Submitted to the Annals of Statistics
arXiv: arXiv:1312.2645
SUPPLEMENT TO SUBSAMPLING BOOTSTRAP OF
COUNT FEATURES OF NETWORKS
By Sharmodeep Bhattacharyya∗,† and Peter J. Bickel∗
University of California, Berkeley∗
Oregon State University†
Appendix.
A1. Variance of P̂b1 (R). The variance of P̂b1 (R) is

1
Varb 
X
m
p |Iso(R)| S⊆Km ,S ∼
=R
X
Varb 
S⊆Km ,S ∼
=R
X
Varb 
m
p |Iso(R)|

1(S ⊆ H)G =

!2
1
=


1(S ⊆ H)G
S⊆Km ,S ∼
=R
Ñ
é2
X
Eb 
Ñ
−
1(S ⊆ H)
S⊆Km ,S ∼
=R

Eb 
X
S⊆Km ,S ∼
=R
Ñ
Eb 
é2
X
1(S ⊆ H)
S⊆Km ,S ∼
=R

G

= Eb 
X
S⊆Km ,S ∼
=R


+ Eb 

X
S,T ⊆Km
S,T ∼
=R,S6=T
Thus,
n−p m−p
n 1(S
m
S⊆Kn ,S ∼
=R
X
1
⊆ G)

G
é2
1(S ⊆ H)G

1(S ⊆ H)G


1(S, T ⊆ H)G

= I + II (Suppose)
I=

1(S ⊆ H)G
2
BHATTACHARYYA AND BICKEL

X

II = Eb 

S,T ⊆Km
S,T ∼
=R,S6=T


1(S, T ⊆ H)G

Now, a host of subgraphs can be formed by the intersection of two copies
of R. The number of intersected vertices can range from 0 to p − 1. Let us
consider, that for number of vertices in intersection as k (k = 1, . . . , (p − 1)),
the number of graph structures that can be formed is gk and we represent
that graph structure by Wjk , where, j = 1, . . . , gk . Thus,
II =
p−1
gk
XX
n−(2p−k) m−(2p−k)
1(S
n
m
X
k=0 j=1 S⊆Kn ,S ∼
=Wjk
⊆ G)
So,
Ñ
é2
X
Eb 
1(S ⊆ H)
S⊆Km ,S ∼
=R

G =
+
n−p m−p
n 1(S
m
S⊆Kn ,S ∼
=R
X
⊆ G)
p−1
gk
XX
k=0 j=1
X
S⊆Kn ,S ∼
=Wjk

Varb 
X
S⊆Km ,S ∼
=R

1(S ⊆ H)G
=
+
n−(2p−k) m−(2p−k)
1(S
n
m
n−p m−p
n 1(S
∼
m
S⊆Kn ,S =R
X
⊆ G)
⊆ G)
p−1
gk
XX
k=0 j=1
X
S⊆Kn ,S ∼
=Wjk
Ñ
−
n−(2p−k) m−(2p−k)
1(S
n
m
n−p m−p
n 1(S
m
S⊆Kn ,S ∼
=R
X
⊆ G)
⊆ G)
é2
3
SUPPLEMENT


X
1
Varb  m
1(S ⊆ H)G
p |Iso(R)| S⊆K ,S ∼
=R
m
1
m
p |Iso(R)|
=
1
m
p |Iso(R)|
−
+
!2 
n−p m−p

n 1(S
m
S⊆Kn ,S ∼
=R
X

⊆ G)
!2 Ñ
n−p m−p
n 1(S
m
S⊆Kn ,S ∼
=R
X


!2 p−1 g
k
X X
X
1


m

p |Iso(R)|
k=0 j=1 S⊆Kn
S∼
=Wjk
é2 
⊆ G)


n−(2p−k) m−(2p−k)
n
m



1(S ⊆ G)
So,
î
Varb P̂b1 (R)
ó
=
−
+
1
m
p |Iso(R)|
1
m
p |Iso(R)|
!2 
n−p m−p

n 1(S
m
S⊆Kn ,S ∼
=R
X
!2 Ñ
⊆ G)
n−p m−p
n 1(S
∼
m
S⊆Kn ,S =R
X



!2 p−1 g
k
X X
X
1


m

p |Iso(R)|
k=0 j=1 S⊆Kn
S∼
=Wjk
é2 
⊆ G)


n−(2p−k) m−(2p−k)
1(S
n
m



⊆ G)
A2. Proof of Theorem 3.1.
Proof.
(i) Now, let us try to try to find the expectation of P̂b1 (R)
4
BHATTACHARYYA AND BICKEL
under the sampling distribution conditional on the given data G.

