Remarks on cutoff phenomena for random walks on Hamming Schemes

0.21 MB·

by  Katsuhiko Kikuchi

Remarks on cutoff phenomena for random walks on
Hamming Schemes
6
1 Katsuhiko Kikuchi
0
2

b
e
F

9
Abstract

]
R The sequence of the simple random walks on Hamming schemes H(n,q) ∞
{ }n=1
P hasa cutoff phenomenonforeach integer q greater than or equalto3. Inthispaper,
h. for the sequence of simple random walks on Hamming schemes {H(n,q)}∞n=1 with
t q 3, we give a simple majorant and a sharp minorant function for total variance
a
≥
m distances between transition distributions and stationary distributions.
[

2
1 Introduction.
v
8
4
For many random walks on finite graphs, the transition distributions converge to
5
3 the distributions of the equilibrium. Moreover, if we have useful majorant and minorant
0 functions for the distance of them, we find the critical behavior of transition distributions,
.
1 for example, the rapidity of decrease of the distances in the small range near the suitable
0
time. Such the phenomenon and the time are called the cutoff phenomenon and the
6
1 time to stationarity, respectively. In this paper, we give majorant and minorant functions
:
v for total variance distances between transition distributions and the distributions of the
i
X equilibrium for the simple random walks on Hamming schemes H(n,q) ∞ with q 3.
{ }n=1 ≥
r Cutoff phenomenon is defined as follows. For a finite set X, we denote by M(X) the
a
vector space of all complex-valued measures on X. Take two measures µ,ν M(X) on
∈
X and define the total varialce distance µ ν by
TV
k − k
µ ν = max µ(S) ν(S) ; S X .
TV
k − k {| − | ⊂ }
Let (X,E ) be a simple connected finite unordered graph without loops. For x,x′ X,
X
∈
we say that x is adjacent to x′ if the (unordered) pair x,x′ belongs to E and write
X
{ }
x x′. The transition probability p( , ) is a function on X X such that (i) p(x,x′) 0,
∼ · · × ≥
(ii) p(x,x′) > 0 if and only if x x′, and (iii) p(x,x′) = 1 for any x X. For a
∼ ∈
x′∈X
X
nonnegative integer, we define transition probability p(k)( , ) after k-steps recursively, by
· ·
p(0)(x,x′) = δx,x′, p(k)(x,x′) = p(k−1)(x,y)p(y,x′), k 1,
≥
y∈X
X
2010 Mathematics Subject Classification. Primary 05C81; Secondary 60C05, 05E18.
1
where δx,x′ is the Kronecker delta. Fix an element x(0) X of X and put ν∗k( ) =
∈ ·
p(k)(x(0), ). Then, we see that ν∗k is a probability measure on X. We say that the
·
transition probability p( , ) is ergodic if there exists an integer k such that p(k)(x,x′) > 0
0
· ·
for any elements x,x′ X and k k . A probability measure π on X is stationary
0
∈ ≥
if π(x′)p(x′,x) = π(x) for any x X. Let (X ,E ) be a sequence of simple
∈ { n Xn }
x′∈X
X
connected finite unordered graphs without loops, p ( , ) the transition probabilities
n
{ · · }
(0)
on X , x the fixted points and π the stationary probabilities. For sequences
n n n
{ } { }
b
n
a , b of positive real numbers with lim = 0, the sequence of the Markov chains
n n
{ } { } n→∞ an
(X ,E ) has an (a ,b )-cutoff if there exist functions f : [0,+ ) R with
{ n Xn } n n ± ∞ −→
lim f (c) = 0, lim f (c) = 1, and for each c > 0, we have
+ −
c→+∞ c→+∞
limsup ν∗⌈an+cbn⌉ π f (c), (1.1)
k n − nkTV ≤ +
n→∞
liminf ν∗⌊an−cbn⌋ π f (c), (1.2)
n→∞ k n − nkTV ≥ −
where α , α denote the least integer greater than or equal to α, the greatest integer
⌈ ⌉ ⌊ ⌋
less than or equal to α, respectively (see [D1], [D2], [DS]). f and f are called the upper
+ −
bound and the lower bound, and we often take as a monotone decreasing, a monotone
increasing function, respectively. We remark that the existence of f does not imply the
+
ergodicity of each random walk on the graph (X ,E ) (see [H]). If we take a majorant
n Xn
function h for ν∗⌊an+cbn⌋ π with n n for some positive integer n , we have the
+ n n TV 0 0
k − k ≥
ergodicity of the random walk on (X ,E ) for any n with n n .
