Table Of ContentOn AVCs with Quadratic Constraints
Farzin Haddadpour∗, Mahdi Jafari Siavoshani†, Mayank Bakshi†, Sidharth Jaggi†
∗School of Electrical Engineering
Sharif University of Technology, Tehran, Iran
Email: [email protected]
†Institute of Network Coding
Chinese University of Hong Kong, Hong Kong
Email: {mahdi,mayank}@inc.cuhk.edu.hk, [email protected]
Abstract—InthisworkwestudyanArbitrarilyVaryingChan- knowledge of Alice’s code, and her message.2
nel (AVC) with quadratic power constraints on the transmitter Even in such an adverse jamming setting we demonstrate
3 andaso-called“oblivious”jammer(alongwithadditionalAWGN) that Alice can communicate with Bob at a rate equalling
1 under a maximum probability of error criterion, and no private (cid:16) (cid:17)
0 randomness between the transmitter and the receiver. This is in 12log 1+ Λ+Pσ2 as long as P >Λ. Note that this equals the
2 contrast to similar AVC models under the average probability of capacity of an AWGN with noise parameter equal to Λ+σ2
error criterion considered in [1], and models wherein common – this means that no “smarter” jamming strategy exists for
n randomness is allowed [2] – these distinctions are important in
a some communication scenarios outlined below. Harry than simply behaving like AWGN with variance Λ. If
J P < Λ no positive rate is possible since Harry can “spoof”
We consider the regime where the jammer’s power constraint
7 is smaller than the transmitter’s power constraint (in the other by transmitting a fake message using the same strategy as
2 regime it is known no positive rate is possible). For this regime Alice – Bob is unable to distinguish between the real and fake
we show the existence of stochastic codes (with no common transmissions3.
] randomness between the transmitter and receiver) that enables
T
reliable communication at the same rate as when the jammer A. Relationship with prior work
I
is replaced with AWGN with the same power constraint. This
.
s matches known information-theoretic outer bounds. In addition The model considered in this work is essentially a special
c
to being a stronger result than that in [1] (enabling recovery of type of Arbitrarily Varying Channel (AVC) for which, to the
[
theresultstherein),ourprooftechniquesarealsosomewhatmore bestofourknowledge,thecapacityhasnotbeencharacterized
1 direct, and hence may be of independent interest. before in the literature. The notion of AVCs was first intro-
v
duced by Blackwell et al. [6], [7], to capture communication
5
models wherein channel have unknown parameters that may
4
3 I. INTRODUCTION vary arbitrarily during the transmission of a codeword. The
6 case when both the transmitter and the jammer operate under
1. Aerial Alice is flying in a surveillance plane high over constraints(analogoustothequadraticconstraintsinthiswork)
0 Hostile Harry’s territory. She wishes to relay her observations has also been considered [3], [4]. For an extensive survey on
3 of Harry’s troop movements back to Base-station Bob over n AVCs the reader may refer to the excellent survey [5] and the
1 channel uses of an AWGN channel with variance σ2. Harry references therein.
:
v obviously wishes to jam Alice’s transmissions. However, both The class of AVCs over discrete alphabets has been studied
i Alice’s transmission energy and Harry’s jamming energy are in great detail in the literature [5]. However, less is known
X
constrained – they have access to energy sources of nP and about AVCs with continuous alphabets. The bulk of the work
r
a nΛ Joules respectively.1 Harry already knows what message on continuous alphabet AVCs (outlined below in this section)
Alice wants to transmit (after all, he knows the movements of focuses on quadratically-constrained AVCs. This is also the
his own troops), and also roughly how she’ll transmit it (i.e., focus of our work.
her communication protocol/code, having recently captured It is important to stress several features of the model
another surveillance drone) but he doesn’t know exactly how considered in this work, and the differences with prior work:
she’lltransmitit(i.e.,hercodeword–forinstance,Alicecould Stochastic encoding: To generate her codeword from her
choose to focus her transmit power on some random subset of •
message, Alice is allowed to use private randomness
the n channel uses). Further, since Alice’s transmissions are (known only to her a priori, but not to Harry or Bob.
very quick, Harry has no time to tune his jamming strategy to Thisisincontrasttothedeterministicencodingstrategies
Alice’s actual codeword – he can only jam based on his prior
2Alternatively,Alicecouldsplitherenergybudgettoconcurrentlytransmit
one symbol on n different frequencies – these together could comprise her
1These are so-called peak power constraints – they must hold for all codeword.Givensuchastrategy,sinceHarrydoesn’tknowAlice’scodeword,
codewords,ratherthanaveragedoverallcodewordsaveragepowerconstraints. heisunabletomakehisjammingstrategydependexplicitlyonthecodeword
If the peak power constraints are relaxed to average power constraints, for Aliceactuallytransmits.
either Alice’s transmissions, or Harry’s jamming (or both), it is known [2] 3Suchajammingstrategyisequivalenttothemoregeneralsymmetrizability
thatstandardcapacityresultsdonothold–only“λ-capacities”exist. conditionintheAVCliterature(see,forinstance[3],[4],and[5]).