Eb 
m
p |Iso(R)| S⊆Km ,S ∼
=R

1
=
=
1
X
E
m
p |Iso(R)|
1
X

S⊆Km ,S ∼
=R
1
X

1(S ⊆ H)G

1(S ⊆ H)G
X
1(S
m
n
p |Iso(R)| H⊆G,|H|=m m S⊆Km ,S ∼
=R
=
1
X
1
X
m
n 1(S
p |Iso(R)| S⊆Kn H⊇S,H⊆G m
S∼
=R
|H|=m
X
1
= m
p |Iso(R)| S⊆K ,S ∼
=R
n
n−p m−p
n 1(S
m
⊆ H)
⊆ G)
⊆ G)
X
1
= n
1(S ⊆ G)
p |Iso(R)| S⊆K ,S ∼
=R
n
So, we have,
Eb [P̄B1 (R)|G] = Eb [P̂b1 (R)|G] = P̂ (R)
(ii) Given G,
Varb [ρ−e P̄B1 (R)|G] = ρ−2e
1
B2
B
X
Varb [P̂b1 (R)]
b=1
é
B
X
+
Covb (P̂b1 (R), P̂b0 1 (R))
b,b0 =1,b6=b0
Now, the formula for Varb [P̂b1 (R)] from A1 we get that
î
Varb ρ−e
n P̂b1 (R)
ó
=
−
+
ρ−e
n
m
p |Iso(R)|
ρ−e
n
m
|Iso(R)|
p
!2 
n−p m−p

n 1(S
∼
m
S⊆Kn ,S =R
X

⊆ G)
!2 Ñ
n−p m−p
n 1(S
m
S⊆Kn ,S ∼
=R
X


!2 p−1 g
k
X X
X
ρ−e

n

m

p |Iso(R)|
k=0 j=1 S⊆Kn
S∼
=Wjk
é2 
⊆ G)


