(t,v∗∗)by the expression ofP4(t, v∗)in (9).
Thus the fourth and fifth term areOp(εn)andOp(a2nε4n).
The third term is somewhat different as the center ofgn∗(t, v∗)in the direction ofv∗degenerates to zero asn→ ∞. Letψkbe thed-dimension unit vector with1at thekth coordinate. Then
ˆ ∞
−∞
vk∗g∗n(t, v∗)dvk∗= ˆ ∞
0
v∗k{g∗n(t, v∗)−gn∗(t, v∗−2vk∗ψk)}dv∗k
= ˆ ∞
0
v∗kN{v∗; 0, A(θ0)}[K{v(v∗)} −K{v(v∗−2v∗kψk)}]dvk∗,
which by a Taylor expansion is bounded by(anεn)−1cfor some constantc. Hence the third term
isOp(εn). Combining the orders of all remainder terms, the expansion ofπeBδ(1)in the lemma
holds.
For any P2(t)∈P2,t, πeBδ{P2(t)} can be expanded similarly toπeBδ(1) in (11), simply by
multplyingP2(t)into every integral in (11). This gives that
e
πBδ{P2(t)}=a
d−p n,ε
n π(θ0)
ˆ
t(Bδ)×Rd
P2(t)gn(t, v)dtdv+Op(a−n,ε1) +Op(a2nε4n)
o .
Then since ´t(B
δ)×Rdgn(t, v)dtdv= Θp(1) by the second result of Lemma 7, e
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Proof of Lemma6. Letrn(s|θ)be the scaled remainderαn{fn(s|θ)−fen(s|θ)}. The error
of usingπeBδ{Pl(t)}to approximateπBδ{Pl(t)}is
πBδ{Pl(t)} −πeBδ{Pl(t)}=α
−1
n
ˆ
Bδ
ˆ
Pl{t(θ)}π(θ)rn(sobs+εnv|θ)K(v)dvdθ.
If this approximation error satisfies
πBδ{Pl(t)} −eπBδ{Pl(t)}
e πBδ(1)
=Op(α−n1), (12)
then, sinceapn,ε−dπeBδ(1) = Θp(1)by Lemma5,
πBδ(1) =eπBδ(1){1 +Op(α
−1
n )},
πBδ{Pl(t)}
πBδ(1)
= eπBδ{Pl(t)} e
πBδ(1)
+Op(α−n1). (13)
By plugging (12) into (1),
πBδ(h)
πBδ(1)
=h(θ0) +a−n,ε1Dh(θ0)T
e πBδ(t)
e πBδ(1)
+Op(α−n1)
+1 2a
−2
n,ε
e πBδ{t
THh(θ t)t}
e πBδ(1)
+Op(α−n1)
.
(14) Verification of (12) is given by the following argument. With the transformationt=t(θ)we have
πBδ{Pl(t)} −πeBδ{Pl(t)}=α
−1
n a−n,εp
ˆ
t(Bδ)
ˆ
Pl(t)π(θ0+a−n,ε1t)rn(sobs+εnv|θ0+an,ε−1t)K(v)dvdt.
LetrWn(w|θ) =αn{fWn(w|θ)−feWn(w|θ)}, and we have
rn(s|θ) =adn|A(θ)|
−1/2r
Wn[anA(θ)
−1/2{s−s(θ)} |θ].
For the value of δ, we choose the smaller value of the one from Lemma5 and the one such thatDs(θ)is lower bounded andA(θ)−1/2 is upper bounded byM IdinBδ for someM >0.
SincerWn(w|θ)is upper bounded byrmax(w)according to Condition5, by applying a Taylor
expansion tos(θ0+a−n,ε1t)we have
|πBδ{Pl(t)} −eπBδ{Pl(t)}| ≤α
−1
n adn,ε−p sup θ∈Bδ
|π(θ)A(θ)−1/2| ˆ
t(Bδ)
ˆ
|Pl(t)|(ana−n,ε1)d
rmax
h
ana−n,ε1M
n
Ds(θ0+t)t−an,εεnv−
1
ana−n,ε1
A(θ0)1/2Tobs
oi
K(v)dvdt,
where t is from the remainder term of the Taylor expansion and satisfies |t| ≤δ. Since
e
πBδ(1) = Θp(a
d−p
n,ε )by Lemma5, it is sufficient to show that the above integral isOp(1). This is
immediate by noting that when eitherlimanεn→ ∞orlimanεn→cε<∞, the above integral
can be written in the form of the first integral in Lemma7and conditions therein are satisfied, whereg1(·)andg2(·)arermax(·)andK(·)rescaled to have identity covariance matrix.