Deterministic Random Realization
often considered in the information theory/coding theory
Scalar a,k,N X,Y x,y
literature, wherein the codeword is a deterministic func- Vector v V v
tion of the message. Set C C∗ C
Public code: Everything Bob knows about Alice’s trans- TableII
•
mission a priori, Harry also knows.4 This is in contrast EXAMPLESOFOURNOTATIONCONVENTIONFORDIFFERENTVARIABLES.
to the randomized encoding model also considered in the
literature (see for instance [2], [9]), in which it is critical
Symbol Meaning
that Alice and Bob share common randomness that is
Ψ(i) Stochasticencoderappliedtothemessagei
unknown to Harry. φ(Y) Deterministicdecoder
Message-aware jamming:Thejammerisalreadyawareof e(s,i) Errorprobability(overthestochasticencoderandthechan-
• nelnoise)forafixedmessageiandjammingvectors
Alice’s message. This is one important difference in our
emax(s) Maximum (over messages) error probability for a fixed
model, from the model in the work closest to ours, that jammingvectors
of [1]. N(a,σ2) Gaussianrandomvariablewithmeanaandvarianceσ2
Oblivious adversary:Thejammerhasnoextraknowledge Bn(c,r) Aballofradiusr inRn whichcenteredatc∈Rn
•
of the codeword being transmitted than what he has al- TableIII
readygleanedfromhisknowledgeofAlice’scodeandher COMMONLYUSEDSYMBOLS.
message. This is in contrast to the omniscient adversary
often considered in the coding theory literature.
These model assumptions are equivalent to requiring public II. NOTATIONANDPROBLEMSTATEMENT
stochastic codes with small maximum error of probability A. Notation
against an oblivious adversary. Several papers also operate
Throughout the paper, we use capital letters to denote ran-
under some of these assumptions, but as far as we know, none
dom variables and random vectors, and corresponding lower-
examines the scenario where all these constraints are active.
case letters to denote their realizations. Moreover, bold letters
The literature on sphere packing focuses on an AVC model are reserved for vectors and calligraphic symbols denote sets.
wherein zero-error probability of decoding is required (or, Random sets are represented by an extra star as superscripts.
equivalently, when the probability (over Alice’s codeword and Some constants are also denoted by capital letters. Our con-
Harry’s jamming actions) of Bob’s decoding error is required vention is summarized in Table II.
toequalzero).InnerandouterboundswereobtainedbyBlach- We use N(a,σ2) to denote for a Gaussian random variable
man [10], [11]. Like several other zero-error communication with mean a and variance σ2. To denote a ball in an n-
problems (including Shannon’s classic work [12]) characteri- dimensional real space of radius r which centered at the point
zation of the optimal throughput possible is challenging, and c Rn, we write (c,r). In Table III, we summarize the
n
in general still an open problem.5 no∈tation used in thisBpaper.
Other related models include:
B. Problem Statement
The vector Gaussian AVC [16]. As in the “usual” vector Inthispaperwestudythecapacityofaquadraticconstrained
•
Gaussian channels, optimal code designs require “water- AVC with stochastic encoder under the attack of a malicious
filling”. adversary who knows the transmitted message but is oblivious
The per-sequence/universal coding schemes in [17]. to the actual transmitted codewords.
•
The correlated/myopic jammers in [18], [19], wherein Let the input and output of the channel are denoted by the
•
jammers obtain a noisy version of Alice’s transmission random variables X and Y where X,Y R. Then, formally,
and base their jamming strategy on this. the channel is defined as follows ∈
The joint source-channel coding, and coding with feed-
• back models considered by Bas¸ar [20], [21]. Y =X+S+V, (1)
SeveralotherAVCvariants,includingdirtypapercoding,
• where S R is the channel state chosen by a malicious
in [22]. ∈
adversary and V N(0,σ2) is Gaussian random variable.
∼
We summarize some of the results mentioned above in HereweassumethatthenoiseV isindependentoverdifferent
Table I. uses of channel (1). The channel input is subjected to a peak
power constraint as follows
n
(cid:88)
4ThisrequirementisananalogueforcommunicationofKerckhoffs’Princi- x 2 = x2 nP, (2)
ple[8]incryptography,whichstatesthatinasecuresystem,everythingabout (cid:107) (cid:107) i ≤
thesystemispublicknowledge,exceptpossiblyAlice’sprivaterandomness. i=1
5TheliteratureonSphericalCodes(see[13],[14],and[15]forsomerela- and the permissible state sequences are those satisfying
tivelyrecentwork)looksattherelatedproblemofpackingunithyperspheres
on the surface of a hypershere. This corresponds to design of codes where (cid:88)n
eachcodewordmeetsthequadraticpowerconstraintwithequality,ratherthan s 2 = s2 nΛ. (3)
allowingforaninequality. (cid:107) (cid:107) i ≤
i=1
ErrorCriterion Capacity
Blachman[10] sup sup [φ(ψ(i)+s+V)(cid:54)=i]≤(cid:15) upperandlowerboundsforthecapacity
s ψ
Hughes&Narayan[2] supsmaxiP(Φ,Ψ),V [Φ(Ψ(i)+s+V)(cid:54)=i]≤(cid:15) C= 21log(1+ Λ+Pσ2)
Csiszar&Narayan[1] sups M1 (cid:80)Mi=1PV [φ(ψ(i)+s+V)(cid:54)=i]≤(cid:15) C=(cid:26) 021log(1+ Λ+Pσ2) iOftPher>wiΛse.
