Exponent Function for Source Coding with Side Information at the Decoder at Rates below the Rate Distortion Function

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by  Yasutada Oohama

Exponent Function for Source Coding with Side
Information at the Decoder at Rates below the Rate
Distortion Function
Yasutada Oohama
University of Electro-Communications, Tokyo, Japan
Email: [email protected]
Abstract—We consider the rate distortion problem with side
informationatthedecoderposedandinvestigatedbyWynerand Xn ϕ(n)(Xn)
Ziv.Theratedistortionfunctionindicatingthetrade-offbetween X ✲ϕ(n) ✲ψ(n) ✲Zn
therateonthedatacompressionandthequalityofdataobtained rate R (cid:0)(cid:0)✒
6 (cid:0)
at thedecoderwasdeterminedbyWynerandZiv. Inthispaper,
1 we study the error probability of decoding at rates below the Yn (cid:0)
0 (cid:0)
ratedistortionfunction.Weevaluate theprobabilityof decoding Y
2
such that the estimation of source outputs by the decoder has a

r distortion not exceeding a prescribed distortion level. We prove
p
that when the rate of the data compression is below the rate
A
distortion function this probability goes to zero exponentially Fig.1. Wyner-Zivsourcecodingsystem.
andderiveanexplicitlowerboundofthisexponentfunction.On
6
theWyner-Zivsourcecodingproblemthestrongconversecoding
2
theorem has not been established yet. We prove thisas a simple
The excess-distortion probability of decoding is
corollary of our result.
]
T 1
I. SOURCE CODING WITH SIDE INFORMATION ATTHE P(n)(ϕ(n),ψ(n);∆)=Pr d(Xn,Zn) ∆ , (3)
.I DECODER e (cid:26)n ≥ (cid:27)
s
6 [c dpiasiLcrere(tXtXetm,aYnemtd)oYtraybkleeessfisnvsiaotleuursecesets.inFanodre{a(cXhtt,,Y=atn)1d},∞t=2h,a1·sb·e·a,atphsretoarbtaiaonbndilaoirmtyy ∆wh(ne)rebeZtwnee=n Xψn(na)(nϕd(nZ)n(Xisn)d,eYfinne)d. bTyhe average distortion
X × Y n
v distribution ∆(n) =△ 1E[d(Xn,Zn)]=△ 1 Ed(X ,Z ).
0 n n t t
5 pXY ={pXY(x,y)}(x,y)∈X×Y Xt=1
056 rWesepewctriivteelynasindependent copies of {Xt}∞t=1 and {Yt}∞t=1, Aofppaaiirr(sR{,(∆ϕ()ni)s,εψ-a(nc)h)i}enva≥b1lesufocrhptXhaYtiffotrhearneyeδxi>sta0seaqnudeanncye
n with n n =n (ε,δ),
1. Xn =X ,X , ,X and Yn =Y ,Y , ,Y . ≥ 0 0
1 2 n 1 2 n
0 ··· ··· 1
We consider a communication system depicted in Fig. 1. log ϕ(n) R+δ, P(n)(ϕ(n),ψ(n);∆) ε.
6 n k k≤ e ≤
1 Data sequences Xn is separately encoded to ϕ(n)(Xn) and
: is sent to the information processing center. At the center the The rate distortion region WZ(εpXY) is defined by
v R |
i decoder function ψ(n) observes ϕ(n)(Xn) and Yn to output
X theestimationZnofXn.Theencoderfunctionϕ(n) isdefined RWZ(ε|pXY)
r by = (R,∆):(R,∆) is ε-achievable for pXY .
a { }
ϕ(n) : n = 1,2, ,M , (1)
X →Mn { ··· n} Furthermore set
where ϕ(n) (= M ) stands for the range of cardinality of
n △
k k (p )= (εp ).
ϕ(n). Let be a reproductionalphabet.The decoderfunction RWZ XY RWZ | XY
ψ(n) is deZfined by ε\>0
Ontheotherhand,wecandefinearatedistortionregionbased
ψ(n) : n n. (2)
Mn×Y → Z ontheaveragedistortioncriterion,aformaldefinitionofwhich
Let d : [0, ) be an arbitrary distortion measure is the following.A pair (R,∆) is achievablefor p if there
X ×Z → ∞ XY
on . The distortion between xn n and zn n is exist a sequence of pairs (ϕ(n), ψ(n)) such that for any
X ×Z ∈X ∈Z { }n≥1
defined by δ >0 and any n with n n =n (δ),
0 0
n ≥
d(xn,zn)=△ d(x ,z ). 1
t t log ϕ(n) R+δ, ∆(n) ∆+δ.
t=1 n k k≤ ≤
X
The rate distortion region ˜ (p ) is defined by We can show that the above functions and sets satisfy the
WZ XY
R
following property.
˜ (p )
RWZ XY Property 2:
△
={(R,∆):(R,∆) is achievable for pXY }. a) The region R(pXY) is a closed convex set of R2+.
b) For any p , we have
We can show that the three rate distortion regions (ε XY
WZ
pXY), ε ∈ (0,1), RWZ(pXY), and R˜WZ(pXY) satRisfy the| R(pXY)=R∗(pXY).
following property.
Proof of Property 2 is given in Appendix C. In Property
Property 1:
2 part b), (p ) is regarded as another expression of
a) RT˜hWeZr(epgXioYn)s aRreWcZl(oεs|epdXcYo)n,vεex∈se(t0s,1o)f,RR2+W, wZ(hpeXreY), and rRes∗u(lptX. YT)h.eTrRahtiesrXeeYgxipornessioWnZ(ispXuYse)fuwlasfodretdeermrivininegdboyurWmynaeinr
R
R2 =△ (R,∆):R 0,∆ 0 . and Ziv [1]. Their result is the following.
+ { ≥ ≥ } Theorem 1 (Wyner and Ziv [1]):
b) (εp ) has another form using (n,ε)-rate distor-
RWZ | XY ˜ (p )= ∗(p )= (p ).
tion region, the definition of which is as follows. We set WZ XY XY XY
R R R
On (p ), Csisza´r and Ko¨rner [4] obtained the fol-
(n,εp ) WZ XY
WZ XY R
R | lowing result.
= (R,∆): There exists (ϕ(n),ψ(n)) such that
{ Theorem 2 (Csisza´r and Ko¨rner [4]):
1
nlog||ϕ(n)||≤R, P(en)(ϕ(n),ψ(n);∆)≤ε}, WZ(pXY)= ˜WZ(pXY)= ∗(pXY)= (pXY).