2. PROOF OFRESULTS FROMSECTION4 2·1. Proof of Proposition2
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Proof of Proposition2. Algorithm 1 generates independent, indentically distributed triples, (φi, θi, s(ni)), where (θi, s(ni)) is generated from gn(θ)f(sn|θ), and, conditional on sn=s(ni),
φiis generated from a Bernoulli distribution with probabilityKεn(sn−sobs).
Nowˆhcan be expressed as a ratio of sample means of functions of these independent, indenti-cally distributed random variables. Thus we can use the standard delta method (Lehmann,2004) for ratio statistics to show that the central limit theorem holds. Further we obtain that the limiting distribution has mean
E{h(θ1)w1φ1}
E(w1φ1)
= E{h(θ1)w1Kεn(s
(1)
n −sobs)}
E{w1Kεn(s
(1)
n −sobs)}
= ´
h´(θ)π(θ)fn(sn|θ)Kεn(sn−sobs)dsndθ
π(θ)fn(sn|θ)Kεn(sn−sobs)dsndθ
,
which is equal tohABC. Its variance is 1
E2(w 1φ1)
var{h(θ1)w1φ1}+
E2{h(θ1)w1φ1}
E4(w 1φ1)
var(w1φ1)−2
E{h(θ1)w1φ1}
E3(w 1φ1)
cov{h(θ1)w1φ1, w1φ1}T =p−acc2,πE{h(θ1)2w21φ1} −h2ABCp2acc,π+h2ABC
E(w21φ1)−p2acc,π
−2hABC
E{h(θ1)w21φ1} −hABCp2acc,π Ti
=p−acc2,πE[{h(θ1)2−2hABCh(θ1)T +h2ABC}w12Kεn(s
(1)
n −sobs)] =p−acc1,πEπABC
(h(θ)−hABC)2
π(θ)
qn(θ)
.
In the above expression we usedpacc,π=E(w1φ1). It is easy to verify that ΣABC,n=p−acc1,πEπABC
(h(θ)−hABC)2
π(θ)
qn(θ)
, (15)
as required.
2·2. Proof of Theorem 2
For simplicity, a consider one-dimensional function h(θ). For multi-dimensional functions, the extension is trivial by considering each element ofΣIS,nseperately. Denote{h(θ)−hABC}2 byGn(θ). In Theorem2(i),ΣIS,nis just the ABC posterior variance ofh(θ), and the derivation
of its order is similar to that ofhABC in Section1of this supplementary material. The result is stated in the following lemma.
LEMMA8. Assume the conditions of Theorem1. Then varπABC{h(θ)}=Op(a−n,ε2).
Proof. Using the notation of Section1, varπABC[h(θ)] =π(Gn)/π(1). It follows immediately from Lemma3that
varπABC{h(θ)}=
πBδ(Gn)
πBδ(1)
{1 +op(1)}.
Applying a first order Taylor expansion ofh(θ)aroundθ=θ0gives
πBδ(Gn)
πBδ(1)
=Gn(θ0) + 2an,ε−1{h(θ0)−hABC}
πBδ{Dh(θt)
Tt}
πBδ(1)
+a−n,ε2πBδ{t
TDh(θ
t)Dh(θt)Tt}
πBδ(1)
,
(16) where θt is from the remainder term and belongs toBδ. In the above decomposition,Gn(θ0)
anda−n,ε1{h(θ0)−hABC}areOp(a−n,ε2)by Theorem1. SinceDh(θt)TtandtTDh(θt)Dh(θt)Tt
577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624
The following lemma states that moments ofK(v)γexist for any postive constantγ.
LEMMA9. Assume Condition2. For any constantγ ∈(0,∞)and coordinates(vi1,· · · , vil) ofvwithl≤p+ 6,´ Ql
k=1vikK(v)
γdv <∞.