n−(2p−k) m−(2p−k)
1(S
n
m



⊆ G)
5
SUPPLEMENT
So,
î
ó
Varb ρ−e
n P̂b1 (R)
= O
Ç
= O
ρen
Ñ
!
1
+O
m
p
1
mp ρen
m eW
pW ρn
m2 2e
p ρn
å
Å
1
+O
m
é
ã
where, W = S ∪ S 0 , pW = |V (W )| and eW = |E(W )|.
(iii) Here, we use properties of the underlying model. Let us condition on
ξ = {ξ1 , . . . , ξn } and the whole graph G separately. Now, conditioning
on ξ, we get the main term of P̂ (R) to be,
(0.1)
Ñ
é
E(ρ−e P̂ (R)|ξ) =
1
X
w(ξi , ξj ) +O(n−1 λn ).
Y
n
p |Iso(R)| S⊆Kn ,S ∼
=R
(i,j)∈E(S)
We shall use the same decomposition as used in [1] of (ρ−e
n P̄B1 (R) −
P̃ (R)) into
Ä
−e
(ρ−e
P̄B1 − Eb [P̂b1 (R)|G]
n P̄B1 (R) − P̃ (R)) = ρn
Ä
ä
ä
+ρ−e
P̂ (R) − E(P̂ (R)|ξ)
n
+E(P̂ (R)|ξ)ρ−e
n − P̃ (R)
Let us define,
U3 = E(P̂ (R)|ξ)ρ−e
n − P̃ (R)
Ä
U2 = ρ−e
P̂ (R) − E(P̂ (R)|ξ)
n
ä
Ä
ä
U1 = ρ−e
P̄B1 − Eb [P̂b1 (R)|G]
n
Now, it is easy to see that
Var(ρ−e P̄B1 (R)) = E(Var(ρ−e P̄B1 (R)|G)) + Var(E(ρ−e P̄B1 (R)|G))
= E(Var(ρ−e P̄B1 (R) − P̂ (R)|G) + Var(P̂ (R))
= E(Var(U1 |G)) + E(Var(P̂ (R)|ξ)) + Var(E(P̂ (R)|ξ))
= E(Var(U1 |G)) + E(Var(U2 |ξ)) + Var(U3 )
−e
We shall try to see the behavior of Var(U1 |G)
= Varb [ρ P̄B1 (R)|G].
î
ó
1
1
From (ii) we get that, Varb ρ−e
n P̂b1 (R) = O mp ρe ∨ m . Similarly,
n
1
−e
Covb [ρ−e
n P̂b1 (R), ρn P̂b0 1 (R)] = O( m ) for acyclic and k-cycle R following similar steps as variance in Appendix A1. If we consider the uniform probability for bootstrap to be γ, then, B = O(γnp ). Note that,
6
BHATTACHARYYA AND BICKEL
if E(Hb ) ∩ E(Hb0 ) = φ, then, Covb (P̂b1 (R), P̂b0 1 (R)) = 0. The number
of pairs such that E(Hb ) ∩ E(Hb0 ) 6= φ is O(m2 γ 2 n2m−2 ). Also, the
number of edges for the leading term in the covariance is equal to or
more than 2e. So,
Ç
å
Ç
å
1
m2 γ 2 n2m−2
E(Varb [ρ P̄B1 (R)|G]) = O
+
O
B(mp ρen ∧ m)
mγ 2 n2m
å
Ç
Ç
å
1
1
m
=O
=O
+
+ o(n−1 )
B(mp ρen ∧ m) n2
B(mp ρen ∧ m)
−e
The second equality follows since we have m/n → 0 as n → ∞. So,
since, B(mp ρen ∧ m) > O(n), we have, E(Var(U1 |G)) = o(n−1 ).
Now, by proof of Theorem 1 in [1], we have,
Var(U2 ) = o(n−1 )
Var(U3 ) = o(n−1 )
−e P̄ (R)) = o(n−1 ). Since, we already know √nSo, we get, Var(ρ
B1
Ä
ä
√
consistency of ρ−e
n-consistency of
n P̂ (R) − P̃ (R)) , this proves the
−e
ρ−e
n P̄B1 (R) to ρn P̂ (R).
A3. Proof of Theorem 3.2. For variance calculation, we also need the
joint inclusion probability of two items, S, S 0 ∈ Sp , which are subgraphs of
G induced by the set of vertices {w1 , . . . , wp } and {w10 , . . . , wp0 } respectively,
0
where, we take that wi+1 wi and wi+1
wi0 , i = 1, . . . , p − 1. So,
πSS 0 ≡ Inclusion Probability of S and S 0 = P[(w1 , . . . , wp ) is selected
and (w10 , . . . , wp0 ) is selected]
=
p
Y
d=1
(qd )z1d
p Ä
Y
qd2
d=1
where,
®
z1d =
®
z2d =
1(wd = wd0 ),
0
1((wd , wd−1 ) = (wd0 , wd−1
)),
for d = 1
for d = 2, . . . , p
1(wd 6= wd0 ),
0
1((wd , wd−1 ) 6= (wd0 , wd−1
)),
for d = 1
for d = 2, . . . , p
äz2d
7
SUPPLEMENT
(i) We know that P̂b2 (R) is a Horvitz-Thompson estimator with inclusion
Q
probability of each population unit to be π = pd=1 qd . So, according
to the sampling theory [2], P̂b2 (R) is an unbiased estimator of P̂ (R)
given the network G, if P(P̂b2 (R) = 0|P̂ (R)) → 0 as n → ∞. Now,
P(P̂b2 (R) = 0|P̂ (R)) ≤ (1 − qd )λn for all d = 1, . . . , p. For all d =
1, . . . , p, (1 − qd )λn → 0 if λn qd → ∞. So, under the condition, λn qd ∞
and qd → 0 as n → ∞, we have, P̂b2 (R) is an asymptotically unbiased
estimator of P̂ (R).
(ii) The variance of P̂b2 (R) coming from the bootstrap sampling only is
given by
î

