Supplement to “Bootstrapping Realized Volatility”
Sı́lvia Gonçalves∗and Nour Meddahi†
This Appendix is organized as follows. First, we introduce some notation. Second, we provide
auxiliary lemmas (and their proofs) used in deriving the cumulants expansions in Appendix A of the
paper. Lastly, we prove Proposition 4.2 and part c) of Proposition 4.3, not included in Appendix B.
Notation
q/2
R ih
P1/h
P1/h
Recall that σ 2i ≡ (i−1)h σ 2u du < ∞, and for any q > 0, σ qh ≡ h1−q/2 i=1 σ 2i
≡ h1−q/2 i=1 σ qi ,
q/2
R1
q
where σ qi ≡ σ 2i
. Note that in general σ qh 6= σ q ≡ 0 σ qu du. We let σ q,p ≡ pσ q/p for any q, p > 0.
σ
( )
Rq
.
We
let µq = E |Z|q , where
When σ q is replaced with σ qh we write σ q,p,h . Similarly, Rq,p ≡
q/p
(Rp )
Z ∼ N (0, 1) and q > 0, and note that µ2 = 1, µ4 = 3, µ6 = 15 and µ8 = 105. Since µ2 = 1, we can
write σ 2 = µ2 σ 2 , which will be convenient for proving the results for the WB.
−1/2
−1/2
√
Write Th = Sh VV̂h
= Sh 1 + hUh
, where
√
Sh ≡
and Vh = V ar
that
√
√
h−1 R2 − µ2 σ 2
h−1 V̂ − Vh
√
,
and Uh ≡
Vh
Vh
h−1 R2 = µ4 − µ22 σ 4h . The proof of Lemma S.2 below relies heavily on the fact
R2 − µ2
σ2
=
1/h
X
i=1
q
− µq σ qi
where for any q > 0, |ri |
where ui ∼ i.i.d. N (0, √
1) .
1/h
ri2
−
µ2 σ 2i
µ − µ22 −1 X 4
h
ri − µ4 σ 4i ,
and V̂ − Vh = 4
µ4
i=1
are (conditionally on σ) independent with zero mean since ri = σ i ui ,
√
h−1 (V̂ ∗ −V ∗ )
, where V ∗ = V ar∗ h−1/2 R2∗ and V̂ ∗ is a
Uh∗ ≡
V∗
−1/2
√
consistent estimator of V ∗ . Then Th∗ = Sh∗ 1 + hUh∗
. For the i.i.d. bootstrap, V ∗ = R4 − R22
∗ ∗2 ∗ = µ4 −µ2
R
and
V̂
and V̂ ∗ = R4∗ − R2∗2 . For the WB, V ∗ = µ∗4 − µ∗2
R4∗ .
4
µ∗4
P2
note that throughout we will use i6=j6=···6=k to denote a sum where all indices differ, e.g.
P Finally,P
i6=j,i6=k,j6=k .
i6=j6=k ≡
Similarly, let Sh∗ ≡
h−1 (R2∗ −E ∗ (R2∗ ))
√
,
V∗
∗
Département de sciences économiques, CIREQ and CIRANO, Université de Montréal. Address: C.P.6128, succ.
Centre-Ville, Montréal, QC, H3C 3J7, Canada. Tel: (514) 343 6556. Email: [email protected].
†
Finance and Accounting Group, Tanaka Business School, Imperial College London. Address: Exhibition Road,
London SW7 2AZ, UK. Phone: +44 207 594 3130. Fax: +44 207 823 7685. E-mail: [email protected].
1
Auxiliary Lemmas
Lemma S.1 Let q, p and s be positive even integers. It follows that
1/h
X
σ qi σ pj = h−2+
q+p
2
i6=j
1/h
X
σ qi σ pj σ sl = h−3+
q+p+s
2
i6=j6=l
+2h−1+
σ qh σ ph − h σ q+p
,
h
σ qh
q+p+s
2
σ ph
σ q+p+s
.
h
(1)
q+p+s
s + σ q+s σ p + σ q σ p+s
σ sh − h−2+ 2
σ q+p
σ
h
h
h
h h
h
(2)
Lemma S.2 Under Assumption H, conditionally on σ,
a1) E |ri |q = µq σ qi .
a2) Vh ≡ V ar h−1/2 R2 = µ4 − µ22 σ 4h .
a3) E
a4) E
R2 − µ2
σ2
R2 − µ2
σ2
3 4 = h2 µ6 − 3µ2 µ4 + 2µ32 σ 6h .
= 3h2 µ4 − µ22
2 σ 4h
2
+ h3 µ8 − 4µ2 µ6 + 12µ22 µ4 − 6µ42 − 3µ24 σ 8h .
µ4 − µ22 (µ6 − µ2 µ4 ) 6
σh.
V̂ − Vh = h
a5) E R2 − µ2
µ4
2 µ − µ22
2
V̂ − Vh = h2 4
a6) E R2 − µ2 σ
µ8 − µ24 − 2µ2 µ6 + 2µ22 µ4 σ 8h .
µ4
3 2 2 (µ − µ µ )
µ
−
µ
4
6
2 4
2
a7) E R2 − µ2 σ 2
V̂ − Vh = 3h2
σ 4h σ 6h + O h3 , as h → 0.
µ4