Proof. By Condition2(iv), for some positive constantM there exists x0 ∈(0,∞)such that whenkvk> x0 ,K(v)< M e−c1kvk
α1
. Then consider the integration in two regions{v:kvk ≤
x0}and{v :kvk> x0}separately. In the first region, sinceK(v)≤1, we have ˆ
kvk≤x0
l
Y
k=1
vikK(v)
γdv≤xl
0Vx0,
whereVx0 is the volume of the d-dimension sphere with radius x0, and is finite. In the second region,
ˆ
kvk>x0
l
Y
k=1
vikK(v)
γdv ≤M
ˆ
kvk>x0
kvkle−c1γkvkα1 dv.
The right hand side of this is proportional toexp{−c1γxα01/(l+d)} by integrating in spherical
coordinates.
Proof of Theorem2. For (i), sincepacc,π =εdnπ(1)andπ(1) = Θp(adn,ε−p)by Lemmas3,5and
6, thenpacc,π = Θp(εdna d−p
n,ε ). Together with Lemma8, (i) holds.
For (ii), if we can show thatpacc,q = Θp(εdnadn,ε), then the order ofΣIS,nis obvious from (15)
and the definition ofΣABC,n. Similar to the expansion ofπ(1)from Lemma3and (13),
pacc,q =εdn
ˆ
πABC(θ|sobs, εn)fABC(sobs |θ)dθ
=εdn (´
Bδπ(θ)feABC(sobs |θ)
2dθ
e πBδ(1)
+Op(α−n1)
)
{1 +op(1)}.
The integral in the above differs from eπBδ(1) by the square power of feABC(sobs|θ) in the
integrand. We will show that this integral has orderΘp(a2n,εd−p), from whichpacc,q = Θp(εdnadn,ε)
trivially holds. Letg∗∗n (t, v;τ1, τ2)be the function
g∗∗n (t, v;τ1, τ2) =N[v; 0,{A(θ0)−1/2+τ2}−2]K
{Ds(θ0) +τ1}t+ 1
anεn
v∗− 1
anεn
A(θ0)1/2Tobs
,
andg∗∗n (t, v;τ1, τ2) =gn∗(t, v+anεnτ1t;τ1, τ2). Here expansions (8) and (9) offen(sobs+εnv| θ0+a−n,ε1t)K(v)are to be used in the form of
adn,ε|A(θ0)|1/2 |A(θ0+a−n,ε1t)|1/2
gn(t, v) +a−n1P3(t, v)gn,r(t, v) , limn→∞anεn<∞,
gn∗(t, v∗) +anε2nP3(t, v∗)gn∗(t, v∗)
+(anε2n)2P4(t, v∗∗)g∗n,r(t, v∗∗) (anεn)d, limn→∞anεn=∞,
(17) where P3(t, v∗)∈P3,(t,v∗), gn,r(t, v) =gn{t, v;en1rs(t, 1), en1rA(t, 2)}, gn,r∗ (t, v∗∗) is
g∗∗n {t, v∗∗;en2rs(t, 3), en2rA(t, 4)} andP4(t, v∗∗) isP4(t, v∗) with the transformationv∗∗=
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By the expression of P4(t, v∗) in (9), it can be seen that P4(t, v∗∗)∈P4,(t,v∗∗).