n Xn ≥ 0
Let n be a positive integer and q an integer with q 2. we denote by [q] =
0
≥
0,1,...,q 1 and [n] = 1,2,...,n . As a graph, the Hamming scheme H(n,q) =
{ − } { }
(X ,E ) is a finite graph with the vertex set X = [q]n and the edge set
n Xn n 0
E = x,x′ X ; ♯ j [n]; x = x′ = 1 ,
Xn {{ } ⊂ n { ∈ j 6 j} }
where x = (x ,...,x ), x′ = (x′,...,x′ ) X , and ♯S is the cardinal number for a finite
1 n 1 n ∈ n
set S. The transition probability p ( , ) for (X ,E ) is defined by
n · · n Xn
1
, x x′,
p (x,x′) = n(q 1) ∼
n  −
0, otherwise.

Put x(0) = (0,...,0) H(n,q) and νn( ) = pn(x(0), ). Then, νn is a probability measure
∈ · ·
on X . The graph H(n,q) is ergodic if and only if q 3. Hora gives in [H] the limit func-
n
e∓c ≥
tion f (c) = Erf 2 of ν∗(an±cbn) π , where Erf : R R is the error func-
± n n TV
2√2 k − k −→
(cid:18) (cid:19)
2 x n(q 1) n(q 1)
tiondefinedbyErf(x) = e−t2dt,and(a ,b ) = − logn(q 1), − .
n n
√π 2q − 2q
Z0 (cid:18) (cid:19)
Candidates of the majorant functions for ν∗k π are given by Diaconis and Hanlon
k n − nkTV
in [DH], and by Diaconis and Ram in [DR]. In those papers, the simple random walk
on a Hamming scheme H(n,q) are regarded as a special case of Metropolis chains on a
hypercube H(n,2). Mizukawa gives in [M1] a majorant function for total variance dis-
n(q 1)
tance with the different time to stationarity − logqn for the case q 3, and in
2q ≥
2
[M2] a majorant and a minorant function for the sequences of random walks on Hamming
schemes with staying.
For each integer q with q 5, we give a simple majorant function of ν∗k π .
≥ k n − nkTV
n(q 1)
Theorem 1.1 Assume that q 5. Let k = − (logn(q 1)+c) be an integer with
≥ 2q −
c > 0. Then, we have
1
ν∗k π 2 (ee−c 1). (1.3)
k n − nkTV ≤ 4 −
The key of the proof of this theorem is that e−x 1 x for any real number x
≥ | − |
5
with x . We cannot adapt the proof for the case q = 3,4. The obstruction is that
≤ 4
4
1 x e−x for a real number x with x . So we replace the majorant function.
| − | ≥ ≥ 3
n
Theorem 1.2 (1) Assume that q = 3 and n 3. For any integer k = (log2n+c)
≥ 3
with c > 0, we have
5
ν∗k π 2 (ee−c 1). (1.4)
k n − nkTV ≤ 2 −
3n
(2) Suppose that q = 4 and n 2. For each integer k = (log3n+c) with c > 0, one
≥ 8
has
9
ν∗k π 2 (ee−c 1). (1.5)
k n − nkTV ≤ 4 −
While, we give minorant function for ν∗k π as follows.