R R R R
which is called the (n,ε)-rate distortion region. Using We are interested in an asymptotic behavior of the error
WZ(n,εpXY), WZ(εpXY) can be expressed as probabilityofdecodingtotendtooneasn for(R,∆) /
R | R | →∞ ∈
(p ). To examine the rate of convergence, we define
WZ XY
R
the following quantity. Set
(εp )=cl (n,εp ) ,
WZ XY WZ XY
R |  R | 
m[≥1n\≥m P(n)(ϕ(n),ψ(n);∆)=△ 1 P(n)(ϕ(n),ψ(n);∆),
  c − e
where cl() stands for the closure operation. G(n)(R,∆p )
· | XY
[k1n]Po.wroTnooftdhoeafstctRrh˜iibWseZpt(rhopepXierYrtr)yeswiusaltsgwidveeetneirnimntrionAdepudpceebnyaduiWxxiylAina.erryItarinasdndwZoeimlvl =△ (ϕ(nm),iψn(n)): (cid:18)−n1(cid:19)logPc(n)(ϕ(n),ψ(n);∆).
(1/n)logkϕ(n)k≤R
variables U and Z, respectively,taking values in finite sets
U By time sharing we have that
and . We assume that the joint distribution of (U,X,Y,Z)
is Z G(n+m) nR+mR′,n∆+m∆′ p
XY
n+m n+m
p (u,x,y,z) (cid:18) (cid:12) (cid:19)
UXYZ nG(n)(R,∆p )+mG(m)(R′,∆(cid:12)′ p )
=pU(u)pX|U(x|u)pY|X(y|x)pZ|UY(z|u,y). ≤ | XY n+m (cid:12)(cid:12)| XY . (4)
The above condition is equivalent to ChoosingR=R′ and∆=∆′ in(4),weobtainthefollowing
subadditivity property on G(n)(R,∆p ) :
U X Y,X (U,Y) Z. XY n≥1
{ | }
↔ ↔ ↔ ↔
Define the set of probability distribution p=p by G(n+m)(R,∆pXY)
UXYZ |
nG(n)(R,∆p )+mG(m)(R,∆p )
(p )=△ p=p : +1, | XY | XY ,
P XY { UXYZ |U|≤|X| ≤ n+m
U X Y,X (U,Y) Z ,
↔ ↔ ↔ ↔ } which together with Fekete’s lemma yields that G(n)(R,
∗(p )=△ p=p : +1, ∆p ) exists and satisfies the following:
XY UXYZ XY
P { |U|≤|X| |
U X Y,Z =φ(U,Y) lim G(n)(R,∆p )= inf G(n)(R,∆p ).
↔ ↔ n→∞ | XY n≥1 | XY
for some φ: .
U ×Y →Z} Set
By definitions it is obvious that ∗(p ) (p ). Set
P XY ⊆P XY G(R,∆p )=△ lim G(n)(R,∆p ),
XY XY
△ | n→∞ |
(p)= (R,∆):R,∆ 0,
R { ≥ △
R≥Ip(X;U|Y),∆≥Epd(X,Z)}, G(pXY)={(R,∆,G):G≥G(R,∆|pXY)}.
△ The exponent function G(R,∆p ) is a convex function of
(p )= (p), XY
XY |
R R (R,∆). In fact, from (4), we have that for any α [0,1]
p∈P[(pXY) ∈
∗(p )=△ (p). G(αR+α¯R′,α∆+α¯∆′ pXY)
XY |
R R αG(R,∆p )+α¯G(R′,∆′ p ).
p∈P[∗(pXY) ≤ | XY | XY
Theregion (p )isalsoa closedconvexset. Ourmainaim where
XY
G
wisetodefirnivdeaannexexpplilciictitchoaurtaecrtebroizuantdioonfofG(p(pXY))w. Ihnostheissepcatipoenr α0 =α0(dmax,|X|)=△ [32log(|X|edmax)+1]−1,
XY
by the plane G=0 coincides with RGWZ(pXY). c1 =c1(dmax,|X|)=△ 4p2lo△g(e|X12d|meadxmax), 2 3  (7)
II. MAIN RESULT c2 =c2(dmax,|X|,|Y|,|Z|)= 8log(|Z||Xed|ma|xY)| .
theInrattheisdsisetcotriotinonwreegstiaotneou(rpmain) creasnulbte.Wexepfirersssteedxpwlaitihntthwaot The two inequalities of (6) implies that fo|Xr e|ach α∈(0,α0],
XY
R α α¯
families of supportinghyperplanes.To describe this result we (p ) c log c (1,1)
sh XY 1 2
define two sets of probability distributions on by R − α¯ α
r
U ×X ×Y ˜(α)(p ) (p (cid:16)). (cid:17)
△ ⊆Rsh XY ⊆Rsh XY
(p )= p : ,U X Y,
Psh XY { UXYZ |U|≤|X| ↔ ↔ Hence we have
X (U,Y) Z .
↔ ↔ } ˜ (p )= (p ). (8)
=△ q =q : . Rsh XY Rsh XY
UXYZ
Q { |U|≤|X||Y||Z|} Proof of Property 3 is given in Appendix D. For (α,µ)
∈
We set [0,1]2, define
R(µ)(pXY) ωq(α||,pµ)(x,y,z|u)
=△ max µ¯Ip(X;U Y)+µEpd(X;Z) , =△ α¯ log qX(x) +logqY|XU(y|x,u)
p∈Psh(pXY){ | } (cid:20) pX(x) pY|X(y|x)
where µ¯=1 µ. Furthermore set qX|UYZ(xu,y,z)
− +log |
q (xu,y)
R˜(α,µ)(p ) X|UY | (cid:21)
XY q (xu,y)
X|YU
△ +4α log | +µd(x,z) ,
=mq∈iQn α¯[D(qX||pX)+D(qY|XU||pY|X|qXU) (cid:20) pX|Y(x|y) (cid:21)
(cid:8)+Iq(X;Z|UY)] f(α,µ,λ)(x,y,z u)=△ exp λω(α,µ)(x,y,z u) ,
+4α[µ¯I (X;U Y)+µE d(X,Z)] , q||p | − q||p |
q | q } Ω(α,µ,λ)(q p ) n o
sh(pXY) | XY
=△ R (R,∆):µ¯R+µ∆ R(µ)(pXY) , =△ −logEq exp −λωq(α||,pµ)(X,Y,Z|U)
{ ≥ }
h n oi
µ∈\[0,1] = log q(u,x,y,z)f(α,µ,λ)(x,y,z u) ,
R˜(sαh)(pXY) − "u,x,y,z q||p | #
=△ (R,∆):µ¯R+µ∆≥ 41αR˜(α,µ)(pXY) , Ω(α,µ,λ)(XpXY)=△ minΩ(α,µ,λ)(q|pXY),
µ∈\[0,1](cid:26) (cid:27) q∈Q
R˜sh(pXY)=△ α∈\(0,1]R˜(sαh)(pXY) =△ FΩ((αα,,µµ,,λλ))((µp¯XRY+)−µ∆4|αpλX(Yµ¯)R+µ∆),
1+4(1 αµ)λ
1
= µα∈∈\[(00,,11]],(cid:26)(R,∆):µ¯R+µ∆≥ 4αR˜(α,µ)(pXY)(cid:27). =△ F(R,s∆u|ppXY) F(−α,µ,λ)(µ¯R+µ∆pXY),
|
(α,µ)∈(0,1]2,λ>0
For R2, we set
R⊆ + (p )=△ (R,∆,G):G F(R,∆p ) .