Ba-sic inequalities (a+εb)2 ≤εa2+ (ε+ε2)b2 and (a+εb+ε2c)2 ≤(ε+ε2)a2+ (ε+ε2+
ε3)b2+ (ε2+ε3+ε4)c2for any real constantsa,b,candε, from the fact that2ab≤a2+b2, are also to be used. Then by the above expansions and inequalities, an expansion of the target in-tegral similar to (11) can be obtained, with the leading terma2n,εd−pπ(θ0)
´
t(Bδ){
´
gn(t, v)dv}2dt
and remainder term with the following upper bound
a
p−2d n,ε
ˆ
Bδ
π(θ)feABC(sobs |θ)2dθ−π(θ0)
ˆ
t(Bδ) nˆ
gn(t, v)dv
o2 dt
≤a−n,ε1
ˆ
t(Bδ)
|A(θ0)|Dθ
π(θ0+6) |A(θ0+6)|
tn
ˆ
gn(t, v)dv
o2 dt
+M
ˆ
t(Bδ) h
a−n1n
ˆ
gn(t, v)dv
o2
+ (a−n1+a−n2)n ˆ
P3(t, v)gn,r(t, v)dv
o2i
dt1{limanεn<∞}
+M
ˆ
t(Bδ) h
{anε2n+ (anε2n)2}
nˆ
gn∗(t, v∗)dv∗o2
+{anε2n+ (anεn2)2+ (anε2n)3}
nˆ
P3(t, v∗)g∗n(t, v∗)dv∗
o2
+{(anε2n)2+ (anε2n)3+ (anε2n)4}
nˆ
P4(t, v∗∗)gn,r∗ (t, v∗∗)dv∗∗ o2i
dt1{limanεn=∞},
where M is the upper bound of π(θ)|A(θ0)|/|A(θ)| for θ∈Bδ with δ chosen so that M
exists. Then if we can show that for any P4(t, v)∈P5,(t,v), d×p matrix function rn1(t)
and d×d matrix function rn2(t) which can be bounded by σId and σIp uniformly in
n and t for any small δ if δ is small enough, (a)´t(B
δ) ´
Rdgn(t, v)dv
2
dt is Θp(1);
(b) ´t(B
δ) ´
RdP4(t, v)gn{t, v;rn1(t), rn2(t)}dv 2
dt is Op(1) when limn→∞anεn<∞; (c)
´
t(Bδ) ´
RdP4(t, v)g ∗∗
n {t, v;rn1(t), rn2(t)}dv
2
dt is Op(1) when limn→∞anεn=∞, the
lemma would hold.
Here δ is selected such that Ds(θ0) +rn1(t) is bounded bounded by m1Ip and m2Id≤
A(θ0)−1/2+rn2(t)≤M2Id, for some positive constantsm1,m2andM2, uniformly innandt. For the purpose of bounding integrals, we can assume thatA(θ0) =Idandrn2(t) = 0without loss of generality by the following inequality whenlimn→∞anεn<∞,
gn{t, v;rn1(t), rn2(t)} ≤
M2d
(2π)d/2 exp
h
−m 2 2
2 kanεnv+A(θ0) 1/2T
obs− {Ds(θ0) +rn1(t)}tk2
i K(v),
and a similar one forg∗∗n {t, v;rn1(t), rn2(t)}.
Consider any P4(t, v)∈P4,(t,v). When limn→∞anεn<∞, let E1 ={v :kanεnvk2≤
673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720
have ˆ
Rd
P4(t, v)gn{t, v;rn1(t), rn2(t)}dv
≤ ˆ
E1 +
ˆ
Ec
1
!
P4(t, v) M
d
2 (2π)d/2exp
−m 2 2
2 kanεnv− {Ds(θ0) +rn1(t)}t+A(θ0) 1/2T
obsk2
K(v)dv
≤P4(t)
exp
−m 2
2(1−β1)
2 k{Ds(θ0) +rn1(t)}t−A(θ0) 1/2T
obsk2
+Kβ2
λ2min(Λ)β1
a2
nε2n
k{Ds(θ0) +rn1(t)}t−A(θ0)1/2Tobsk2
, (18)
whereP4(t)∈P4,tand the above inequality uses Lemma9. Then using(a+b)2 ≤2(a2+b2),
ˆ
t(Bδ) ˆ
Rd
P4(t, v)gn{t, v;rn1(t), rn2(t)}dv
2 dt
≤ ˆ
t(Bδ)
P8(t) exp
h
−m22(1−β1)k{Ds(θ0) +rn1(t)}t−A(θ0)1/2Tobsk2
i dt
+ ˆ
t(Bδ)
P8(t)K2β2
λ2min(Λ)β1
a2
nε2n
k{Ds(θ0) +rn1(t)}t−A(θ0)1/2Tobsk2
dt,
whereP8(t)∈P8,t.