k n − nkTV
Theorem 1.3 Fix a positive real number c > 0. For any positive real number b >
0
0, there exists a positive integer n such that c logn (q 1), and for any integer
0 0 0
≤ −
n(q 1)
k = − (logn(q 1) c) with 0 c c , we have
0
2q − − ≤ ≤
ν∗k π 1 (4q +b)e−c. (1.6)
k n − nkTV ≥ −
The above theorem says that we can take a function 1 4qe−c as a lower function of
−
ν∗k π .
k n − nkTV
2 Hamming schemes.
In this section, we give notations of Hamming schemes, refering to [BI], [CST], [D1].
For a positive integer m, we denote by [m] = 1,2,...,m , [m] = 0,1,...,m 1
0
{ } { − }
and by ♯S the cardinal number for a finite set S.
Let n be a positive integer and q an integer with q 2. Put
≥
H(n,q) = [q]n = x = (x ,...,x ); x [q] (1 j n) . (2.1)
0 { 1 n j ∈ 0 ≤ ≤ }
Take x = (x ,...,x ), y = (y ,...,y ) H(n,q) and define d(x,y) by
1 n 1 n
∈
d(x,y) = ♯ j [n]; x = y . (2.2)
j j
{ ∈ 6 }
3
For x,x′ H(n,q), we say that x′ is adjacent to x if d(x,x′) = 1 and write x x′. We fix
∈ ∼
an integer q with q 2 and denote by X = H(n,q) for simplicity. We call an unordered
n
≥
pair x,x′ X such that x x′ the edge of X . Put
n n
{ } ⊂ ∼
E = x,x′ X ; x x′ . (2.3)
Xn {{ } ∈ n ∼ }
Then thepair(X ,E )is asimple undirected finitegraphwithout loops. Wecall H(n,q)
n Xn
the Hamming scheme. X and E are called the vertex set and the edge set of H(n,q),
n Xn
respectively. We see that the distance d( , ) on X coincides that derived from E .
· · n Xn
We denote by S the symmetric group on [m] or [m] for a positive integer m. Let
m 0
G = S S = Sn ⋊S denote the wreath product of S by S with the product
n q ≀ n q n q n
(τ ,...,τ ;σ)(τ′,...,τ′;σ′) = (τ τ′ ,...,τ τ′ ;σσ′), (2.4)
1 n 1 n 1 σ−1(1) n σ−1(n)
where τ ,τ′ S (1 j n), σ,σ′ S , and we regard S and S as symmetric groups
j j ∈ q ≤ ≤ ∈ n q n
acting on [q] and on [n], respectively. G acts on X by
0 n n
g x = (τ1(xσ−1(1)),...,τn(xσ−1(n))), (2.5)
·
where g = (τ ,...,τ ;σ) G and x = (x ,...,x ) X . The action of G on X is
1 n n 1 n n n n
∈ ∈
transitive. Put x(0) = (0,...,0) X . Then the stabilizer H of G at x(0) is given by
n n n
∈
H = S S = (τ ,...,τ ;σ) G ; τ S for all j [n] , (2.6)
n q−1 n 1 n n j q−1
≀ { ∈ ∈ ∈ }
where we regard S as a symmetric group acting on [q 1]. For each integer j
q−1
− ∈
0,1,...,n we put
{ }
j
x(j) = (1,...,1,0,...,0) X . (2.7)
n
∈
Then, X = G /H and we have thezH}|-or{bit decomposion
n n n n
n
X = H x(j). (2.8)
n n
·
j=0
[
We see that x x′ implies that g x g x′ for any x,x′ X and g G . Hence,
n n
∼ · ∼ · ∈ ∈
for any j 0,1,...,n , we have that x H x(j) if and only if d(x(0),x) = j. For
n
∈ { } ∈ ·
j 0,1,...,n , put
∈ { }
j
g(j) = ((0,1),...,(0,1),1 ,...,1 ;1 ), (2.9)
Sq Sq Sn
z }| {
where (0,1) S is the transposition of 0 and 1, and 1 S , 1 S are the
∈ q Sq ∈ q Sn ∈ n
identity permutations. We see that g(j) x(0) = x(j). Hence, we have the decomposition
n ·
G = H g(j)H of G into H -double cosets. Let L1(G ) denote the algebra of all
n n n n n n
j=0
[
functions on G with the convolution
n
f f (g) = f (g(g′)−1)f (g′) = f (g′)f ((g′)−1g), (2.10)
1 2 1 2 1 2
∗
gX′∈Gn gX′∈Gn
4
where f ,f L1(G ) and g G . We see that (G ,H ) is a Gelfand pair, that is, the
1 2 n n n n
∈ ∈
subalgebra L1(H G /H ) L1(G ) of all H -biinvariant functions on G is a commu-
n n n n n n
\ ⊂
tative algebra since (g(j))−1 = g(j) for any j 0,1,...,n (see [CST], Example 4.3.2).