2
4 2
3 (µ − µ µ ) σ 6
µ
−
µ
4
µ
−
3µ
µ
+
2µ
2 
6
2 4
6
2 4
2
h
a8) E R2 − µ2 σ 2
V̂ − Vh = h3 4
4 8 +
µ4
2
2
2
+6 µ8 − µ4 − 2µ2 µ6 + 2µ2 µ4 µ4 − µ2 σ h σ h
4
O h , as h → 0.
2
2 µ4 − µ22
2
2
µ10 − 2µ4 µ6 − µ2 µ8 + 2µ2 µ24 h2 σ 10
=
a9) E R2 − µ2 σ
V̂ − Vh
h = O h , as
2
µ4
h → 0.
2 2 2 2 2 4 8
2
2 µ4 − µ2
2
2
2
=h
+
µ4 − µ2 µ8 − µ4 σ h σ h + 2 (µ6 − µ2 µ4 ) σ 6h
V̂ − Vh
a10) E R2 − µ2 σ
2
µ
4
O h3 , as h → 0.
2 3 = O h3 + O h4 , as h → 0.
a11) E R2 − µ2 σ 2
V̂ − Vh
h
a12) E
σ2
R2 − µ2 σ 2
as h → 0.
4 i
V̂ − Vh
2 = h3
µ4 − µ22
µ24
2 