When anεn→ ∞, let E2 ={v :k(anεn)−1vk2 ≤β1k{Ds(θ0) +rn1(t)}t−
(anεn)−1A(θ0)1/2Tobsk2}for someβ1∈(0,1). Then for anyβ2 ∈(0,1)we have ˆ
Rd
P4(t, v)g∗∗n {t, v;rn1(t), rn2(t)}dv
≤ ˆ
E2 +
ˆ
Ec
2
!
P4(t, v)K
1
anεn
v+{Ds(θ0) +rn1(t)}t− 1
anεn
A(θ0)1/2Tobs
(19)
× M
d
2 (2π)d/2exp
−m 2 2 2 kvk
2
dv
≤P4(t)
K
λ2min(Λ)(1−β1)k{Ds(θ0) +rn1(t)}t−
1
anεn
A(θ0)1/2Tobsk2
+ exp
−a 2
nε2nβ1m22β2
2 k{Ds(θ0) +rn1(t)}t− 1
anεn
A(θ0)1/2Tobsk2
, (20)
whereP4(t)∈P4,t. Then using(a+b)2 ≤2(a2+b2),
ˆ
t(Bδ) ˆ
Rd
P4(t, v)gn∗∗{t, v;rn1(t), rn2(t)}dv
2 dt
≤ ˆ
t(Bδ)
P8(t)K
2λ2min(Λ)(1−β1)
2 k{Ds(θ0) +rn1(t)}t− 1
anεn
A(θ0)1/2Tobsk2
dt
+ ˆ
t(Bδ)
P8(t) exp
−a 2
nε2nβ1m22β2
2 k{Ds(θ0) +rn1(t)}t− 1
anεn
A(θ0)1/2Tobsk2
dt
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For (a), to see that the limit of´t(B
δ){
´
Rdgn(t, v)dv}
2dtis lower bounded away from zero, just use the positivity of the limit of the integrand and Fatou’s lemma to interchange the order of
limit and integral.
2·3. Proof of Theorem 3
Now let wn(θ) be the importance weight π(θ)/qn(θ), define πBδ,IS(h) =
´
Bδh(θ)π(θ)fABC(sobs |θ)wn(θ)dθ and define πB c
δ,IS(h)correspondingly. Then by (15), we
have
ΣABC,n=p−acc1,π
πBδ,IS(Gn) +πBδc,IS(Gn)
πBδ(1) +πBδc(1)
. (21)
Proof of Theorem3. For pacc,qn, we only need to consider the case when β = 0.
Re-call that t(θ) =an,ε(θ−θ0). By the transformation t=t(θ), since an,εσn= 1, qn(θ) =
apn,ε|Σ|−1/2q{Σ−1/2(t−cµ)}. Then, similar to the expansion ofπ(1)from Lemma3,
pacc,qn =ε
d n
ˆ
qn(θ)fABC(sobs |θ)dθ
=εdn|Σ|−1/2 ˆ
t(Bδ)
q{Σ−1/2(t−cµ)}feABC(sobs |θ0+a−n,ε1t)dt{1 +op(1)}.
The above integral differs fromπeBδ(1)by replacingπ(θ0+a
−1
n,εt)with the densityq{Σ−1/2(t−
cµ)} which does not degenerate to a constant as n→ ∞. We will show that this integral has
order Θp(1). Plugging in the expansion (17) of feABC(sobs |θ0+a−n,ε1t) into pacc,qn, we can
obtain an expansion similar to (11), differing in that parts from expandingπ(θ0+an,ε−1t)/|A(θ0+
a−n,ε1t)|1/2 are replaced by the Taylor expansion
q{Σ−1/2(t−c
µ)}
|A(θ0+a−n,ε1t)|1/2
=q{Σ−1/2(t−c
µ)}
1 +a−n,ε1Dθ
1
|A{θ0+6(t)}|1/2
t
,
where k6(t)k ≤δ. The explicit form is ommitted here to avoid repetition. It can be seen that pacc,qn = Θp(a
d
n,εεdn) if (a)
´
Rd×t(Bδ)q{Σ
−1/2(t−c
µ)}gn(t, v)dvdt=
Θp(1); (b)
´
Rd×t(Bδ)P3(t, v)q{Σ
−1/2(t−c
µ)}gn{t, v;rn1(t), rn2(t)}dvdt=
Op(1) when limn→∞anεn<∞; and (c)
´
Rd×t(Bδ)P3(t, v)q{Σ
−1/2(t−
cµ)}gn∗∗{t, v;rn1(t), rn2(t)}dvdt=Op(1) when limn→∞anεn=∞, where rn1(t) and
rn2(t) are defined as in the proof of Theorem 2. Since q{Σ−1/2(t−cµ)} is uniformly upper
bounded for t∈Rp, (b) and (c) hold and the integral in (a) isOp(1)following the arguments
for the similar cases in the proof of Theorem 2. By the positivity of the limit of the integrand and Fatou’s lemma, the limit of the integral in (a) is lower bounded away from 0. Therefore
pacc,qn = Θp(a
d
n,εεdn)holds.