∈ { }
We denote by L(X ) the Hilbert space of all functions on X with the inner product
n n
f ,f = f (x)f (x), (2.11)
h 1 2iL(Xn) 1 2
xX∈Xn
1
where f ,f L(X ), and write f = f,f 2 for f L(X). G acts on L(X )
1 2 ∈ n k kL(Xn) h iL(Xn) ∈ n n
by
(g f)(x) = f(g−1 x), (2.12)
· ·
where g G , f L(X ) and x X . It is easy to show that the action is unitary.
n n n
∈ ∈ ∈
Let W be a G -module. We denote by W the subspace of all H -invariant elements
n Hn n
in W, that is,
W = w W ; h w = w for all h H . (2.13)
Hn { ∈ · ∈ n}
The condition that (G ,H ) is a Gelfand pair indicates the properties of irreducible
n n
components appearing in L(X ).
n
Lemma 2.1 Let L(X ) = V be an irreducible decomposition of L(X ).
n λ n
λ∈Λ
M
(1) L(Xn) is multiplicity-free, that is, Vλ is not equivalent to Vλ′ if λ = λ′.
6
(2) For any λ Λ, we have dim(V ) = 1.
∈ λ Hn
Proof. See [CST], Theorem 4.4.2, 4.6.2 for example.
Each irreducible component V is called the spherical representation for (G ,H ).
λ n n
We construct irreducible components in L(X ). Take an integer a [q] and define a
n 0
∈
function χ : [q] C by
a 0
−→
χ (x) = ζax,
a q
2πi
where x [q] and ζ = exp C is a primitive q-th root of 1 in C. For x [q] , we
0 q 0
∈ q ∈ ∈
have
q−1
q, x = 0,
χ (x) =
a
(0, x = 0.
a=0 6
X
For a = (a ,...,a ) X , we define a function χ : X C by
1 n n a n
∈ −→
n
χ (x) = χ (x ) = ζa1x1+···+anxn, (2.14)
a aj j q
j=1
Y
where x = (x ,...,x ) X . Then, we see that χ L(X ); a X is an orthogonal
1 n n a n n
∈ n { ∈ ∈ }
basis for L(X ) and χ = q for a X .
n k akL(Xn) 2 ∈ n
For j 0,1,...,n , we put
∈ { }
V = Cχ . (2.15)
j a
♯{l∈[nM];al6=0}=j
5
Then, V is G -invariant for each j with 0 j n and we have the orthogonal desompo-
j n
≤ ≤
sition
n
L(X ) = V . (2.16)
n j
j=0
M
For j 0,1,...,n , we put
∈ { }
ω = χ . (2.17)
j a
♯{l∈[n]X;al6=0}=j
Then, ω is nonzero H -invariant element in V , and any H -invariant element in V is the
j n j n j
scalar multiple of ω . Hence, all V ’s are irreducible and L(X ) is multiplicity-free.