2
3 µ4 − µ22
2
µ8 − µ24
σ 4h
2
+12 µ4 − µ22 (µ6 − µ2 µ4 )2 σ 6h
σ 8h
2
σ 4h

+O h4 ,
Lemma S.3 Under Assumption H, conditionally on σ,
E (Sh )
E Sh2
E Sh3
E Sh4
= 0;
= 1;
√
=
hB1 σ 6,4,h ;
= 3 + hB2 σ 8,4,h ;
E (Sh Uh ) = A1 σ 6,4,h ;
√
E Sh2 Uh =
h (A2 σ 8,4,h ) ;
and as h → 0,
E Sh3 Uh
= A3 σ 6,4,h + O (h) ;
√ h D1 σ 8,4,h + D2 σ 26,4,h + O h3/2 ;
E Sh4 Uh =
E Sh Uh2 = O h1/2 ; E Sh3 Uh2 = O h1/2 ;
E Sh2 Uh2 = C1 σ 8,4,h + C2 σ 26,4,h + O (h) ;
E Sh4 Uh2 = E1 σ 8,4,h + E2 σ 26,4,h + O (h) .
The constants A1 , A2 , B1 , B2 , and C1 are as in Theorem A.1, and A3 = 3A1 , C2 = 2A21 , D1 = 6A2 ,
D2 = 4A1 B1 , E1 = 3C1 , and E2 = 12A21 .
Remark 1 The WB analogue of Lemma S.2 replaces σ qh with Rq and µq with µ∗q = E ∗ |η i |q . The
WB analogue of S.3 replaces σ q,p,h with Rq,p (and µq with µ∗q = E ∗ |η i |q ), yielding e.g. E ∗ Sh∗3 =
√
µ∗ −3µ∗ µ∗ +2µ∗3
h (B1∗ R6,4,h ), where B1∗ = 6 ∗ 2 ∗24 3/22 .
(µ4 −µ2 )
Lemma S.7 below is the i.i.d. bootstrap analogue of Lemma S.3. The next results are auxiliary in
proving Lemma S.7.
Lemma S.4 Let ri∗ ∼ i.i.d. from {ri : i = 1, . . . , 1/h}. Under Assumption H, conditionally on σ, for
any q > 0 and for any i = 1, . . . , 1/h,
a1) E ∗ (|ri∗ |q ) = hq/2 Rq and E ∗ Rq∗ = Rq = OP (1) .
h
2 i
a2) E ∗ ri∗2 − hR2
= h2 R4 − R22 .
a3)
E∗
a4) E ∗
a5) E ∗
a6) E ∗
h
ri∗2
h
ri∗2 − hR2
− hR2
h
ri∗2 − hR2
h
ri∗2 − hR2
3 i
= h3 R6 − 3R4 R2 + 2R23 .
5 i
= h5 R10 − 5R8 R2 + 10R6 R22 − 10R4 R23 + 4R25 .
4 i
6 i
q = h4 R8 − 4R6 R2 + 6R4 R22 − 3R24 .
= h6 R12 − 6R10 R2 + 15R8 R22 − 20R6 R23 + 15R4 R24 − 5R26 .
= OP (hq ), for any q ≥ 7, as h → 0.
h
2 i
= h4 R8 − R42 .
a8) E ∗ ri∗4 − h2 R4
a7) E ∗
ri∗2 − hR2
3
ri∗2 − hR2 ri∗4 − h2 R4 = h3 (R6 − R4 R2 ) .
h
i
2 ∗4
ri − h2 R4 = h4 R8 − R42 − 2R6 R2 + 2R4 R22 .
a10) E ∗ ri∗2 − hR2
a9) E ∗
a11) E ∗
h
ri∗2 − hR2
E∗
h
ri∗2
a13) E ∗
ri∗2 − hR2
a12)
− hR2
3
4
ri∗4 − h2 R4
ri∗4
−
h2 R4
ri∗4 − h2 R4
a14) For any q, p > 0, E ∗
i
i
2 = h5 R10 − R4 R6 − 3R8 R2 + 3R42 R2 + 3R6 R22 − 3R4 R23 .
=
ri∗2 − hR2
h6
R12 − R4 R8 − 4R10 R2 + 4R4 R6 R2 + 6R8 R22
−6R42 R22 − 4R6 R23 + 4R4 R24
.
= h5 R10 − 2R4 R6 − R8 R2 + 2R42 R2 .