AsΣIS,nis equal topacc,qnΣABC,n, by (15) we have
ΣIS,n =
pacc,qn
pacc,π
πBδ,IS(Gn) +πBδc,IS(Gn)
πBδ(1) +πBδc(1)
= pacc,qn
pacc,π
πBδ,IS(Gn)
πBδ(1)
{1 +op(1)},
where the second equality holds by noting thatωn(θ)≤β−1. Given the obtained orders ofpacc,qn
andpacc,π,ΣIS,n =Op(a−n,ε2)ifπBδ,IS(Gn)/πBδ(1) =Op(a
−p−2
769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816
following expansion
πBδ,IS(Gn)
πBδ(1)
=G(θ0)
πBδ,IS(1)
πBδ(1)
+ 2a−n,ε1{h(θ0)−hABC}
πBδ,IS{Dh(θt)
Tt}
πBδ(1)
+a−n,ε2πBδ,IS{t
TDh(θ
t)Dh(θt)Tt}
πBδ(1)
,
and we only needπBδ,IS{P2(t)}/πBδ(1) =Op(a
−p
n,ε)for anyP2(t)∈P2,t. Sincewn(θ)≤(1−
β)−1wn,0(θ), wherewn,0(θ)is the weight whenβ = 0, it is sufficient to consider the caseβ = 0. Similar to the proof of Theorem 1, first the normal counterpart eπBδ,IS{P2(t)}/πeBδ(1) of
πBδ,IS{P2(t)}/πBδ(1), wherefABC(sobs |θ)is replaced byfeABC(sobs |θ), is considered, then
it is shown that their difference can be ignored. Using the transformationt=t(θ)and plugging in expansion (17) offeABC(sobs |θ0+a−n,ε1t)into
e
πBδ,IS{P2(t)}, we obtain an expansion similar
to (11), differing in that parts from expandingπ(θ0+a−n,ε1t)/|A(θ0+a−n,ε1t)|1/2are replaced by
the Taylor expansion 1
qn(θ)
π(θ0+a−n,ε1t)2
|A(θ0+a−n,ε1t)|1/2
= 1
apn,ε|Σ|−1/2q{Σ−1/2(t−cµ)}
π(θ0)2+a−n,ε1Dθ
π{θ0+7(t)}2 |A{θ0+7(t)}|1/2
t
,
wherek7(t)k ≤δ. The explicit form is omitted here to avoid repetition. Then it can be seen that if we can show that
(d) ˆ
t(Bδ)
´
RdP5(t, v)gn{t, v;rn1(t), rn2(t)}dv q{Σ−1/2(t−c
µ)}
dt=Op(1)when lim
n→∞anεn<∞,
(e) ˆ
t(Bδ)
´
RdP5(t, v)g ∗∗
n {t, v;rn1(t), rn2(t)}dv
q{Σ−1/2(t−c
µ)}
dt=Op(1)when lim
n→∞anεn=∞,
wherern1(t)andrn2(t)are defined as in the proof of Theorem2,πeBδ,IS{P2(t)}=Op(a
d−2p n,ε )
and eπBδ,IS{P2(t)}/eπBδ(1) =Op(a
−p
n,ε) by Lemma 5. By (18) and the following equality for
d×pfull column-rank matrixAand vectorc,
kAt−ck=kP1/2(t−P−1Ac)k2+cT(I−AP−1AT)c,
whereP =ATAandP1/2P1/2=P, for (d) we have ´
RdP5(t, v)gn{t, v;rn1(t), rn2(t)}dv q{Σ−1/2(t−c
µ)}
≤P5(t)
expn−m21m22γ
2 kt−P(θ0, t)Tobsk 2o
q{Σ−1/2(t−c
µ)}
exp
−m 2 2∆