XY XY
G { ≥ | }
κ(1,1)=△ (a κ,b κ) R2 :(a,b) . We can show that the above functions and sets satisfy the
R− { − − ∈ + ∈R}
following property.
Then we have the following property. Property 4:
Property 3: For any pXY, we have a) The cardinality bound appearing in the
|U|≤ |X||Y||Z|
definition of Ω(α,µ,λ)(p ) is sufficient to describe this
(p )= (p ). (5) XY
Rsh XY R XY quantity.
For any α (0,α0], we have b) Define a probability distribution q(λ) =qU(λX)YZ by
∈
q(λ)(u,x,y,z)
α α¯
R(µ)(pXY)−c1 α¯ log c2α q(u,x,y,z)exp λω(α,µ)(x,y,z u)
1 R˜(α,µ)(p )r R(µ)((cid:16)p )(cid:17), (6) =△ E exp λωn(−α,µ)(qX||p,Y,Z U)| o.
≤ 4α XY ≤ XY q − q||p |
h n oi
Then we have It follows from Theorem 3 and Property 4 part d) that if
d (R,∆) is outside the rate distortion region, then the error
Ω(α,µ,λ)(q p )=E ω(α,µ)(X,Y,Z U) ,
dλ | XY q(λ) q||p | probability of decoding goes to one exponentially and its
ddλ22Ω(α,µ,λ)(q|pXY) h i expItonimenmteisdinaotetlybefloolwlowFs(Rfr,o∆m|pTXhYeo).rem 3 that we have the
following corollary.
= Var ω(α,µ)(X,Y,Z U) .
− q(λ) q||p | Corollary 1: For any R,∆>0 and any pXY, we have
h i
The second equality implies that Ωq(α,µ,λ)(q|pXY) is a G(R,∆|pXY)≥F(R,∆|pXY). (10)
concave function of λ>0.
Furthermore for any p , we have
c) Define XY
(p ) (p )
ρ=ρ(pXY) G XY ⊆G XY
△
=△ mq∈aQx(α,µm)∈a[x0,1]2Varq ωq(α||,pµ)(X,Y,Z|U) . ={(R,∆,G):G≥F(R,∆|pXY)}. (11)
h i Proof of Theorem 3 will be given in the next section. The
Since exponent function in the case of ∆ = 0 can be obtained
2
0 ω(α,µ)(x,y,z u) < as a corollary of the result of Oohama and Han [3] for the
≤ q||p | ∞
separate sourcecodingproblemof correlatedsources[2]. The
h i
for(u,x,y,z) ,wehaveρ(p )< .
∈U×X×Y×Z XY ∞ techniques used by them is a method of types [4], which is
Then for any λ (0,1], we have
∈ not useful for proving Theorem 3. In fact when we use this
Ω(α,µ,λ)(q p ) method, it is very hard to extract a condition related to the
XY
|
1 Markov chain condition U X Y, which the auxiliary
λE ω(α,µ)(X,Y,Z U) ρ(p )λ2 ↔ ↔
≥ q q||p | − 2 XY random variable U must satisfy when (R,∆) is on the
∈ U
= α¯[Dh(qX||pX)+D(qY|XiU||pY|X|qXU) obnoutnhdeariynfoofrmthaetiosnetspRe(cptrXuYm).mSeothmoedninotvreolduteccehdnibqyueHsabnas[e5d]
+I (X;Z UY)]+4α[µ¯I (X;U Y)
q | q | are necessary to prove this theorem.
+µ¯D(q p q )+µE d(X,Z)]
X|UY X|Y UY q From Theorem 3 and Property 4 part d), we obtain an
|| |
1ρ(p )λ2 explicit outer bound of WZ(εpXY) with an asymptotically
XY R |
−2 vanishing deviationfrom (p ) = (p ). The strong
WZ XY XY
R R
α¯[D(qX pX)+D(qY|XU pY|X qXU) converse theorem immediately follows from this corollary.
≥ || || |
+I (X;Z UY)] FromTheorem3andProperty4partd)wehavethefollowing
q
|
corollary.
+4α[µ¯I (X;U Y)+µE d(X,Z)]
q q
| Corollary 2: For each fixed ε (0,1) and for any δ > 0,
1ρ(p )λ2. there exists a positive integer n w∈ith
−2 XY 0
Specifically, we have n0 =n0(ε,δ,dmax, , , ,ρ(pXY))
|X| |Y| |Z|
Ω(α,µ,λ)(pXY) λR˜(α,µ)(pXY) 1ρ(pXY)λ2. such that for n≥n0, we have
≥ − 2
(εp ) (p ) κ (1,1),
d) For any δ > 0, there exists a positive number ν = RWZ | XY ⊆R XY − n
ν(δ,d , , , ) (0,1] such that for every where
max
|X| |Y| |Z| ∈
τ (0,ν], the condition (R + τ,∆ + τ) / (pXY) 1
∈ ∈ R 3+δ
implies κ =△ ρ(pXY)log 5 + 2 log 5 .
n 2n 1−ε n 1−ε
ρ(p ) (r )
F(R,∆p ) XY g2 τ3+δ >0, (cid:16) (cid:17) (cid:16) (cid:17)
| XY ≥ 2 · ρ(pXY) It immediately follows from the above result that for each
(cid:16) △(cid:17) fixed ε (0,1), we have
where g is the inverse function of ϑ(a) = (1/2)a + ∈
a2,a>0. (εp )= (p )= (p ).
WZ XY WZ XY XY
R | R R
Proof of Property 4 parta) is given in AppendixB. Proof
Proofofthiscorollary willbegiveninthenextsection..The
of Property 4 parts b), c) and d) are given in Appendix E.
direct part of coding theorem, i.e., the inclusion of (p )
Our main result is the following. R XY
(εp ) was established by Csisza´r and Ko¨rner [4].