q
ri∗4 − h2 R4
p = OP hq+2p , as h → 0.
Lemma S.5 Let ri∗ ∼ i.i.d. from {ri : i = 1, . . . , 1/h}. Under Assumption H, conditionally on σ,
a1) V ∗ ≡ V ar∗ h−1/2 R2∗ = R4 − R22 .
i
h
a2) V̂ ∗ − V ∗ = R4∗ − R4 − (R2∗ − R2 )2 + 2R2 (R2∗ − R2 ) .
h
i
a3) E ∗ (R2∗ − R2 )3 = h2 R6 − 3R4 R2 + 2R23 .
h
i
h
2 i
+ h3 R8 − 4R6 R2 + 12R4 R22 − 6R24 − 3R42 .
a4) E ∗ (R2∗ − R2 )4 = h2 3 R4 − R22
i
h
a5) E ∗ (R2∗ − R2 )5 = h3 10 R6 − 3R4 R2 + 2R23 R4 − R22 + OP h4 , as h → 0.
h
i
h
3 i
+ OP h4 , as h → 0.
a6) E ∗ (R2∗ − R2 )6 = h3 15 R4 − R22
a7) E ∗ [(R2∗ − R2 )q ] = OP h4 for q = 7, 8, as h → 0.
a8) E ∗ [(R2∗ − R2 ) (R4∗ − R4 )] = h (R6 − R4 R2 ) .
h
i
a9) E ∗ (R2∗ − R2 )2 (R4∗ − R4 ) = h2 R8 − R42 − 2R6 R2 + 2R4 R22 .
h
i
3
∗
∗
a10) E (R2 − R2 ) (R4 − R4 ) = 3h2 (R6 − R4 R2 ) R4 − R22 + OP h3 , as h → 0.
∗
h
i
4
∗
∗
3
a11) E (R2 − R2 ) (R4 − R4 ) = h
∗
3
4 R6 − 3R
4 R2 + 2R2 (R6 − R4 R2 )
+ OP h4 , as
2
2
2
+6 R4 − R2 R8 − R4 − 2R6 R2 + 2R4 R2
h → 0.
i
h
h
i
2
a12) E ∗ (R2∗ − R2 )5 (R4∗ − R4 ) = h3 15 R4 − R22 (R6 − R4 R2 ) + OP h4 , as h → 0.
h
i
a13) E ∗ (R2∗ − R2 )6 (R4∗ − R4 ) = OP h4 , as h → 0.
h
i
a14) E ∗ (R2∗ − R2 ) (R4∗ − R4 )2 = h2 R10 − 2R4 R6 − R8 R2 + 2R42 R2 .
i
h
h
i
a15) E ∗ (R2∗ − R2 )2 (R4∗ − R4 )2 = h2 R4 − R22 R8 − R42 + 2 (R6 − R4 R2 )2 + OP h3 , as h →
0.
h
i
a16) E ∗ (R2∗ − R2 )3 (R4∗ − R4 )2 = OP h3 , as h → 0.
4
h
i
a17) E (R2∗ − R2 )4 (R4∗ − R4 )2 = h3
∗
"
#
2
R8 − R42 3 R4 − R22
+ OP h4 , as h → 0.
2
2
+12 (R6 − R4 R2 ) R4 − R2
Lemma S.6 Let ri∗ ∼ i.i.d. from {ri : i = 1, . . . , 1/h}. Under Assumption H, conditionally on σ, as
h → 0,
h
i
a1) E ∗ (R2∗ − R2 ) V̂ ∗ − V ∗ = h R6 − 3R4 R2 + 2R23 + OP h2 .
a2) E
∗
h
(R2∗
2
− R2 )
∗
V̂ − V
∗
i
2
=h
2 2
−
R
R8 − R42 − 2R6 R2 + 2R4 R22 − 3 R
4
2
+ OP h3 .
3
−2R2 R6 − 3R4 R2 + 2R2
h
i
a3) E ∗ (R2∗ − R2 )3 V̂ ∗ − V ∗ = h2 3 R4 − R22 R6 − 3R4 R2 + 2R23 + OP h3 .
a4)
E
∗
h
(R2∗
4
− R2 )
∗
V̂ − V
∗
i
3
4 R6 − 3R
4 R2 + 2R2 (R6 − R4 R2 )
= h
+6 R4 − R22 R8 − R42 − 2R6 R2 + 2R4 R22
h
3 i
−h3 15 R4 − R22
−h3 20R2 R6 − 3R4 R2 + 2R23 R4 − R22 + OP h4 .
3
2 ∗
∗
∗
= OP h2 .
a5) E (R2 − R2 ) V̂ − V
∗