2 k{Ds(θ0) +rn1(t)}t−A(θ0) 1/2T
obsk2
+P5(t)
Kαnλ2min(Λ)(1−γ−∆)m21
a2
nε2n kt−P(θ0, t)Tobsk
2o
q{Σ−1/2(t−c
µ)}
×K∆
λ2min(Λ)(1−γ−∆)
a2
nε2n
k{Ds(θ0) +rn1(t)}t−A(θ0)1/2Tobsk2
,
where P(θ0, t) = [{Ds(θ0) +rn1(t)}T{Ds(θ0) +rn1(t)}]−1{Ds(θ0) +rn1(t)}TA(θ0)1/2,
817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864
andαin Condition7. Then since both ratios on the right hand side of the above inequality are
Op(1)by Condition7, by Lemma7and Lemma9, (d) holds. Similarly by (20), for (e) we have
´
RdP5(t, v)g ∗∗
n {t, v;rn1(t), rn2(t)}dv
q{Σ−1/2(t−c
µ)}
≤P5(t) exp
n
−m21m22γ 2 kt−
1
anεnP(θ0, t)Tobsk
2o
q{Σ−1/2(t−c
µ)}
×exp
−(a 2
nε2nβ1β2−γ)m22
2 k{Ds(θ0) +rn1(t)}t−A(θ0) 1/2T
obsk2
+P5(t)
Kα n
λ2min(Λ)(1−β1)m21kt−a1
nεnP(θ0, t)Tobsk
2o
q{Σ−1/2(t−c
µ)}
×K1−αhλ2min(Λ)(1−β1)k{Ds(θ0) +rn1(t)}t−A(θ0)1/2Tobsk2
i ,
where bothP5(t)belong toP5,t. Thus by Condition7, Lemma7and Lemma9, (e) holds.
There-foreeπBδ,IS{P2(t)}/eπBδ(1) =Op(a
−p n,ε).
To show thatπBδ,IS{P2(t)}/πBδ(1) =Op(a
−p
n,ε), similar to the discussion of (13), it is
suffi-cient to show that
πBδ,IS{P2(t)} −eπBδ,IS{P2(t)}
e πBδ(1)
=Op(αn−1a
−p
n,ε). (22)
With the transformationt=t(θ)we haveπBδ,IS{P2(t)} −eπBδ,IS{P2(t)}is equal to
α−n1a−n,ε2p
ˆ
t(Bδ)
ˆ
P2(t)π(θ0+a−n,ε1t)2
rn(sobs+εnv|θ0+a−n,ε1t)K(v)
|Σ|−1/2q{Σ−1/2(t−c
µ)}
dvdt.
Then by following the arguments of the proof of Lemma6, we have |πBδ,IS{P2(t)} −eπBδ,IS{P2(t)}| ≤α
−1
n ad
−2p n,ε sup
θ∈Bδ
|π(θ)2A(θ)−1/2|
× ˆ
t(Bδ)
ˆ
|P2(t)|
(ana−n,ε1)drmax
h
ana−n,ε1M
n
Ds(θ0+t)t−an,εεnv− a 1
na−n,ε1A(θ0)
1/2T obs
oi K(v)
q{Σ−1/2(t−c
µ)}
dvdt.
The ratio above is similar to the ratio ofgn{t, v;r1(t), r2(t)}/q{Σ−1/2(t−cµ)}except that the
normal density is replaced byrmax(·). Then by Condition7, previous arguments for proving (iv) and (v) can be followed. HenceπBδ,IS{P2(t)} −πeBδ,IS{P2(t)}=Op(αna
d−2p
n,ε )and (22) holds.
ThereforeΣIS,n =Op(a−n,ε2).
REFERENCES
CREEL, M. & KRISTENSEN, D. (2013). Indirect likelihood inference (revised). UFAE and IAE working papers, Unitat de Fonaments de l’Analisi Economica (UAB) and Institut d’Analisi Economica (CSIC).
