and this is conclusion i. To prove ii we use similar type of argument. Here we define new random variables ˜
Λ
n
with P ˜ Λ
n
= λ
n ,k
, λ
n ,l
= 1n
2
for 1 ≤ k, l ≤ n. Similarly define ˜E
n
and ˜ Y
n
. Again ˜
Λ
n
= ˜ E
n
+ ˜ Y
n
and P
kY
n
k ε = 1
n
2 n
X
k ,l=1
P k y
n ,k
, y
n ,l
k ε → 0, as n → ∞. So ˜
Λ
n
and ˜ E
n
will have same limiting distribution and hence conclusion ii holds. We use normal approximation heavily in our proofs. Lemma 7 is a fairly standard consequence
of normal approximation and follows easily from Bhattacharya and Ranga Rao [1976] Corollary 18.1, page 181 and Corollary 18.3, page 184. We omit its proof. Part i will be used in Section
4.1– 4.4 and Part ii will be used in Section 4.4.
Lemma 7. Let X
1
, . . . , X
k
be independent random vectors with values in R
d
, having zero means and an average positive-definite covariance matrix V
k
= k
−1
P
k j=
1
C ovX
j
. Let G
k
denote the distribution of k
−12
T
k
X
1
+ . . . + X
k
, where T
k
is the symmetric, positive-definite matrix satisfying T
2 k
= V
−1 k
, n ≥ 1.
If for some δ 0, EkX
j
k
2+δ
∞, then there exists C 0 depending only on d, such that i
sup
B ∈C
|G
k
B − Φ
d
B| ≤ C k
−δ2
[λ
min
V
k
]
−2+δ
ρ
2+δ
ii for any Borel set A, |G
k
A − Φ
d
A| ≤ C k
−δ2
[λ
min
V
k
]
−2+δ
ρ
2+δ
+ 2 sup
y ∈R
d
Φ
d
∂ A
η
− y where Φ
d
is the standard d dimensional normal distribution function, C is the class of all Borel mea-
surable convex subsets of R
d
, ρ
2+δ
= k
−1
P
k j=
1
E kX
j
k
2+δ
and η = Cρ
2+δ
n
−δ2
.
4.1 Proof of Theorem 1
The proof for circulant matrix mainly depends on Lemma 7 which helps to use normal approximation and, Lemma 8 given below which allows us to approximate the eigenvalues by appropriate partial
sums of independent random variables. The latter follows easily from Fan and Yao [2003] Theorem 2.14ii, page 63. We have provided a proof in Appendix for completeness. For k = 1, 2,
· · · , n, define
ξ
2k −1
= 1
p n
n −1
X
t=
ε
t
cosω
k
t , ξ
2k
= 1
p n
n −1
X
t=
ε
t
sinω
k
t .
Lemma 8. Suppose Assumption B holds and
{ε
t
} are i.i.d random variables with mean 0, variance 1. For k =
1, 2, · · · , n, write
λ
n ,k
= 1
p n
n −1
X
l=
x
l
e
i ω
k
l
= ψe
i ω
k
[ξ
2k −1
+ iξ
2k
] + Y
n
ω
k
, then we have
max
≤kn
E |Y
n
ω
k
| → 0 as n → ∞. 2476
Proof of Theorem 1: We first assume LebC
= 0. Note that we may ignore the eigenvalue λ
n
and also λ
n 2
whenever n is even since they contribute atmost 2n to the ESD F
n
x, y. So for x, y ∈ R, E
[F
n
x, y] s n
−1 n
−1
X
k= 1,k
6=n2
P b
k
≤ x, c
k
≤ y. Define for k = 1, 2,
· · · , n, η
k
= ξ
2k −1
, ξ
2k ′
, Y
1n
ω
k
= R[Y
n
ω
k
], Y
2n
ω
k
= I [Y
n
ω
k
], where Y
n
ω
k
are same as defined in Lemma 8. Then b
k
, c
k ′
= Bω
k
η
k
+ Y
1n
ω
k
, Y
2n
ω
k ′
. Now in view of Lemma 6 and Lemma 8, to show E[F
n
x, y] → F
C
x, y it is sufficient to show that 1
n
n −1
X
k= 1,k
6=n2
P Bω
k
η
k
≤ x, y
′
→ F
C
x, y. To show this, define for 1
≤ k ≤ n − 1, except for k = n2 and 0 ≤ l ≤ n − 1, X
l ,k
= p
2ε
l
cosω
k
l ,
p 2ε
l
sinω
k
l
′
. Note that
E X
l ,k
= 0 4.2
n
−1 n
−1
X
l=
C ovX
l ,k
= I 4.3
sup
n
sup
1 ≤k≤n
[n
−1 n
−1
X
l=
E k X
l k
k
2+δ
] ≤ C ∞. 4.4
For k 6= n2
B ω
k
η
k
≤ x, y
′
= B
ω
k
n
−12 n
−1
X
l=
X
l ,k
≤ p
2x, p
2 y
′
. Since
{r, s : Bω
k
r, s
′
≤ p
2x, p
2 y
′
} is a convex set in R
2
and {X
l ,k
, l = 0, 1, . . . n − 1}
satisfies 4.2–4.4, we can apply Part i of Lemma 7 for k 6= n2 to get
P Bω
k
n
−12 n
−1
X
l=
X
l ,k
≤ p
2x, p
2 y
′
− P Bω
k
N
1
, N
2 ′
≤ p
2x, p
2 y
′
≤ C n
−δ2
[n
−1 n
−1
X
l=
E kX
l k
k
2+δ
] ≤ C n
−δ2
→ 0, as n → ∞. Therefore
lim
n →∞
1 n
n −1
X
k= 1,k
6=n2
P B
ω
k
η
k
≤ x, y
′
= lim
n →∞
1 n
n −1
X
k= 1,k
6=n2
H
C
2πk n
, x, y = Z
1
H
C
2πs, x, yds.
2477
Hence E
[F
n
x, y] ∼ n
−1 n
−1
X
k= 1,k
6=n2
P b
k
≤ x, c
k
≤ y → Z
1
H
C
2πs, x, yds. 4.5
Now, to show V [F
n
x, y] → 0, it is enough to show that 1
n
2 n
X
k 6=k
′
;k,k
′
=1
C ovJ
k
, J
k
′
= 1
n
2 n
X
k 6=k
′
;k,k
′
=1
E J
k
, J
k
′
− EJ
k
EJ
k
′
→ 0. 4.6
where for 1 ≤ k ≤ n, J
k
is the indicator that {b
k
≤ x, c
k
≤ y}. Now as n → ∞, 1
n
2 n
X
k 6=k
′
;k,k
′
=1
E J
k
EJ
k
′
= 1
n
n
X
k= 1
E J
k 2
− 1
n
2 n
X
k= 1
E J
k 2
→ Z
1
H
C
2πs, x, yds
2
. So to show 4.6, it is enough to show as n
→ ∞, 1
n
2 n
X
k 6=k
′
;k,k
′
=1
E J
k
, J
k
′
→ Z
1
H
C
2πs, x, yds
2
. Along the lines of the proof used to show 4.5 one may now extend the vectors of two coordinates
defined above to ones with four coordinates and proceed exactly as above to verify this. We omit the routine details. This completes the proof for the case LebC
= 0. When LebC
6= 0, we have to show 4.1 only on D
c 1
of Lemma 1. All the above steps in the proof will go through for all x, y in D
c 1
. Hence if LebC 6= 0, we have our required LSD. This completes
the proof of Theorem 1.
4.2 Proof of Theorem 2