R4 R2 )2
R4 − R22 R8 − R42 + 2 (R6 −
2 

3
−12 (R6 − R4hR2 ) R4 − R22 i(R2 )
+
O
h
.
a6) E ∗ (R2∗ − R2 )2 V̂ ∗ − V ∗
= h2 

P
2
2
2
+4 (R2 ) 3 R4 − R2
2 3
∗
∗
∗
= OP h3 .
a7) E (R2 − R2 ) V̂ − V
∗
a8)
2 h
i
2
4
∗
∗
∗
E (R2 − R2 ) V̂ − V
= h3 3 R4 − R22
R8 − R42 + 12 (R6 − R4 R2 )2 R4 − R22
h
i
3
2 2
−h 60 R4 − R2 (R6 − R4 R2 ) (R2 )
i
h
3
+h3 60 R4 − R22 (R2 )2 + OP h4 .
∗
Lemma S.7 Let ri∗ ∼ i.i.d. from {ri : i = 1, . . . , 1/h}. Under Assumption H, conditionally on σ,
E ∗ (Sh∗ )
E ∗ Sh∗2
E ∗ Sh∗3
E ∗ Sh∗4
= 0;
= 1;
√
=
hB̃1 ;
= 3 + hB̃2 ;
5
and as h → 0,
E ∗ (Sh∗ Uh∗ )
E ∗ Sh∗2 Uh∗
E ∗ Sh∗3 Uh∗
E ∗ Sh∗4 Uh∗
E ∗ Sh∗ Uh∗2
E ∗ Sh∗2 Uh∗2
= Ã1 + OP (h) ;
√
hÃ2 + OP h3/2 ;
=
= Ã3 + OP (h) ;
√
hD̃ + OP h3/2 ;
=
= OP h1/2 ; E ∗ Sh∗3 Uh∗2 = OP h1/2 ;
= C̃ + OP (h) ; and E ∗ Sh∗4 Uh∗2 = Ẽ + OP (h) .
The bootstrap constants Ã1 , Ã2 , B̃2 , C̃, D̃ and Ẽ are as in Theorem A.2, and Ã3 and B̃1 are such that
Ã3 = 3Ã1 and B̃1 = Ã1 .
Proof of Lemmas S.1−S.7
Proof of Lemma S.1. For (1), note that


 

1/h
1/h
1/h
1/h
X
X
X
X

σ qi σ pj = 
σ qi  
σ pj  − 
σ q+p
i
i=1
i6=j
−1+ 2q
= h
j=1

h
−2+ q+p
2
= h
1− q2
1/h
X
i=1
i=1

−1+ p2
σ qi  h
.
σ qh σ ph − h σ q+p
h

1− p2
h
1/h
X
j=1

−1+ q+p
2
σ pj  − h

1− q+p
2
h
1/h
X
i=1

σ iq+p 
For (2), note that




1/h
1/h
1/h
1/h
X q p
X
X
X
X
X q+p
X q+s p X q p+s
q 
p 
s
s
s

σi σj σk =
σi
σj
σk −
σ q+p+s
−
σ
σ
−
σi σj −
σi σj ,
j
i
i
i=1
i6=j6=k
j=1
k=1
i=1
i6=j
i6=j
i6=j
and then proceed as for (1).
Proof of Lemma S.2. a1) follows from ri = σ i ui , where ui ∼ i.i.d. N (0, 1). For a2), note
2
P1/h
that R2 = i=1 ri2 , where ri2 is (conditional on σ) independent with V ar ri2 = E ri4 − E ri2
=
2
4
2
4
2
2
4
2
µ4 σ i − µ2 σ i = µ4 − µ2 σ i , with σ i ≡ σ i . To prove the remaining results we use the multinomial
formula to compute the coefficients in the expansions. In particular, we have that
(a1 + a2 + · · · + ad )n =
X
n1 ,n2 ,...,nd ≥0
n1 +···+nd =n
n!
an1 an2 · · · and d .
n1 !n2 ! · · · nd ! 1 2
Proof of a3): Write
I1 ≡ E
R2 − µ2 σ 2
3 