Theorem 3: ForanyR,∆>0,anyp ,andforany(ϕ(n), ⊆ RWZ | XY
XY They proveda weak converse theoremto obtain the inclusion
ψ(n)) satisfying
(p ) (p ). Until now we have had no result
1 WZ XY XY
R ⊆ R
logMn R, on the strong converse theorem. The above corollary stating
n ≤
the strong converse theorem for the Wyner-Ziv source coding
we have
problem implies that a long standing open problem since
P(n)(ϕ(n),ψ(n);∆) 5exp nF(R,∆p ) . (9) Csisza´r and Ko¨rner [4] has been resolved.
c ≤ {− | XY }
III. PROOF OF THEMAIN RESULT we have the following:
In this section we prove Theorem 3. We first present a
lemmawhichupperboundsthecorrectprobabilityofdecoding
P(n)(ϕ(n),ψ(n);∆) p
by the information spectrum quantities. We set c ≤ SnXnYnZn
(cid:26)
Sn =△ ϕ(n)(Xn),Zn =△ ψn(ϕ(n)(Xn),Yn). η 1 n log qX(i)t(Xt),
≥ n p (X )
It is obvious that Xt=1 Xt t
Sn Xn Yn,Xn (Sn,Yn) Zn. η 1 n logqY(iti|)UtXt(Yt|Ut,Xt),
↔ ↔ ↔ ↔ ≥ n p (Y X )
t=1 Yt|Xt t| t
Then we have the following. X
Lemma 1: For any η >0 and for any (ϕ(n), ψ(n)) satisfy- η 1 n logqX(iiti|)UtYtZt(Xt|Ut,Yt,Zt),
ing ≥ n p (X U ,Y )
1 Xt=1 Xt|UtYt t| t t
nlogMn ≤R, R+η 1 n logqX(ivt|)UtYt(Xt|Ut,Yt),
we have ≥ n p (X Y )
t=1 Xt|Yt t| t
X
n
P(cn)(ϕ(n),ψ(n);∆)≤pSnXnYnZn ∆ 1 loged(Xt,Zt) +4e−nη, (18)
(cid:26) ≥ n )
1 q(i) (Xn) Xt=1
η log Xn , (12)
≥ n pXn(Xn) where for each t = 1,2, ,n, the following probability
η 1 logqY(iin)|SnXn(Yn|Sn,Xn), (13) distribution and stochastic m·a·t·rices:
≥ n p (Yn Xn)
Yn|Xn
η 1 logqX(iini)|SnYnZn(X|n|Sn,Yn,Zn), (14) qX(i)t,qY(iti|)UtXt,qX(iiti|)UtYtZt, and qX(ivt|)UtYt
≥ n p (Xn S ,Yn)
Xn|SnYn | n appearing in the first term in the right members of (18) have
R+η 1 logqX(ivn)|SnYn(Xn|Sn,Yn), (15) a property that we can choose their values arbitrary.
≥ n pXn|Yn(Xn Yn) Proof:Ontheprobabilitydistributionsappearingintheright
|
1 members of (16), we take the following choices. In (12), we
∆≥ nlogexp{d(Xn,Zn)}(cid:27)+4e−nη. (16) choose qX(i)n so that
The probability distribution and stochastic matrices appearing
n
in the right members of (16) have a property that we can
q(i) (Xn)= q(i)(X ). (19)
select them arbitrary. In (12), we can choose any probability Xn Xt t
distributionq(i) on n.In(13),wecanchooseanystochastic tY=1
Xn X
matrix q(ii) : n n. In (14), we can choose
Yn|SnXn Mn× X →Y In (13), we choose q(ii) so that
any stochastic matrix q(iii) : n n n. Yn|SnXn
Xn|SnYnZn Mn× Y × Z →X
In (15), we can choose anystochastic matrix q(iv) :
× Yn →Xn. Xn|SnYn Mn qY(iin)|SnXn(Yn|Sn,Xn)
Proof of this lemma is given in Appendix F. n
Lemma 2: Suppose that for each t = 1,2,···,n, the joint = qY(iti|)SnXtYtn+1(Yt|Sn,Xt,Ytn+1)
distribution pSnXtYn of the random vector SnXtYn is a tY=n1
mwearhgainvaeltdhiestfroiblluotwioinngofMpaSrnkXovnYcnha.iTn:hen, for t = 1,2,···,n, = qY(iti|)XtUt(Yt|UtXt). (20)
t=1
Y
X S Xt−1Yn Yt−1 (17)
t ↔ n t ↔
In (14), we choose q(iii) so that
or equivalently that I(Xt;Yt−1|SnXt−1Ytn)=0. Xn|SnYnZn
Proof of this lemma is given in Appendix G. For t =
Y1,n2,·)·b·,ena, rsaentduotm=△ve(csto,rxtt−ak1i,nygtn+v1a)l.ueLsetinUt =△ (Sn,tX−1t−1, qX(iini)|SnYnZn(Xn|Sn,Yn,Zn)
t+1 Mn × X × n
n . From Lemmas 1 and 2, we have the following. = q(iii) (X S Xt−1,Yn,Z )
Yt+Le1mma 3: For any η >0 and for any (ϕ(n), ψ(n)) satisfy- Xt|SnXt−1YtnZt t| n t t
t=1
Y
ing n
1 = q(iii) (X U Y Z ). (21)
logMn R, Xt|UtYtZt t| t t t
n ≤ t=1
Y
n
In (14), we note that 1
∆ loged(Xt,Zt) +4e−nη, (24)
p (Xn S ,Yn) ≥ n t=1 )
Xn|SnYn | n X
n where for each t = 1,2, ,n, the following probability
= pXt|SnXt−1Yn(Xt|Sn,Xt−1,Yn) distribution and stochastic m·a·t·rices:
t=1
(=a) Yn pXt|SnXt−1Ytn(Xt|Sn,Xt−1,Ytn) appearing iqnXtth,eqYfitr|UsttXtetr,mqXitn|UtthYteZtri,ghantdmqeXmtb|UetrYstof (24) are
t=1
Y chosen so that they are induced by the joint distribution q =
n t
= pXt|UtYt(Xt|Ut,Yt). (22) qUtTXotYetvZatlu∈atQe ta.n upper bound of (24) in Lemma 4. We use
t=1
Y
the following lemma, which is well known as the Crame´r’s
Step(a)followsfromLemma2.In(15),wechooseq(iv)
Xn|SnYn bound in the large deviation principle.
so that
Lemma 5: For any real valued random variable Z and any
q(iv) (Xn S ,Yn) θ >0, we have
Xn|SnYn | n
n Pr Z a exp[ (θa logE[exp(θZ)])].