j j n
For j 0,1,...,n , we define a function φ : G C by
j n
∈ { } −→
ω ω 1
j j
φ (g) = , g = ω ,g ω . (2.18)
j ω · ω ω 2 h j · jiL(Xn)
(cid:28)k jkL(Xn) k jkL(Xn)(cid:29)L(Xn) k jkL(Xn)
Then, φ is H -biinvariant and φ (1 ) = 1, where 1 G is the unit element. More-
j n j Gn Gn ∈ n
over, φ is real-valued since (g(j))−1 H g(j)H for any j 0,1,...,n (see [CST], The-
j n n
∈ ∈ { }
orem 4.8.2). φ is called the spherical function on G . We regard φ as an H -invariant
j n j n
function on X . φ is calculated as
n j
j r
1 l n l 1
φ (g(l)) = − , (2.19)
j
n r j r −q 1
r=0 (cid:18) (cid:19)(cid:18) − (cid:19)(cid:18) − (cid:19)
X
j
(cid:18) (cid:19)
wherel 0,1,...,n (see[CST], Theorem5.3.2). Wegiveanotherrealizationofφ (g(l)).
j
∈ { }
For a complex number α and a nonnegative integer m, put
α(α+1) (α+m 1), m 1,
(α) = ··· − ≥
m
(1, m = 0.
(α) is called the Pochhammer symbol. Take complex numbers α,β,γ C, a variable x,
m
∈
and define the Gauss hypergeometric series
∞
α,β (α) (β)
F ;x = m mxm.
γ (γ) m!
(cid:18) (cid:19) m=0 m
X
We write φ (l) = φ (g(l)) for simplicity. The polynomial φ is called the Krawtchouk
j j j
polynomial. Using a Gauss hypergeometric series, φ (l) is realized as
j
j r
j, l q ( j) ( l) q
r r
φ (l) = F − − ; = − − , (2.20)
j n q 1 ( n) r! q 1
(cid:18) − − (cid:19) r=0 − r (cid:18) − (cid:19)
X
where j,l 0,1,...,n .
∈ { }
For f L1(H G /H ), we define
n n n
∈ \
f(φ ) = f(g)φ (g) = f(g)φ (g), (2.21)
j j j
gX∈Gn gX∈Gn
b
6
where j 0,1,...,n . f is called the spherical transform of f L1(H G /H ). For
n n n
∈ { } ∈ \
f ,f L1(H G /H ), We have
1 2 n n n
∈ \
b
(f f ) = f f . (2.22)
1 2 1 2
∗
Take f ∈ L(Xn)Hn, j ∈ {0,1,...,n} andbdefibneb
F(f)(φ ) = f(x)φ (x) = f(x)φ (x). (2.23)
j j j
xX∈Xn xX∈Xn
F(f) is called the spherical transform of f L(X ) .
∈ n Hn
For an H -invariant function f L(X ) on X , we denote by f the H -biinvariant
n ∈ n Hn n n
function on G corresponding to f. We see that
n
e
1
F(f)(φ ) = f(x)φ (x) = f(g x(0))φ (g)
j j j
♯H ·
n
xX∈Xn xX∈Xn g·xX(0)=x
1 1
= f(g)φ (g) = (f) (φ ).
j j
♯H ♯H
n n
gX∈Gn
e e
b
Take f ,f L(X ) and define f f L(X ) such that
1 2 ∈ n Hn 1 ∗ 2 ∈ n Hn
1
(f f ) = (f f ). (2.24)
1 2 1 2
∗ ♯H ∗
n
e e
Lemma 2.2 Let f ,f L(X ) be tewo elements in L(X ) . For j 0,1,...,n ,
1 2 ∈ n Hn n Hn ∈ { }
we have
F(f f ) = F(f )F(f ). (2.25)
1 2 1 2
∗
Proof. By (2.22) and (2.23), we have
1 1
F(f f )(φ ) = ((f f ) ) (φ ) = ((f ) (f )) (φ )
1 ∗ 2 j ♯H 1 ∗ 2 j (♯H )2 1 ∗ 2 j
n n
1
= ((f ) (efb) )(φ ) = F(f )(φe)F(fe)(φb).