1/h 1/h 1/h
XX
X
ri2 − µ2 σ 2i rj2 − µ2 σ 2j rk2 − µ2 σ 2k  .
=E
i=1 j=1 k=1
6
h
3 i
The only non zero contribution to I1 is when i = j = k, in which case we get E ri2 − µ2 σ 2i
=
3
6
µ6 − 3µ2 µ4 + 2µ32 σ 6i and I1 = h2 µ6 − 3µ2 µ4 +
2µ2 σ h , proving a3). Proof of a4): Using the inde2
2
pendence and zero mean property of ri − µ2 σ i , we have that
E
R2 − µ2
σ2
4 1/h
X
=
E
i=1
= E
=
h
ri2
4
µ2 σ 2i
−
+3
1/h
X
E
i6=j
h
ri2 − µ2 σ 2i
2 i
E
h
rj2 − µ2 σ 2j
1/h
h
2 i2 X 4 4
σi σj
σ 8i + 3 E u2i − µ2
1/h
h
i
2 i
u2i − µ2
4 i X
3µ42
6µ22 µ4
−
4µ2 µ6 h3 σ 8h
2 2
+ h3 µ8 − 4µ2 µ6 + 12µ4 µ22 − 6µ42 − 3µ24
µ8 −
i=1
+
= 3h2 µ4 − µ22
σ 4h
i6=j
+ 3 µ4 −
2
µ22
2
2
4
h
σ h − h σ 8h
σ 8h ,
where we have made use of Lemma S.1 and of the following results:
h
4 i
E u2i − µ2
= E u8i − 4u2i µ32 + 6u4i µ22 − 4u6i µ2 + µ42 = µ8 − 3µ42 + 6µ22 µ4 − 4µ2 µ6 .
h
2 i
= E u4i − 2u2i µ2 + µ22 = µ4 − µ22 .
E u2i − µ2
Proof of a5):
1/h
E
h
R2 − µ2 σ 2
V̂ − Vh
i
=
=
µ4 − µ22 −1 X
h
E ri6 − ri2 µ4 σ 4i − µ2 σ 2i ri4 + µ2 µ4 σ 6i
µ4
i=1
2
µ4 − µ22 (µ6 − µ2 µ4 ) 6
µ4 − µ2 −1 2
6
σh.
h h (µ6 − µ2 µ4 ) σ h = h
µ4
µ4
Proof of a6):
1/h
i
2 4
µ4 − µ22 −1 X h 2
=
V̂ − Vh
E
R2 − µ2
h
ri − µ4 σ 4i
E ri − µ2 σ 2i
µ4
i=1
2
µ
−
µ
µ8 − µ24 − 2µ2 µ6 + 2µ22 µ4 8
4
2
2
σh.
= h
µ4
3 µ −µ2
R2 − µ2 σ 2
V̂ − Vh
Proof of a7): Write E
= 4µ 2 h−1 I2 , where by the independence and
σ2
2 4
mean zero property of |ri |q − µ2 σ qi ,
I2 =
1/h
X
i=1
= M1
E
h
1/h
X
ri2
−
3
µ2 σ 2i
σ 10
i + 3E
i=1
3
= 3h
µ4 −
µ22
h
ri4
−
µ4 σ 4i
u2i − µ2
(µ6 −
2 i
µ2 µ4 ) σ 4h
i
E
σ 6h
+3
1/h
X
E
i6=j
u2j − µ2
4
+O h
h
ri2 − µ2 σ 2i
2
rj2 − µ2 σ 2j
1/h
,
u4j − µ4
X
rj4 − µ4 σ 4j
i
σ 4i σ 6j
i6=j
h
i
2 − µ 3 u4 − µ
given Lemma S.1, the fact that σ 10
=
O
(1)
under
our
assumptions,
and
where
M
=
E
u
1
4
2
i
i
h
h
h
2 i
i
2
2
2
4
is a constant, and E ui − µ2
= µ4 − µ2 and E uj − µ2 ui − µ4 = µ6 − µ2 µ4 .
7
Proof of a8): Write E
q
R2 − µ2
mean zero property of |ri | −
I3 =
1/h
X
E
i=1
+6
h
1/h
X
ri2
E
i6=j
=
M1 h5 σ 12
h
4
µ2 σ 2i
ri4
ri2 − µ2 σ 2i
2 i
−
h
µ2 σ qi ,
4
+ 4M2 h
σ2
4 V̂ − Vh
=
µ4 −µ22 −1
µ4 h I3 ,
h
ri2 − µ2 σ 2i
where by the independence and
and Lemma S.1,
−
µ4 σ 4i
E
σ 6h
h
2
i
+4
1/h
X
i6=j
rj2 − µ2 σ 2j
h5 σ 12
h
−
E
2
rj4 − µ4 σ 4j
3 i
E
rj2 − µ2 σ 2j
rj4 − µ4 σ 4j
i
+ 6M3 h4 σ 4h σ 8h − h5 σ 12
h
2
= h 4M2 σ 6h + 6M3 σ 4h σ 8h + O h5 ,
4
i
h
3 i h 2
i
4 4
E uj − µ2 u4j − µ4
ui − µ4 , M2 ≡ E u2i − µ2
u2i − µ2
h
2 i h 2
2 4
i
= µ6 − 3µ2 µ4 + 2µ32 (µ6 − µ2 µ4 ) and M3 ≡ E u2i − µ2
E u i − µ2
ui − µ4
= µ8 − µ24 − 2µ2 µ6 + 2µ22 µ4 µ4 − µ22 , and given the fact that σ qh = O (1) under our assumptions.
2 2
2 i
P1/h h
(µ −µ2 )
2
= 4 µ2 2 h−2 i=1 E ri2 − µ2 σ 2i ri4 − µ4 σ 4i
Proof of a9): Write E
V̂ − Vh
R2 − µ2 σ
=
4
O h2 .
2 2 2
(µ −µ2 )
V̂ − Vh
= 4 µ2 2 h−2 I4 , where by the independence
R2 − µ2 σ 2
Proof of a10): Write E
where M1 ≡ E
h
4