ó
1 1 − π X
Varb ρ−e
1(S ∼
= R)
n P̂b2 (R) =
N2
π S∈S
p

X
+
S,S 0 ∈Sp ,S6=S 0
πSS 0 − π 2
1(S ∼
= R, S 0 ∼
= R)
π2
where,
Ç å
N=
ρen
n
|Iso(R)|
p
From the formula of Varb [P̂b2 (R)|G], we see that, the covariance terms
vanishes when πSS 0 = π 2 . Now, if q1 = 1, then, πSS 0 = π 2 if E(S) ∩
E(S 0 ) = φ. The number of pairs such that E(S) ∩ E(S 0 ) 6= φ is
O(p2 n2p−2 ).
Now, the condition of q1 = 1 is a bit restrictive. In stead, if we have
q1 → 1 as n → ∞, then, the highest order term of covariance term
comes from the case when E(S) ∩ E(S 0 ) 6= φ but the root nodes are
same that is w1 = w10 . So, for some constant C > 0,
1
N2
≤
X
S,S 0 ∈Sp ,S6=S 0
C
N2
πSS 0 − π 2
1(S ∼
= R, S 0 ∼
= R)
π2
X
S,S 0 ∈Sp ,S6=S 0
q1 − q12
1(S ∼
= R, S 0 ∼
= R)
q12
ÇÅ
ã
å
1
n2p−1
=O
−1
q1
n2p
ÅÅ
ã ã
1
1
−1
=O
q1
n
8
BHATTACHARYYA AND BICKEL
Now, for the variance term to vanish we need the conditions q1 = 1 or
q1 → 1 and qd → 0 and λn qd → ∞ for d = 2, . . . , p as n → ∞. Since,
we know that O(1) ≤ λn O(n), we get nqd → ∞ for d = 2, . . . , p as
n → ∞. So, we have
Å
ã Å
ã
1 1−π X
1
1
∼
1(S
R)
=
−
1
O
=
N 2 π S∈S
π
np ρe
p
Ç
= O
1
p
n ρen π
å
p
Y
1
= O
nρne−p+1
!
1
·
λ q
d=2 n d
So,
Ä
E
î
óä
Varb ρ−e
n P̂b2 (R)
ÅÅ
=O
ã
ã
1
1
−1
+O
q1
n
1
nρe−p+1
n
p
Y
1
·
λ q
d=2 n d
!
(iii) We shall use the same decomposition as used in [1] of (ρ−e
n P̄B2 (R) −
P̃ (R)) into
Ä
−e
(ρ−e
P̄B2 − Eb [P̂b2 (R)|G]
n P̄B2 (R) − P̃ (R)) = ρn
Ä
ä
ä
+ρ−e
P̂ (R) − E(P̂ (R)|ξ)
n
+E(P̂ (R)|ξ)ρ−e
n − P̃ (R)
Let us define,
U3 = E(P̂ (R)|ξ)ρ−e
n − P̃ (R)
Ä
ä
Ä
ä
U2 = ρ−e
P̂ (R) − E(P̂ (R)|ξ)
n
U1 = ρ−e
P̄B2 − Eb [P̂ (R)|G]
n
Now, it is easy to see that
Var(ρ−e P̄B2 (R)) = E(Var(ρ−e P̄B2 (R)|G)) + Var(E(ρ−e P̄B2 (R)|G))
= E(Var(ρ−e P̄B2 (R) − P̂ (R)|G) + Var(P̂ (R))
= E(Var(U1 |G)) + E(Var(P̂ (R)|ξ)) + Var(E(P̂ (R)|ξ))
= E(Var(U1 |G)) + E(Var(U2 |ξ)) + Var(U3 )
We shall try to see the behavior of Varb [ρ−e
n P̂b2 (R)|G].
Ä
E
î
óä
Varb ρ−e
n P̂b2 (R)
ÅÅ
=O
ã
ã
1
1
−1
+O
q1
n
1
nρe−p+1
n
p
Y
1
·
λ
q
d=2 n d
!
9
SUPPLEMENT
Now, since the bootstrap samples for subgraph sampling are selected
independently, we have that,
Ä
E
î
Varb ρ−e
n P̄B2 (R)
ÅÅ
óä
=O
ã
ã
1
1
−1
+O
q1
nB
p
Y
1
Bnρne−p+1
ä
Ä
1
·
λ q
d=2 n d
Now, under the condition B1 q11 − 1 → 0, qd → 0 for all d = 1, . . . , p
Q
1
, we have
and B pd=2 qd ≥ np−1
ρe
n
Ä
ä
−1
E(Var(U1 |G)) = E Varb [ρ−e
n P̂b2 (R)|G] = o(n )
Now, by proof of Theorem 1 in [1], we have,
Var(U2 ) = o(n−1 )
Var(U3 ) = o(n−1 )
−e P̄ (R)) = o(n−1 ). Since, we already know √nSo, we get, Var(ρ
B2
Ä
ä
√
consistency of ρ−e
P̂
(R)
− P̃ (R)) , this proves the n-consistency of
n
−e
ρ−e
n P̄B2 (R) to ρn P̂ (R).
B1. Proof of Proposition 6.
î
ó
σ 2 (R; ρ) = Var ρ−e P̂ (R)