(♯H )2 1 2 j 1 j 2 j
n
e e
b b
For f L(X ) and a nonnegative integer k, we define f∗k recursively by
∈ n Hn
f∗0 = δ , f∗k = f∗(k−1) f, k 1. (2.26)
x(0)
∗ ≥
Lemma 2.3 For j 0,1,...,n and a nonnegative integer k, we have
∈ { }
F(f∗k)(φ ) = F(f)(φ )k. (2.27)
j j
Proof. We prove it by induction in k. We see that
F(f∗0)(φ ) = δ (x)φ (x) = φ (x(0)) = 1.
j x(0) j j
xX∈Xn
7
We assume k 1 and the claim satisfies for any integer less than k. Then we have
≥
F(f∗k)(φ ) = F(f∗(k−1) f) = F(f∗(k−1))F(f) = F(f)k−1F(f) = F(f)k.
j
∗
Take x,x′ X and define the transition probability p (x,x′) by
n n
∈
1
, x x′,
p (x,x′) = n(q 1) ∼ (2.28)
n  −
0, x x′.

6∼
Since G preserves adjacency on X , forx,x′ X and g G , we see that
n n n n
∈ ∈
p (g x,g x′) = p (x,x′). (2.29)
n n
· ·
Put
ν (x) = p (x(0),x). (2.30)
n n
Then, ν is an H -invariant probability measure on X .
n n n
Lemma 2.4 For j 0,1,...,n , we have
∈ { }
jq
F(ν )(φ ) = 1 . (2.31)
n j
− n(q 1)
−
Proof. Since x(0) x if and only if x H x(1), we see that
n
∼ ∈ ·
j, 1 q ( j) ( 1) q jq
φ (x(1)) = F − − ; = 1+ − · − = 1 .
j n q 1 ( n) 1 · q 1 − n(q 1)
(cid:18) − − (cid:19) − · − −
Hence, we have
F(ν )(φ ) = ν (x)φ (x) = ν (x)φ (x) = ν (x)φ (x(1))
n j n j n j n j
xX∈Xn xX(0)∼x x∈HXn·x(1)
1 jq jq
= n(q 1) 1 = 1 .
− · n(q 1) − n(q 1) − n(q 1)
− (cid:18) − (cid:19) −
3 Upper bounds.
In this section, we give a majorant function for the distances between the k-step
transitions distributions and the distributions of equilibrium for the simple random walks
on the Hamming schemes H(n,q) with q 3.
{ } ≥
We denote by M(X ) the vector space of all complex-valued measure on X . For a
n n
measure µ M(X ) on X , we put
n n
∈
µ = max µ(S) ; S X . (3.1)
TV n
k k {| | ⊂ }
8
For two measure µ,ν M(X ), we define the total variance distance by µ ν . We
n TV
∈ k − k
regarda measure µ M(X ) onX asa functionµ : X C defined by µ(x) = µ( x )
n n n
∈ −→ { }
for x X . Take two probability measures µ,ν M(X ) on X . Then we have an
n n n
∈ ∈
equality
2
1 ♯X
µ ν 2 = µ(x) ν(x) n µ ν 2 .
k − kTV 2 | − | ≤ 4 k − kL(Xn)
!
xX∈Xn
We denote by π the uniform probability measure on X , that is,
n n
♯S ♯S
π (S) = = , (3.2)
n ♯X qn
n
where S X is a subset of X . We find the upper bound of the total variance distance
n n
⊂
ν∗k µ with the Fourier transforms of spherical functions.
k n − nkTV
Lemma 3.1 (The upper bound lemma) For a nonnegative integer k, we have
n
1
ν∗k π 2 d F(ν )(φ ) 2. (3.3)
k n − nkTV ≤ 4 j| n j |
j=1
X
Proof. See [D1], Chapter 3B, Lemma 1 or [CST], Corollary 4.9.2.