= q(iv) (X S ,Xt−1,Yn) { ≥ }≤ − −
Xt|SnXt−1Ytn t| n t Here we define a quantity which serves as an exponential
t=1
Yn upper bound of P(n)(ϕ(n), ψ(n)). Let (n)(p ) be a set of
c XY
= qX(ivt|)UtYt(Xt|Ut,Yt). (23) all probability distributions pSnXnYnZPn on Mn ×Xn ×Yn
t=1 n having the form:
Y
×Z
From Lemma 1 and (19)-(23), we have the bound (18) in
p (s,xn,yn,zn)
Lemma 4. SnXnYnZn
n
ofFaollrperaocbhabtil=ity1,d2is,t·ri·b·u,tnio,nlsetqQUt(XUttYt×ZtXo×n Y ×Z) be a set =pSn|Xn(s|xn)(t=1pXtYt(xt,yt))pZn|YnSn(zn|yn,s).
Y
t = n t n−t+1 For simplicity of notation we use the notation p(n) for
U ×X ×Y ×Z M ×X ×Y ×Z
p (n) (p ). We assume that p =
isnucclhudtehdatinthtehastuopfppoXrtYo.fFothretm=a1rg,i2n,a·l··d,inst,riwbuetisoimnpqlXytwYtrities p1SS,2nn,XXntYYtn,nZnZt,nwi∈se aPsimmpalrygiXnwaYrlitedisptrib=utiopn of p(n.).UFtoXFrotYrpt(Zntt) =
Qt=Q(Ut ×X ×Y ×Z). Similarly, for t = 1,2,···,n, we (n)·(p·· ) and qn n, we detfine UtXtYtZt ∈
simply write qt =qUtXtYtZt ∈Qt. Set P XY ∈Q
n n Ω(α,µ,θ)(p(n),qn pXY)
n =△ t = ( t ), n |
Q Q Q U ×X ×Y×Z △ (α,µ)
From Lemqmna=△3,t{Yw=q1te}intm=1m∈etYd=Qia1nte.lyobtainthe followinglemma. ==−llooggEEp(n)"tY=1nexppαX¯nθt(−Xθtω)pt||qptαY(¯Xtθ|Xt,tY(Yt,t|ZXt|tU)t)o#
− p(n)"(tY=1qXα¯θt(Xt)qYα¯tθ|XtUt(Yt|Xt,Ut))
Lemma 4: For any η >0, for any (ϕ(n), ψ(n)) satisfying n pα¯θ (X U ,Y )
Xt|UtYt t| t t
1 logMn R, ×(tY=1qXα¯θt|UtYtZt(Xt|Ut,Yt,Zt))
n ≤ n p4αθ (X Y )
and for any qn n, we have the following: Xt|Yt t| t e4αµθd(Xt,Zt) ,
∈Q ×(tY=1qX4αtθ|YtUt(Xt|Ut,Yt) )#
P(cn)(ϕ(n),ψ(n);∆)≤pSnXnYnZn where for each t = 1,2,···,n, the following probability
(cid:26) distribution and stochastic matrices:
n
1 q (X )
η log Xt t , q ,q ,q ,q
≥ n t=1 pXt(Xt) Xt Xt|UtYt Xt|UtYtZt Yt|XtUt
η 1 Xn logqYt|UtXt(Yt|Ut,Xt), acphpoeseanrinsgo itnhatthethedyefianrietioinnduocfedΩ(bαy,µ,tθh)e(p(jno)in,qtnd|pisXtrYib)utiaorne
≥ n p (Y X )
η 1 Xt=n1logqXt|YUtt|YXttZt(Xt|t|Ut t,Yt,Zt), qtH=eqreUtwXetYgtZivte∈aQretm. ark on an essential difference between
R+η ≥ n1 Xt=n1logqXtp|XUttY|Ut(tYXt(t|XUtt|,UYtt,)Y,t) wpp(rinoth)bap∈b(inli)Pt,ys(indn)ic(septrXtihbYeu)ytioaanrnesdmpqatn,rgti∈na=lQd1ins,.t2riF,b·ou·rt·io,tnhnse,oafforeprm(nce)or.nOstihnsetethnnet
≥ n t=1 pXt|Yt(Xt|Yt) otherhand,forthelatter,qn isjustasequenceofnprobability
X
distributions.Hence,wemaynothavetheconsistencybetween Solving (26) with respect to η, we have
the n elements qt, t=1,2, ,n, of qn. 1Ω(α,µ,θ)(p(n),qn p ) 4αθ(µ¯R+µ∆)
By Lemmas 4 and 5, we·h·a·ve the following proposition. η = n | XY − .
1+(3+α 4αµ)θ
Proposition 1: For any α, µ, θ >0, any qn n, and any −
∈Q For this choice of η and (25), we have
(ϕ(n), ψ(n)) satisfying
P(n)(ϕ(n),ψ(n);∆) 5e−nη
1 c ≤
logM R,
n n ≤ =5exp n[1+(3+α 4αµ)θ]−1
− −
(cid:26)
we have
1
Ω(α,µ,θ)(p(n),qn p ) 4αθ(µ¯R+µ∆) ,
XY
P(n)(ϕ(n),ψ(n);∆) ×(cid:20)n | − (cid:21)(cid:27)
c
completing the proof.
5exp n[1+(3+α)θ]−1 Set
≤ −
1(cid:26) Ω(α,µ,θ)(pXY)
× nΩ(α,µ,θ)(p(n),qn|pXY)−4αθ(µ¯R+µ∆) . =△ inf min max 1Ω(α,µ,θ)(p(n),qn p ).
(cid:20) (cid:21)(cid:27) XY
n≥1p(n)∈P(n)(pXY)qn∈Qn n |
Proof:ByLemma4,forα,µ,θ 0,wehavethefollowing
By Proposition 1 we have the following corollary.
≥
chain of inequalities: Corollary 3: For any (α,µ) [0,1]2, for any θ > 0, and
∈
for any (ϕ(n), ψ(n)) satisfying
1
P(n)(ϕ(n),ψ(n);∆) p log ϕ(n) R,
c ≤ SnXnYnZn n || ||≤
(cid:26)
n we have
α¯ q (X )
α¯η log Xt t ,
≥ n p (X ) P(n)(ϕ(n),ψ(n);∆)
t=1 Xt t c
α¯η ≥ αn¯ Xt=n1logqYtp|UYttX|Xtt((YYtt|U|Xt,tX) t), ≤5exp(−n"Ω(α,µ,θ1)(+pX(3Y+)−α4−αθ4(αµ¯µR)θ+µ∆)#).