and mean zero property of |ri |q − µ2 σ qi ,
I4 =
1/h
X
E
i=1
+2
h
ri2
1/h
X
E
i6=j
=
−
2
µ2 σ 2i
ri2 − µ2 σ 2i
D1 h5 σ 12
h
+ D2
σ 8h
4
= h
D2 σ 4h
ri4
h4 σ 4h
+ 2D3
−
i
ri4 − µ4 σ 4i
σ 8h
2
µ4 σ 4i
σ 6h
−
h5 σ 12
h
2 +
i6=j
1/h
X
E
E
h
ri2 − µ2 σ 2i
rj2 − µ2 σ 2j
4
+ 2D3 h
+ O h5 ,
σ 6h
2
2 i
E
h
rj4 − µ4 σ 4j
−
h5 σ 12
h
rj4 − µ4 σ 4j
2 i
h
2 2 i
2 i
2
4
given Lemma S.1, and where D1 = E
− µ2
− µ4 , D2 = E ui − µ2
E uj − µ4
=
2
= (µ6 − µ2 µ4 )2 .
µ4 − µ22 µ8 − µ24 and D3 = E u2i − µ2 u4i − µ4
h
u2i
2
u4i
8
Proof of a11): Write E
1/h
X
I5 =
h
E
i=1
+3
ri2 − µ2 σ 2i
1/h
X
1/h
X
R2 − µ2
3
σ2
3 ri4 − µ4 σ 4i
2 i
h
ri2 − µ2 σ 2i
2 i
E
E
h
ri2 − µ2 σ 2i
2
ri4 − µ4 σ 4i
i6=j
h
V̂ − Vh
E
i6=j
+6
1/h
X
+
2 E
i6=j
rj2 − µ2 σ 2j
i
E
h
=
(µ4 −µ22 )
2
h−2 I5 , with
µ24
ri2 − µ2 σ 2i
rj4 − µ4 σ 4j
rj2 − µ2 σ 2j
3 i
E
h
rj4 − µ4 σ 4j
2 i
2 i
rj4 − µ4 σ 4j
5 4 10
5 8 6
5 6 8
6 14
6 14
6 14
= K1 h6 σ 14
h
h
+
3K
+
6K
h
σ
σ
σ
σ
σ
σ
σ
σ
σ
+
K
−
h
−
h
−
h
3
4
2
h
h h
h
h h
h
h h
h
6
8 6
14
= h5 K2 σ 6h σ 8h + 3K3 σ 4h σ 10
h + 6K4 σ h σ h + h (K1 − K2 − 3K3 − 6K4 ) σ h ,
where we have used the independence and mean zero property of |ri |q − µ2 σ qi , Lemma S.1, and where
K1 through K4 are constants depending on µq . Since σ qh = O (1), the result follows.
2 4 2
(µ −µ2 )
2
= 4 µ2 2 h−2 I6 , with
Proof of a12): Write E
R2 − µ2 σ
V̂ − Vh
4
I6 =
1/h
X
E
i=1
+8
h
1/h
X
ri2
+6
1/h
X
+6
i
E
h
ri2 − µ2 σ 2i
2
ri4 − µ4 σ 4i
2 i
E
h
ri2 − µ2 σ 2i
3 i
E
E
h
ri2 − µ2 σ 2i
2
ri4 − µ4 σ 4i
E
h
1/h
X
1/h
X
i6=j6=k
E
ri2 − µ2 σ 2i
h
ri2
−
2 i
2
µ2 σ 2i
h
E
i
E
E
i
h
ri2 − µ2 σ 2i
rj2 − µ2 σ 2j
2 i
h
rj2 − µ2 σ 2j
2
2 i
rj2 − µ2 σ 2j
E
h
E
h
rj4 − µ4 σ 4j
2 i
2 i
rj4 − µ4 σ 4j
E
4 i
rj4 − µ4 σ 4j
rj2 − µ2 σ 2j
h
rj2 − µ2 σ 2j
E
rj2 − µ2 σ 2j
h
E
i6=j
ri4 − µ4 σ 4i
i6=j6=k
+12
+
1/h
X
3
i6=j
+3
i
ri2 − µ2 σ 2i
i6=j
1/h
X
−
2
µ4 σ 4i
h
i6=j
+4
ri4
E
i6=j
1/h
X
−
4
µ2 σ 2i
rj4 − µ4 σ 4j
rk4 − µ4 σ 4k
rj4 − µ4 σ 4j
2 2
= h5 3J7 σ 4h σ 8h + 12J8 σ 6h σ 4h + O h6 + O h7 ,
E
i
2 i
rk2 − µ2 σ 2k
rk4 − µ4 σ 4k
given the independence and mean zero property of |ri |q − µ2 σ qi , Lemma S.1, and where
h
2 i
2
2 i2 h 4
and
µ8 − µ24
=
µ4 − µ22
E ui − µ4
J7
=
E u2i − µ2
h
i
2
2
J8 = E u2i − µ2
E u2i − µ2 u4i − µ4
= µ4 − µ22 (µ6 − µ2 µ4 )2 .
9
Proof of Lemma S.3. The first two results are obvious given Sh . The remaining results follow from
the definition of Sh and Lemma S.2.
Proof of Lemma S.4. Part a1) follows from the properties of the i.i.d. bootstrap. The remaining results follow from a1), given the binomial expansions. Note in particular hthat since Riq =
2
q = OP (hq ) . For instance, for a2), E ∗ ri∗2 − hR2
=
OP (1), it follows that E ∗ ri∗2 − hR2
E ∗ ri∗4 − 2ri∗2 hR2 + (hR2 )2 = h2 R4 − R22 . The other results follows similarly.
Proof of Lemma S.5. For a1), since ri∗ are i.i.d. from {ri : i = 1, . . . , 1/h}, it follows that