Before estimating the total variance distance νk π , we give two inequalities.
k n − nkTV
5
Lemma 3.2 (1) For a real number x such that x , we have e−x 1 x .
≤ 4 ≥ | − |
4
(2) If the real number x satisfies the condition x , one has e−x 1 x .
≥ 3 ≤ | − |
Proof. First, we see that
3 ∞ ∞
8 1 1 5 1 5 1 11
= e = + = + = .
3 j! ≤ j! ≤ 2 2 3j−2 2 4 4
j=0 j=0 j=3 ·
X X X
(1) By Taylor’s theorem, for x R, there exists a real number θ R with 0 < θ < 1
∈ ∈
such that
x2
e−x = 1 x+ e−θx.
− 2
5
If x 1, we have e−x 1 x = 1 x . Hence, we only consider the case 1 x . We
≤ ≥ − | − | ≤ ≤ 4
see that
5
11 161051
e5 = 256 = 44.
≤ 4 1024 ≤
(cid:18) (cid:19)
1 5 5
So e−54 = 1. Therefore, for any real number x such that 1 x , we have
≥ 4 4 − ≤ ≤ 4
5
e−x e−54 1 x 1 = 1 x .
≥ ≥ 4 − ≥ − | − |
9
(2) We see that
4
8 4096
e4 = 27 = 33.
≥ 3 81 ≥
(cid:18) (cid:19)
1 4 4
Hence, e−43 = 1. This implies that for any real number x with x ,
≤ 3 3 − ≥ 3
4
e−x e−43 1 x 1 = 1 x .
≤ ≤ 3 − ≤ − | − |
Here, we give a majorant function for the total variance distance νk π with a
k n − nkTV
large integer q.
Theorem 3.3 Let q be an integer with q 5. Take a positive integer k such that
≥
n(q 1)
k = − (logn(q 1)+c) with c > 0. Then, we have
2q −
1
ν∗k π 2 (ee−c 1).
k n − nkTV ≤ 4 −
jq q 5
Proof. The condition q 5 implies that for any integer j
≥ n(q 1) ≤ q 1 ≤ 4 ∈
− −
0,1,...,n . Hence, by Lemma 3.1, we have
{ }
n n 2k
1 1 n jq
ν∗k π 2 d F(ν )(φ ) 2k = (q 1)j 1
k n − nkTV ≤ 4 j| n j | 4 j − − n(q 1)
j=1 j=1 (cid:18) (cid:19) (cid:12) − (cid:12)
X X (cid:12) (cid:12)
1 n nj(q −1)je−n2(qj−kq1) = 1 n 1ej(logn(q−(cid:12)(cid:12)1)−n(2qk−q1)). (cid:12)(cid:12)
≤ 4 j! 4 j!
j=1 j=1
X X
n(q 1)
Since k = − (logn(q 1)+c), we have
2q −
1 n e−cj 1 ∞ e−cj 1
ν∗k π 2 = (ee−c 1).
k n − nkTV ≤ 4 j! ≤ 4 j! 4 −
j=1 j=1
X X
It remains to show the case q = 3,4. For any real number α R, we denote by
∈
α = max m Z; m α ,
⌊ ⌋ { ∈ ≤ }
α = min m Z; m α .
⌈ ⌉ { ∈ ≥ }
In order to estimate a total variance distances, where q = 3,4, we give a lemma.
Lemma 3.4 (1) Let m be an integer with m 2 and l an integer such that 0 l m.
≥ ≤ ≤
x m+2 2m 2
Put f (x) = − = 1 − . Then, we have
m
x+m − x+m
2m−l−1
3m l
f (p) 2log − log9. (3.4)
m
≤ l+m ≤
p=l
X
10