X We shall call Ω(α,µ,θ)(p ) the communication potential.
α¯η α¯ n logqXt|UtYtZt(Xt|Ut,Yt,Zt), The above corollary impliXesY that the analysis of Ω(α,µ,θ)(
≥ n t=1 pXt|UtYt(Xt|Ut,Yt) pXY) leads to an establishment of a strong converse theo-
X
rem for Wyner-Ziv source coding problem. In the following
4αµ¯ n q (X U ,Y )
4αµ¯(R+η) log Xt|UtYt t| t t , argumentwe drive an explicitlower boundof Ω(α,µ,θ)(pXY).
≥ n p (X Y )
t=1 Xt|Yt t| t We use a new techique we call the recursive method. The
X
recursivemethodisapowerfulltooltodriveasingleletterized
n
4αµ
4αµ∆ loged(Xt,Zt) +4e−nη exponentfunction for rates below the rate distortion function.
≥ n ) Thismethodisalsoapplicabletoprovetheexponentialstrong
t=1
X
conversetheoremsforothernetworkinformationtheoryprob-
p 4α(µ¯R+µ∆)+(3+α 4αµ)η
≤ SnXnYnZn − lems [6], [7], [8].
(cid:26) For each t=1,2, ,n, set
n
1 ···
≥ n t=1ωq(αt|,|µp)t(Xt,Yt,Zt|Ut))+4e−nη fq(tα||,pµt,θ)(xt,yt,zt|ut)
= pSnXXnYnZn(−n1 Xt=n1θωq(αt|,|µp)t(Xt,Yt,Zt|Ut) ==△ epxαX¯pθt(nx−t)θωq(αtp|,|αY¯µptθ)t|(Xxtt(,yytt|x,zt)t|ut)opαX¯θt|UtYt(xt|ut,yt)
qα¯θ(x )qα¯θ (y x ,u )qα¯θ (x u ,y ,z )
θ[4α(µ¯R+µ∆)+(3+α 4αµ)η] +4e−nη Xt t Yt|XtUt t| t t Xt|UtYtZt t| t t t
(a)≥− − (cid:27) p4Xαtθ|Yt(xt|yt) e4αµθd(xt,zt).
≤ exp n 4αθ(µ¯R+µ∆)+(3+α−4αµ)θη ×qX4αtθ|YtUt(xt|ut,yt)
(cid:20) (cid:26) By definition we have
1
Ω(α,µ,θ)(p(n),qn p ) +4e−nη. (25)
−n | XY exp Ω(α,µ,θ)(p(n),qn p )
(cid:27)(cid:21) − | XY
Step (a) follows from Lemma 5. We choose η so that = np (s,yn) p o (xn,zn s,yn)
SnYn XnZn|SnYn |
s,yn xn,zn
η =4αθ(µ¯R+µ∆)+θ(3+α 4αµ)η X n X
− 1 − f(α,µ,θ)(x ,y ,z u ). (27)
Ω(α,µ,θ)(p(n),qn pXY). (26) × qt||pt t t t| t
−n | t=1
Y
Foreacht=1,2, ,n,we definetheconditionalprobability where C˜ is a constant for normalization given by
t
···
distribution
t
p(α,µ,θ;qt) C˜t = pSnYn(s,yn) Φi(,αq,iµ,θ)(s,yn).
XtZt|SnYn s,yn i=1
=△ p(α,µ,θ;qt) (xt,zt s,yn) X Y
XtZt|SnYn | (xt,zt,s,yn)∈Xt×Zt×Mn×Yn For t=1,2,···,n, define
n o
by Λ(α,µ,θ) =△ C˜C˜−1, (34)
t,qt t t−1
pX(αt,Zµ,tθ|;Sqnt)Yn(xt,zt|s,yn) where we define C˜0 =1. Then we have the following.
=△ C−1(s,yn)p (xt,zt s,yn) Lemma 7:
t XtZt|SnYn | n
t exp Ω(α,µ,θ)(p(n),qn p ) = Λ(α,µ,θ), (35)
f(α,µ,θ)(x ,y ,z u ) − | XY t,qt
× qi||pi i i i| i n o tY=1
iY=1 Λ(α,µ,θ) = p(α,µ,θ;qt−1)(s,yn)Φ(α,µ,θ)(s,yn)
where t,qt SnYn t,qt
s,yn
X
Ct(s,yn)=△ pXtZt|SnYn(xt,zt|s,yn) = pS(αn,Yµ,nθ;qt−1)(s,yn)
xXt,zt s,yn
X
× t fq(iα||,pµi,θ)(xi,yi,zi|ui) (28) ×xt,ztpX(αt,−µ1,θZ;qt−t−11|S)nYn(xt−1,zt−1|s,yn)
i=1 X
Y p (x ,z xt−1,zt−1,s,yn)
are constants for normalization. For t=1,2, ,n, define × XtZt|Xt−1Zt−1SnYn t t|
··· f(α,µ,θ)(x ,y ,z u ). (36)
Φ(α,µ,θ)(s,yn)=△ C (s,yn)C−1(s,yn), (29) × qt||pt t t t| t
t,qt t t−1
Proof: By the equality (32) in Lemma 6, we have
wherewedefineC (s,yn)=1for(s,yn) n. Then
0 n
∈M ×Y
we have the following lemma. exp Ω(α,µ,θ)(p(n),qn p )
XY
Lemma 6: For each t = 1,2, ,n, and for any (s, yn − |
··· n n n o
xt,zt)∈Mn ×Yn ×Xt ×Zt, we have =C˜n = C˜tC˜t−−11 (=a) Λt(,αq,tµ,θ). (37)
p(α,µ,θ;qt) (xt,zt s,yn)=(Φ(α,µ,θ)(s,yn))−1 tY=1 tY=1
XtZt|SnYn | t,qt Step(a)followsfromthedefinition(34)ofΛ(α,µ,ν,θ).Wenext
p(α,µ,θ;qt−1) (xt−1,zt−1 s,yn) t,qt
××pXXttZ−t1|YStn−X1|tS−n1YYnn(xt,zt|s,xt−| 1,zt−1,yn) bporothvesi(d3e6s)oinf (L3e3m),mwae7h.avMeultiplying Λt(,αq,tµ,θ) = C˜t/C˜t−1 to
Φt(×,αq,tfµ,q(θtα)||,(pµst,,θy)(nx)t,yt,zt|ut). (30) Λt(,αq,tµ,θ)pS(αn,Yµ,nθ;qt)(s,tyn) (38)
= p(α,µ,θ;qt−1) (xt−1,zt−1 s,yn) =C˜t−−11pSnYn(s,yn) Φi(,αq,iµ,θ)(s,yn),
Xt−1Zt−1|SnYn | iY=1
xXt,zt =p(α,µ,θ;qt−1)(s,yn)Φ(α,µ,θ)(s,yn). (39)
×pXtZt|SnXt−1Yn(xt,zt|s,xt−1,zt−1,yn) SnYn t,qt
f(α,µ,θ)(x ,y ,z u ). (31) Taking summations of (38) and (39) with respect to (s,yn),
× qt||pt t t t| t we have (36) in Lemma 7.
