The proof of the above lemma is omitted.
Theorem 3. Suppose Assumptions B and C hold. Then the ESD of RC
n
converges in L
2
to F
R
given in 3.7–3.8.
Remark 3. If
{x
i
} are i.i.d, with finite 2 + δ moment, then f ω = 12π for all ω ∈ [0, 2π] and the LSD F
R
· agrees with 1.2 given earlier.
3.4 k
-Circulant matrix
As mentioned before, it appears difficult to prove general results for all possible pairs k, n. We investigate two subclasses of the k-circulant.
3.4.1 n = k
g
+ 1 for some fixed g ≥ 2
For any d ≥ 1, let
G
d
x = P
d
Y
i= 1
E
i
≤ x ,
where {E
i
} are i.i.d. E x p1. Note that G
d
is continuous. For any integer d
≥ 1, define H
d
ω
1
, . . . , ω
d
, x on [0, 2π]
d
× R
≥0
as
H
d
ω
1
, . . . , ω
d
, x =
G
d x
2d
2π
d
Q
d i=
1
f ω
i
if Q
d i=
1
f ω
i
6= 0 1
if Q
d i=
1
f ω
i
= 0.
Lemma 4. i For fixed x, H
d
ω
1
, . . . , ω
d
, x is bounded continuous on [0, 2π]
d
. ii F
d
defined below is a valid continuous distribution function. F
d
x = Z
1
· · · Z
1
H
d
2πt
1
, . . . , 2πt
d
, x
d
Y
i= 1
d t
i
for x ≥ 0.
3.9 The proof of lemma is omitted.
Theorem 4. Suppose Assumptions B and C hold. Suppose n = k
g
+ 1 for some fixed g ≥ 2. Then as n
→ ∞, F
n
−12
A
k ,n
converges in L
2
to the LSD U
1
Q
g i=
1
E
i 12g
where {E
i
} are i.i.d. with distribution function F
g
given in 3.9 and U
1
is uniformly distributed over the 2gth roots of unity, independent
of the {E
i
}.
Remark 4. If
{x
i
} are i.i.d, then f ω = 12π for all ω ∈ [0, 2π] and the LSD is U
1
Q
g i=
1
E
i 12g
where {E
i
} are i.i.d. E x p1, U
1
is as in Theorem 4 and independent of {E
i
}. This limit agrees with Theorem
3 of Bose, Mitra and Sen [2008].
Remark 5. Using the expression 2.8 for the characteristic polynomial, it is then not difficult to man- ufacture
{k = kn} such that the LSD of n
−12
A
k ,n
has some positive mass at the origin. For example,
2472
suppose the sequences k and n satisfy k
g
= −1 + sn where g ≥ 1 is fixed and s = on
13
. Fix primes p
1
, p
2
, . . . , p
t
and positive integers β
1
, β
2
, . . . , β
t
. Define e
n = p
β
1
1
p
β
2
2
. . . p
β
t
t
n .
Suppose k = p
1
p
2
. . . p
t
m → ∞. Then the ESD of en
−12
A
k ,
e n
converges weakly in probability to the LSD which has
1 −
Π
t s=
1
p
β
s
s −1
mass at zero, and rest of the probability mass is distributed as U
1
Q
g i=
1
E
i 12g
where U
1
and {E
i
} are as in Theorem 4.
3.4.2 n = k
g
− 1 for some g ≥ 2
For z
i
, w
i
∈ R, i = 1, 2, .., g, and with {N
i
} i.i.d. N0, 1, define H
g
ω
i
, z
i
, w
i
, i = 1, . . . , g = P Bω
1
, ω
2
, .., ω
g
N
1
, ..., N
2g ′
≤ z
i
, w
i
, i = 1, 2, .., g
′
.
Lemma 5. i H
g
is a bounded continuous in ω
1
, . . . , ω
g
for fixed {z
i
, w
i
, i = 1, . . . , g }.
ii F
g
defined below is a proper distribution function. F
g
z
i
, w
i
, i = 1, . . . , g = Z
1
· · · Z
1
H
g
2πt
i
, z
i
, w
i
, i = 1, . . . , g Y
d t
i
. 3.10
iii If LebC
= 0 then F
g
is continuous everywhere and may be expressed as F
g
z
i
, w
i
, i = 1, .., g =
Z · · ·
Z I
{t≤z
k
,w
k
,k=1,..,g }
h Z
1
· · · Z
1
I
{ Q
f 2πu
i
6=0}
2π
g
Q
g i=
1
[π f 2πu
i
]
g
Y
i= 1
e
−
1 2
t2 2i
−1 +t2
2i π f 2πui
Y du
i
i
dt .
where t = t
1
, t
2
, . . . , t
2g −1
, t
2g
and dt =
Q d t
i
. Further F
g
is multivariate with independent com- ponents if and only if f is constant almost everywhere Lebesgue.
iv If LebC
6= 0 then F
g
is discontinuous only on D
g
= {z
i
, w
i
, i = 1, . . . , g : Q
g i=
1
z
i
w
i
= 0}. The proof of lemma is omitted.
Theorem 5. Suppose Assumptions B and C hold. Suppose n = k
g
− 1 for some g ≥ 2. Then as n
→ ∞, F
n
−12
A
k ,n
converges in L
2
to the LSD Q
g i=
1
G
i 1g
where RG
i
, I G
i
; i = 1, 2, . . . g has the distribution
F
g
given in 3.10.
Remark 6. If {x
i
} are i.i.d, with finite 2 + δ moment, then f ω ≡ 12π and the LSD simpli- fies to U
2
Q
g i=
1
E
i 12g
where {E
i
} are i.i.d. E x p1, U
2
is uniformly distributed over the unit circle independent of
{E
i
}. This agrees with Theorem 4 of Bose, Mitra and Sen [2008].
3.5 Simulations
