3.1 Circulant matrix
Define for x, y ∈ R
2
and ω ∈ [0, 2π],
H
C
ω, x, y = ¨
P B
ωN
1
, N
2 ′
≤ p
2x, y
′
if f
ω 6= 0, I
x ≥ 0, y ≥ 0 if
f ω = 0,
where N
1
and N
2
are i.i.d. standard normal distributions.
Lemma 1. i For fixed x, y, H
C
is a bounded continuous function in ω.
ii F
C
defined as follows is a proper distribution function. F
C
x, y = Z
1
H
C
2πs, x, yds. 3.1
iii If LebC
= 0 then F
C
is continuous everywhere and can be expressed as F
C
x, y = ZZ
I
{v
1
,v
2
≤x, y}
h Z
1
1 2π
2
f 2πs
e
−
v2 1
+v2 2
2π f 2πs
ds i
d v
1
d v
2
. 3.2
Further, F
C
is bivariate normal if and only if f is constant almost everywhere Lebesgue. iv If
LebC 6= 0 then F
C
is discontinuous only on D
1
= {x, y : x y = 0}. The proof of the Lemma is easy and we omit it. The normality claim in iii follows by applying
Cauchy Schwartz inequality to compare fourth moment and square of the variance and using the fact that for the normal distribution their ratio equals 3. We omit the details.
Theorem 1. Suppose Assumptions B and C hold. Then the ESD of C
n
converges in L
2
to F
C
· given in 3.1–3.2.
Remark 1. If {x
i
} are i.i.d with finite 2 + δ moment, then f ω ≡ 12π, and F
C
reduces to the bivariate normal distribution whose covariance matrix is diagonal with entries
12 each. This agrees with Theorem
15, page 57 of Sen [2006] who proved the result under Assumption A.
3.2 Symmetric Circulant Matrix
For x ∈ R and ω ∈ [0, π] define,
H
S
ω, x = ¨
P p
2π f ωN 0, 1 ≤ x if f ω 6= 0,
I x ≥ 0
if f
ω = 0. 3.3
As before, we now have the following Lemma. We omit the proof.
Lemma 2. i For fixed x, H
S
is a bounded continuous function in ω and H
S
ω, x + H
S
ω, −x = 1. ii F
S
defined below is a proper distribution function and F
S
x + F
S
−x = 1. F
S
x = 2 Z
12
H
S
2πs, xds. 3.4
2470
iii If LebC
′
= 0 then F
S
is continuous everywhere and may be expressed as F
S
x = Z
x −∞
h Z
12
1 π
p f
2πs e
−
t2 4π f 2πs
ds i
d t .
3.5 Further, F
S
is normal if and only if f is constant almost everywhere Lebesgue. iv If
LebC
′
6= 0 then F
S
is discontinuous only at x = 0.
Theorem 2. Suppose Assumptions B and C hold and
lim
n →∞
1 n
2 [np2]
X
k= 1
f 2πk
n
−32
−→ 0 for all 0 p 1. 3.6
Then the ESD of SC
n
converges in L
2
to F
S
given in 3.4–3.5. The same limit continues to hold for P T
n
.
Remark 2. i 3.6 is satisfied if inf
ω
f ω 0.
ii It is easy to check that the variance, µ
2
and the fourth moment µ
4
of F
S
equal R
12
4π f 2πsds and
R
12
24π
2
f
2
2πsds respectively. By Cauchy-Schwartz inequality it follows that
µ
4
µ
2 2
≥ 3 and equal to
3 iff f ≡
1 2π
. In the latter case, F
S
is standard normal. This agrees with Remark 2 of Bose and Mitra
[2002] under Assumption A.
3.3 Reverse circulant matrix
Define H
R
ω, x on [0, 2π] × R as H
R
ω, x = G
x
2
2π f ω
if f
ω 6= 0 1
if f
ω = 0, where Gx = 1
− e
−x
for x 0, is the standard exponential distribution function.
Lemma 3. i For fixed x, H
R
ω, x is bounded continuous on [0, 2π]. ii F
R
defined below is a valid symmetric distribution function. F
R
x =
1 2
+ R
12
H
R
2πt, xd t if x
1 2
− R
12
H
R
2πt, xd t if x
≤ 0. 3.7
iii If LebC
′
= 0 then F
R
is continuous everywhere and can be expressed as
F
R
x =
1 −
R
12
e
−
x2 2π f 2πt
d t if x
R
12
e
−
x2 2π f 2πt
d t if x
≤ 0. 3.8
Further, F
R
is the distribution of the symmetric version of the square root of chisquare variable with two degrees of freedom if and only if f is constant almost everywhere Lebesgue.
iv If LebC
′
6= 0 then F
R
is discontinuous only at x = 0.
2471
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
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