Furthermore, we have The following proposition is a mathematical core to prove
our main result.
exp Ω(α,µ,θ)(p(n),qn p )
− | XY Proposition 2: For θ (0,1/α¯), we choose the parameter
∈
n n o λ such that
= p (s,yn) Φ(α,µ,θ)(s,yn). (32) θ λ
SnYn t,qt λ= θ = . (40)
s,yn t=1 1 α¯θ ⇔ 1+α¯λ
X Y −
The equality (32) in Lemma 6 is obvious from (27), (28), Then for any α,µ>0 and for any θ (0,1/α¯), we have
∈
and (29). Proofs of (30) and (31) in this lemma are given in
1
Appendix H. Next we define a probability distribution of the Ω(α,µ,θ)(pXY) Ω(α,µ,λ)(pXY). (41)
≥ 1+λα¯
random pair (S ,Yn) taking values in n by
n n
M ×Y Proof: Set
p(α,µ,θ;qt)(s,yn)
SnYn ˆ =△ q =q : n−1 n−1 ,
t Qn { UXYZ |U|≤|Mn||X ||Y |}
=C˜t−1pSnYn(s,yn)Yi=1Φi(,αq,iµ,θ)(s,yn), (33) Ωˆn(α,µ,λ)(pXY)=△ qm∈Qiˆnn Ω(α,µ,λ)(q|pXY).
θ
Set qα¯ (X U ,Y ) 1−α¯θ
Xt|UtYt t| t t
pS(αn,Xµ,tθY;qtnt−Z1t)(s,xt,ytn,zt) ×(qXα¯t|UtYtZt(Xt|Ut,Yt,Zt))
=pU(αt,Xµt,θY;tqZt−t1)(ut,xt,yt,zt) p4Xαtµ¯|Yt(Xt|Yt) e4αµd(Xt,Zt) 1−θα¯θ 1−α¯θ
=△ pS(αn,Yµ,nθ;qt−1)(s,yn) ×(qX4αtµ|¯YtUt(Xt|Ut,Yt) ) 
yt−X1,zt−1 p (X U ,Y ) α¯θ 
××ppXX(αtt,Z−µt1,θ|ZX;qt−tt−−111|SZ)nt−Y1nS(nxYt−n(1x,zt,t−zt1||xst,−y1n,)zt−1,s,yn). (42) = e×x(cid:18)pEqt((cid:26)1qXXαtt¯||UUθ)ttYYΩtt((αX,µtt,||1U−θttα¯,θY)tt()q(cid:27)t p(cid:19)XY)
− − |
Then by Lemma 7, we have
(=b)expn 1 Ω(α,µ,λ)(q p ) o
Λ(α,µ,θ) = p(α,µ,θ;qt−1)(u ,x ,y ,z ) −1+α¯λ t| XY
t,qt UtXtYtZt t t t t (cid:26) (cid:27)
ut,xXt,yt,zt (c) exp 1 Ωˆ(α,µ,λ)(p )
f(α,µ,θ)(x ,y ,z u ). ≤ −1+α¯λ n XY
× qt||pt t t t| t (cid:26) (cid:27)
For each t = 1,2, ,n, we recursively choose q = (=d)exp 1 Ω(α,µ,λ)(p ) . (43)
··· t −1+α¯λ XY
qUtXtYtZt so that (cid:26) (cid:27)
Step (a) follows from Ho¨lder’s inequality. Step (b) follows
qUtXtYtZt(ut,xt,yt,zt)=pU(αt,Xµt,θY;tqZt−t1)(ut,xt,yt,zt) from (40). Step (c) follows from the definition of Ωˆn(α,µ,λ)
and choose q , q , q , and q appearing (pXY). Step (d) follows from that by Property 4 part a), the
in Xt Yt|XtUt Xt|UtYtZt Xt|YtUt bound is sufficient to describe Ωˆn(α,µ,λ)(
|U| ≤ |X||Y||Z|
p ). Hence, we have the following:
fq(tα|,|µpt,θ)(xt,yt,zt|ut)= pqXαXα¯θθtt((xxtt))qYα¯tθp|XαY¯tθt|UXtt((yytt||xxtt,)ut) XY qmn∈aQxn n1Ω(α,µ,θ)(p(n),qn|pXY)
×qXα¯θpt|αX¯Uθtt|YUttZYtt((xxtt||uutt,,yytt,)zt)qX4αtpθ|4XYαttUθ|Ytt((xxtt||uytt,)yt)e4αµθd(xt,zt) ≥ n1Ω(α,µ,θ)(p(n),qn|pXY)(=a)−n1 t=n1logΛt(,αq,tµ,θ)
X
fsourchetahcahttthe=yar1e,2th,edis,trnib,uwtieonhsaivnedutcheedfboyllqoUwtXintgYtcZhta.iTnheonf (b) 1 Ω(α,µ,λ)(pXY). (44)
··· ≥ 1+α¯λ
inequalities:
Step(a)followsfrom(35)in Lemma7.Step(b)followsfrom
Λ(α,µ,θ) (43). Since (44) holds for any n 1 and any p(n) (n)
t,qt ≥ ∈ P
(p ), we have
= Eqt"(pqXαXα¯¯θθtt((XXtt))qYα¯tθp|XαY¯ttθ|UXtt((YYtt|X|Xtt,)Ut)) XY Ω(α,µ,θ)(pXY)≥ 1+1α¯λΩ(α,µ,λ)(pXY).
pα¯θ (X U ,Y )
Xt|UtYt t| t t Thus we have (41) in Proposition 2.