= Var 
1(S ⊆ G) 
e
ρ np |Iso(R)|
,S ∼
=R
X
S⊆Kn
2

1
Ä
X
=Ä
1(S ⊆ G) − P̃ (R)
ä2 E 
n
e
∼
ρ p |Iso(R)|
S⊆Kn ,S =R

1

ä2 E 

=Ä
ρe np |Iso(R)|
ä2

X
S,T ⊆Kn
S,T ∼
=R,S∩T 6=∅

1(S, T ⊆ G)
− 1 −
n−p !
Ä
ä2
p
P̃ (R)
n
p
If R is a connected subgraph, then, we can write,

Ä
ρe
1
n
p
|Iso(R)|

ä2 E 


X
S,T ⊆Kn
S,T ∼
=R,S∩T 6=∅

1(S, T ⊆ G)

1
=Ä
ä2
n
e
ρ p |Iso(R)|

X
W :W =S∪T
S,T ∼
=R,S∩T 6=∅
E

X
W ⊆Kn
1(W ⊆ G)
!
10
BHATTACHARYYA AND BICKEL
Ä
ä
σ(R1 , R2 ; ρ) = Cov ρ−e1 P̂ (R1 ), ρ−e2 P̂ (R2 )
1
ä×
=Ä
n
n
ρe1 +e2 p1 p2 |Iso(R1 )||Iso(R2 )|
Ü
êÜ
X

E

ê
X
1(S ⊆ G)
S⊆Kn ,
S∼
=R1
1(S ⊆ G)
S⊆Kn ,
S∼
=R2

 − P̃ (R1 )P̃ (R2 )


=Ä
ρe1 +e2


1
äE 

n n
p1 p2 |Iso(R1 )||Iso(R2 )|
X

1(S, T ⊆ G)

S,T ⊆Kn
S∼
=R1 ,T ∼
=R2 ,S∩T 6=∅
n−p1 !
p
P̃ (R1 )P̃ (R2 )
− 1 − n2 p2
If R is a connected subgraph, then, we can write,

1
Ä
ρe1 +e2
=Ä
ρe1 +e2
n n
p1 p2 |Iso(R1 )||Iso(R2 )|


äE 

X

S,T ⊆Kn
S∼
=R1 ,T ∼
=R2 ,S∩T 6=∅
1(S, T ⊆ G)


1
ä
n n
p1 p2 |Iso(R1 )||Iso(R2 )|
X
E
W :W =S∪T
S∼
=R1 T ∼
=R2 ,S∩T 6=∅

X
1(W ⊆ G)
W ⊆Kn
B2. Proof of Lemma 7. Let us define
1/(1 − x)
σ̃ 2 (R) = Ä
ä2
ρen np |Iso(R)|
where, x = 1 −
((n−p)!)2
n!(n−2p)!
σ̃(R1 , R2 ) = Ä e1 +e2
ρn
X
1(W ⊆ G) −
W ⊆Kn :W =S∪T,
S,T ∼
=R,|S∩T |=1,p
2
xρ−2e
n P̂ (R)
(1 − x)
and
1/(1 − y)
ä
n n
p1 p2 |Iso(R1 )||Iso(R2 )|
−(e1 +e2 )
X
W ⊆Kn ,W =S∪T,
S∼
=R1 ,T ∼
=R2 ,|S∩T |=1
1(W ⊆ G) −
yρn
P̂ (R1 )P̂ (R2 )
.
(1 − y)
11
SUPPLEMENT
where, y = 1 −
Now,
(n−p1 )!(n−p2 )!
n!(n−p1 −p2 )!
.