1/h
1/h
X
X
V ∗ = h−1 V ar∗ 
ri∗2  = h−1
V ar∗ ri∗2 = h−2 V ar∗ r1∗2 .
i=1
i=1
∗2 2
= h2 R4 − (hR2 )2 . Thus, V ∗ = R4 − R22 . Part a2) follows
P1/h
because V ∗ = R4 −R22 and V̂ ∗ = R4∗ −R2∗2 . For the remaining of the proof, note that i6=j 1 = h−2 −h−1 ,
P
P1/h
−3 + 2h−1 − 3h−2 , and
−4 − 6h−3 + 11h−2 − 6h−1 . In addition, note that
i6=j6=k 1 = h
i6=j6=k6=m 1 = h
∗2
But V ar∗ r1
∗4
= E ∗ r1
R2∗
− R2 =
− E ∗ r1
1/h
X
i=1
ri∗2
− hR2
and
R4∗
−1
− R4 = h
1/h
X
i=1
ri∗4 − h2 R4 ,
where for any q > 0 |ri∗ |q − hq/2 Rq are (conditionally on the sample) i.i.d. with zero mean, and
Rq = OP (1). Using this independence property, we evaluate the bootstrap expectations of the sums
of products and cross products of |ri∗ |q − hq/2 Rq by relying on Lemma S.4 to compute the appropriate
bootstrap moments of products and cross products of |ri∗ |q − hq/2 Rq . We proceed as in the proof of
Lemma S.2 and use the multinomial expansions to compute the number of coefficients in each sum.
Proof of Lemma S.6. Using part a2) of Lemma S.5, for q = 1, . . . , 4, we can write
i
h
i
h
= E ∗ [(R2∗ − R2 )q (R4∗ − R4 )] − E ∗ (R2∗ − R2 )2+q
E ∗ (R2∗ − R2 )q V̂ ∗ − V ∗
i
h
−2 (R2 ) E ∗ (R2∗ − R2 )1+q
≡ I1q − I2q − I3q .
(3)
Similarly, for q = 1, . . . , 4, note that
2 h
i
h
i
q
∗
∗
∗
∗
= E ∗ (R2∗ − R2 )q (R4∗ − R4 )2 − 2E ∗ (R2∗ − R2 )2+q (R4∗ − R4 )
E (R2 − R2 ) V̂ − V
h
i
i
h
−4 (R2 ) E ∗ (R2∗ − R2 )1+q (R4∗ − R4 ) + E ∗ (R2∗ − R2 )4+q
h
i
h
i
+4 (R2 ) E ∗ (R2∗ − R2 )3+q + 4 (R2 )2 E ∗ (R2∗ − R2 )2+q
For a1), set q = 1 in (3). We have that
I11 = E ∗ [(R2∗ − R2 ) (R4∗ − R4 )] = h (R6 − R4 R2 )
h
i
I21 = E ∗ (R2∗ − R2 )3 = h2 R6 − 3R4 R2 + 2R23
h
i
I31 = 2R2 E ∗ (R2∗ − R2 )2 = 2R2 h R4 − R22 ,
by Lemma S.5. a8), a3), a1), respectively. Thus
h
i
E ∗ (R2∗ − R2 ) V̂ ∗ − V ∗
= h (R6 − R4 R2 ) − 2R2 R4 − R22 − h2 R6 − 3R4 R2 + 2R23
= h R6 − 3R4 R2 + 2R23 + OP h2 .
10
The remaining results follow similarly.
Proof of Lemma S.7. The proof follows the proof of Lemma S.3, given the definition of V ∗ and
given Lemmas S.5 and S.6.
Proof of Propositions 4.2 and 4.3.c) in Section 4
Proof of Proposition 4.2. When g (z) = log z, we have that
√
"
00
2
σ6
(σ 4 )1/2 glog (σ 2 )
1 (σ 4 )1/2 2
2
q1,log (x) =
(2x + 1) + √
=√
x
0 (σ 2 )
3 (σ 4 )3/2
2 glog
2 σ2
!
#
4 σ6 σ2
2 σ6 σ2
−1 +
,
3 (σ 4 )2
3 (σ 4 )2
whereas for g (z) = z , g 00 (z) = 0 and
√
"
2
σ6
1 (σ 4 )1/2 2
2
q1 (x) =
(2x + 1) = √
x
3 (σ 4 )3/2
2 σ2
Since (σ 4 )2 ≤ σ 6 σ 2 by the Cauchy-Schwartz inequality, it follows that
4 σ6 σ2
3 (σ 4 )2
1
3
≤
!
#
2 σ6 σ2
+
.
3 (σ 4 )2
4 σ6 σ2
3 (σ 4 )2 −1
<
4 σ6 σ2
3 (σ 4 )2 ,
which implies
that q1,log (x) < q1 (x) for x fixed and non-zero. When x = 0, it follows trivially that q1,log (0) = q1 (0) .
Proof of Proposition 4.3.c). Define C =
C ∗|
6
√ 4σ
2(σ 4 )3/2
and C ∗ =
15σ 6 −9σ 4 σ 2 +2(σ 2 )3
,
(3σ 4 −(σ 2 )2 )3/2
and note that
C > 0. It suffices to prove that |C −
≤ |C|, which in turn is equivalent to proving 0 ≤ C ∗ ≤ 2C.
∗
Next we show that C ≥ 0. The Jensen’s inequality implies that σ 4 ≥ (σ 2 )2 , and since σ 4 > 0, it
follows that the denominator of C ∗ is positive. For the numerator of C ∗ , note we can write
15σ 6 − 9σ 4 σ 2 + 2(σ 2 )3 ≥ 15σ 6 − 9(σ 4 )3/2 + 2(σ 2 )3 ≥ 9((σ 4 )3/2 − (σ 4 )3/2 ) + 6σ 6 + 2(σ 2 )3 ,
using −(σ 2 )2 ≥ −σ 4 . Since the function ψ(x) = x3/2 for x > 0 is convex, we have that (σ 4 )3/2 −
(σ 4 )3/2 ≥ 0, which implies 15σ 6 − 9σ 4 σ 2 + 2(σ 2 )3 ≥ 6σ 6 + 2(σ 2 )3 > 0, proving that the numerator of
∗
C ∗ is also positive. Next we prove CC ≤ 2. We can write
√
C∗
15σ 6 − 9σ 4 σ 2 + 2(σ 2 )3 2 2(σ 4 )3/2
≡ C1 × C2 .
=
C
8σ 6
(3σ 4 − (σ 2 )2 )3/2
We show that C1 ≤ 2 and C2 ≤ 1. First, note that
15σ 6 − 9σ 4 σ 2 + 2(σ 2 )3
8σ 6
≤ 2 ⇐⇒ 15σ 6 −9σ 4 σ 2 +2(σ 2 )3 ≤ 16σ 6 ⇐⇒ 0 ≤ σ 6 +7σ 4 σ 2 +2σ 2 σ 4 − (σ 2 )2 ,
which proves the result since σ 4 − (σ 2 )2 ≥ 0 and 0 ≤ σ 6 + 7σ 4 σ 2 . Finally, we have that
√
2 2(σ 4 )3/2
(3σ 4 − (σ 2 )2 )3/2
3
3
≤ 1 ⇐⇒ 8(σ 4 )3 ≤ 3σ 4 − (σ 2 )2 ⇐⇒ 8(σ 4 )3 ≤ 2σ 4 + σ 4 − (σ 2 )2
,
which holds true since σ 4 − (σ 2 )2 ≥ 0.
11