×(qXα¯θt|UtYtZt(Xt|Ut,Yt,Zt)) Proof of Theorem 3: For θ ∈(0,1/α¯), set
p4αµ¯θ(X Y ) θ λ
Xt|Yt t| t e4αµθd(Xt,Zt) λ= θ = . (45)
×(qX4αtµ|¯YθtUt(Xt|Ut,Yt) )# Then we have the fol1lo−wiα¯nθg:⇔ 1+α¯λ
= Eqt"(qpqαXαXα¯¯¯tt((XXtt))(XqYα¯tp|UXαY¯tt|U,XYtt((Y)Ytt|X|Xtt,θ)Ut))θ n1 log(Pc(n)(ϕ(n)5,ψ(n);∆))
Xt|UtYt t| t t (a) Ω(α,µ,θ)(p ) 4αθ(µ¯R+µ∆)
×(qXα¯t|UtYtZt(Xt|Ut,Yt,Zt)) ≥ 1+Xθ(Y3+−α 4αµ)
p4Xαtµ¯|Yt(Xt|Yt) e4αµd(Xt,Zt) θ (b) 1+1α¯λΩ(α,µ,λ)(pXY−)− 14+αα¯λλ(µ¯R+µ∆)
×(qX4αtµ|¯YtUt(Xt|Ut,Yt) ) ≥ 1+ 1+λα¯λ(3+α−4αµ)
p (X U ,Y ) α¯θ Ω(α,µ,λ)(pXY) 4αλ(µ¯R+µ∆)
Xt|UtYt t| t t = 1+α¯λ+λ(−3+α 4αµ)
×(cid:26)qXt|UtYt(Xt|Ut,Yt)(cid:27) # = F(α,µ,λ)(µ¯R+µ∆p −).
(a) E pαX¯t(Xt) pαY¯t|Xt(Yt|Xt) 1−θα¯θ Step (a) follows from Corolla|ryXY3. Step (b) follows from
≤  qt(qXα¯t(Xt)qYα¯t|XtUt(Yt|Xt,Ut)) Proposition 2 and (45). Since the above bound holds for any
 
positive α, µ, and λ, we have Then we have the following chain of inequalities:
5exp[ n∗F(R,∆p )]
XY
1 5 − |
log F(R,∆p ). (a) n∗ρ(p )
n (P(cn)(ϕ(n),ψ(n);∆))≥ | XY ≤ 5exp − 2XY ·g2 ρ(τp3X+Yδ)
Thus (9) in Theorem 3 is proved. (<b)5exp(cid:20) n∗ρ(pXY) g2(cid:16) κ3n+∗δ (cid:17)(cid:21)
− 2 · ρ(pXY)
Proof of Corollary 2: To prove this corollary we use the (cid:20) (cid:18) (cid:19)(cid:21)
followingexpressionof (εp statedinProperty1part (c) n∗ρ(pXY) 2 5
WZ XY) = 5exp log
b): R | (cid:20)− 2 n∗ρ(pXY) (cid:18)1−ε(cid:19)(cid:21)
= 1 ε. (51)
−
Step (a) follows from Property 4 part d). Step (b) follows
(εp )=cl (n,εp ) . (46)
RWZ | XY  RWZ | XY  from κn∗ < τ. Step (c) follows from (50). The bound (51)
m[≥1n\≥m contradicts (48). Hence we have (R+κ ,∆+κ ) (p )
n n XY
  ∈R
or equivalent to
We assume that
(R,∆) (p ) κ (1,1)
XY n
∈R −
(R,∆) (n,εp ). (47)
∈ RWZ | XY forn n0. Recallingthe firstassumption(47) on(R,∆), we
m≥1n≥m ≥
[ \ obtain
Then there exists a positive integer m0 = m0(ε) and some WZ(n,εpXY) (pXY) κn(1,1). (52)
(ϕ(n), ψ(n)) such that for n m (ε), we have R | ⊆R −
{ }n≥m0 ≥ 0 m[≥1n\≥m
Taking the closure of both sides of (52), using (46), and
1
log ϕ(n) R, P(n)(ϕ(n),ψ(n);∆) ε. considering that (p ) κ (1,1) is a closed set, we have
n || ||≤ e ≤ R XY − n
that for n n ,
0
≥
Then by Theorem 3, we have
(εp )
WZ XY
R |
1 ε P(n)(ϕ(n),ψ(n);∆)
− ≤ c =cl WZ(n,εpXY) (pXY) κn(1,1),
 R | ⊆R −
5exp nF(R,∆p ) (48)
XY m≥1n≥m
≤ {− | } [ \
completing the proof. 
foranyn m (ε).Fix anyδ >0.We takeapositivenumber
0
≥
ν = ν(δ,d , , , ) (0,1] appearing in Property 4
max
|X| |Y| |Z| ∈
part d) and set APPENDIX
A. Properties of the Rate Distortion Regions
1
3+δ
In this appendix we prove Property 1. Property 1 part a)
κ = ρ(p )ϑ 2 log 5 (49)
n ( XY rnρ(pXY) 1−ε !) can easily be proved by the definitions of the rate distortion
(cid:16) (cid:17)
regions. We omit the proofs of this part. In the following
1
3+δ
= ρ(pXY)log 5 + 2 log 5 . argument we prove part b).
2n 1−ε n 1−ε Proof of Property 1 part b: We set
(r )
(cid:16) (cid:17) (cid:16) (cid:17)
(m,εp )= (n,εp ).
Since g is an inverse function of ϑ, (49) is equivalent to RWZ | XY RWZ | XY
n≥m
\
By the definitions of (m,εp ) and (εp ), we
g κ3n+δ = 2 log 5 . (50) havethat (m,εpRW)Z | X(Yεp )RfoWrmZ | 1X.YHence
(cid:16)ρ(pXY)(cid:17) rnρ(pXY) (cid:16)1−ε(cid:17) we have tRhaWtZ | XY ⊆RWZ | XY ≥
We take a sufficiently large positive integer m so that we
1 (m,εp ) (εp ). (53)
haveκ <ν forn m . Setn =max m , m .We claim RWZ | XY ⊆RWZ | XY
that fornn n , w≥e hav1e (R+0κ ,∆+{κ 0) 1}(p ). To m[≥1
0 n n XY
≥ ∈ R We next assume that (R,∆) (εp ). Set
prove this claim we suppose that (R+κn∗,∆+κn∗) does ∈RWZ | XY
notbelongto (pXY) forsomen∗ n0. Since (pXY) is a (δ) (εp )=△ (R+δ,∆):(R,∆) (εp )
closed set, theRre exists a positivenum≥berτ >κ Rsufficiently RWZ | XY { ∈RWZ | XY }
n∗
close to κn∗ such that Then, by the definitions of RWZ(n,ε |pXY) and RWZ(
εp ), we have that for any δ > 0, there exists n (δ) such
XY 0
|
κn∗ <τ ≤ν, that for any n≥n0(δ),
(R+τ,∆+τ) / (p ). (R+δ,∆) (n,εp ),
XY WZ XY
∈R ∈R |