î
ó

 xρ−2e E P̂ (R)2
X
n


1(W ⊆ G)−
E σ̃ 2 (R) = Ä
ä2 E 


(1 − x)
ρen np |Iso(R)|
W ⊆Kn :W =S∪T,
î
ó
1/(1 − x)
S,T ∼
=R,|S∩T |=1,p


î
Ä
ä
ó
 xρ−2e Var P̂ (R) + P (R)2

X
n


1(W ⊆ G)−
=Ä
ä2 E 
n


(1 − x)
e
ρn p |Iso(R)|
W ⊆Kn :W =S∪T,
1/(1 − x)
S,T ∼
=R,|S∩T |=1,p

=
1
Ä
1 − x ρe

1
n
n p


ä2 E 

|Iso(R)|
X



1(W ⊆ G) −
W ⊆Kn :W =S∪T,
S,T ∼
=R,|S∩T |=1,p
Ä
ä
xρ−2e
n Var P̂ (R)
−
1
1
=
Ä
ä
1 − x ρe n |Iso(R)| 2
p
xP̃ (R)2
1−x
(1 − x)

X
E

X
1(W ⊆ G) −
W ⊆Kn
W :W =S∪T
S,T ∼
=R,S∩T 6=∅

xP̃ (R)2
1−x

Ä
ä

 xρ−2e Var P̂ (R)
X
n


1(W ⊆ G)−
−
ä2 E 
Ä
n


(1 − x)
e
(1 − x) ρn p |Iso(R)|
W ⊆Kn :W =S∪T,
1
Ä
=
Var ρ−e
n P̂ (R)
(1 − x)
S,T ∼
=R,1<|S∩T |<p
Ä
ä
−
xρn−2e Var P̂ (R)
(1 − x)
ä
Ä
Ä
ä
Ä
ää
− o Var ρ−e
n P̂ (R)
Ä
Ä
ää
−e
= Var ρ−e
n P̂ (R) − o Var ρn P̂ (R)
Similarly, we get,
Ä
ä
Ä
Ä
ää
−e2
−e1
−e2
1
E [σ̃(R1 , R2 )] = Cov ρ−e
n P̂ (R1 ), ρn P̂ (R2 ) − o Cov ρn P̂ (R1 ), ρn P̂ (R2 )
Now, from the Theorem 1(a) in [1], we know that as λn → ∞, if ρ̂n =
as defined in (2.6),
ρ̂n P
→1
ρn
So, using the estimate ρ̂n , we get that,
σ̂ 2 (R) P
→ 1,
σ 2 (R; ρ)
σ̂(R1 , R2 ) P
→1
σ(R1 , R2 ; ρ)
D̄
n−1
12
BHATTACHARYYA AND BICKEL
B3. Proof of Lemma 8. Given G,
î
2
E σ̂Bi
(R)|G
=
ó
Ä
X
W =S∪T,S,T ∼
=R,
|S∩T |=1,p
=
X
W =S∪T,S,T ∼
=R,
|S∩T |=1,p
(1 −
Ä
ä
−2e
n xE ρ̂n
P̄Bi (R)2 |G
pW |Iso(R)|
Ä
ä2 E P̄Bi (W )|G −
(1 − x)
x) ρ̂en np |Iso(R)|
ρ̂enW
ä
(1 −
Ä
n xρ̂n−2e P̂ (R)2 xVar
pW |Iso(R)|
−
Ä
ä2 P̂ (W )−
(1 − x)
1
x) ρ̂en np |Iso(R)|
ρ̂enW
ä
Ä
2
= σ̂ (R) −
xVar T̂Bi (R)
−x
= σ̂ 2 (R) − o σ̂ 2 (R)
1−x
where the last inequality follows since x = O
Theorem 3.2 for i = 1, 2.
Similarly, we get,
Ä
Ä ä
1
n
Ä
ä
T̂Bi (R)
ä
and Theorem 3.1 and
E [σ̂Bi (R1 , R2 )|G] = σ̂(R1 , R2 ) − o σ̂ 2 (R)
ä
So using Lemma 7, we have that,
2 (R)
σ̂Bi
P
→ 1,
2
σ (R; ρ)
σ̂Bi (R1 , R2 ) P
→1
σ(R1 , R2 ; ρ)
for i = 1, 2
References.
[1] Bickel, P. J., Chen, A. and Levina, E. (2011). The method of moments and degree
distributions for network models. Ann. Statist. 39 2280–2301. . MR2906868
[2] Thompson, S. K. (2012). Sampling, third ed. Wiley Series in Probability and Statistics. John Wiley & Sons Inc., Hoboken, NJ. . MR2894042 (2012k:62024)
Department of Statistics
44 Kidder Hall
Corvallis, OR, 97331
E-mail: bhattash@science.oregonstate.edu
Department of Statistics
367 Evans Hall
Berkeley, CA, 94720
E-mail: bickel@stat.berkeley.edu