(cid:26) 1log(1+ P ) ifP >Λ
OurSetup supsmaxiPΨ,V [φ(Ψ(i)+s+V)(cid:54)=i]≤(cid:15) C= 02 Λ+σ2 Otherwise.
TableI
COMPARISONOFEXISTINGRESULTSONQUADRATIC-CONSTRAINEDAVCSWITHAWGN.
sn(i) Vn III. MAINRESULTS
The main results of the paper, stated in Theorem 1 and its
i St. Enc. + + Dec. ˆi corollary.
xn(i,t) yn
t Theorem 1. The capacity of a quadratic-constrained AVC
channelunderthemaximumprobabilityoferrorcriterionwith
n� transmit constraint P and jamming constraint Λ and additive
0
Gaussian noise of power σ2 is given by
Figure1. ApowerconstraintAVCwithstochasticencoder.Hereweassume
that the adversary has access to the transmitted message i but not to the (cid:26) 1log(1+ P ) if P >Λ,
transmittedcodewordxn(i,t). C = 2 Λ+σ2
0 Otherwise.
Remark 1. The result of Theorem 1 matches the result of
The problem setup is depicted pictorially in Figure 1. stochastic encoder over discrete alphabets [23], [5, Theo-
A code with stochastic encoder (Ψ,φ) of block-length n rem 7], in which it is shown that for the average probability of
consists of a set of encoders that are denoted by a random errorcriterion,usingastochasticencoderdoesn’tincreasethe
variable Ψ : 1,...,M Rn and a deterministic decoder capacity. Because the number of possible adversarial actions
φ: Rn 0,{...,M w}h(cid:55)→ere 0 denote for an error and M (cid:44) here is uncountably large, the technique of [23], which relies
(cid:55)→{ }
enRisthenumberofmessages6.Eachencoderψisconstructed on taking a union bound over at most exponential-sized set of
by a set of codewords x ,...,x from Rn. possible adversarial actions, does not work.
1 M
{ }
Here in this paper, we focus on the maximum probability
Corollary 1. The capacity of a quadratic-constrained AVC
of error. First, for a fixed jamming vector s, let us define the
underthemaximumprobabilityoferrorcriterionwithtransmit
probability of error given that the message i has been sent as
constraint P and jamming constraint Λ is given by
follows
(cid:26) 1log(1+ P) if P >Λ,
C = 2 Λ
Sunday, January 20, 13 e(s,i)(cid:44)P [φ(Ψ(i)+s+V)=i]. (4) 0 Otherwise.
Ψ,V
(cid:54)
IV. PROOFOFMAINRESULTS
Thenthemaximumprobabilityoferrorforafixedsisdefined
In this section, we present the proof of Theorem 1 and its
by
corollary. The proof of the converse parts of Theorem 1 is
e (s)(cid:44) max e(s,i). (5)
max stated in Section IV-B.
i 1,...,M
∈{ } For the achievability part of Theorem 1, we claim that the
Now the capacity for the above channel can be stated as in sameminimumdistancedecoderproposedin[1]toachievethe
Definition 1. capacityfortheaverageprobabilityoferrorcriterion,whichis
given by
Definition 1. The capacity C of an AVC with stochastic en-
(cid:26)
coder under the quadratic transmit constraint P and jamming φ(y)= i if (cid:107)y−xi(cid:107)2 <(cid:107)y−xj(cid:107)2, forj (cid:54)=i, (6)
constraint Λ is the supremum over the set of real numbers 0 if nosuchi:1 i M exists,
≤ ≤
such that for every δ > 0 and sufficiently large n there exist alsoachievesthecapacityforthemaximumprobabilityoferror
codeswithstochasticencoder(Ψ,φ)thatsatisfiesthefollowing criterion.
conditions.First,forthenumberofmessagesM encodedbythe Note that in order to show the suprimum over s subject to
code we have M >exp(n(C δ)). Moreover, each codeword (3)ofe (s)goestozeroitissufficienttoshowthatforevery
− max
satisfies the quadratic constraint (2) and finally for the code message i the suprimum over s subject to (3) of e(s,i) goes
we have to zero.
lim sup e (s)=0. To communicate, Alice (the transmitter) randomly picks a
max
n→∞s:(cid:107)s(cid:107)2≤nΛ codebook C and fixes it. The codebook C comprises en(δ0+R)
codewords x(i,t), 1 i enR and 1 t enδ0, each
≤ ≤ ≤ ≤
6FornotationalconvenienceweassumethatenR isaninteger. chosen uniformly at random and independently from a sphere
(cid:16) (cid:17)
1 2 en�0 sufficiently large n if R< 1log 1+ P we have
··· 2 σ2+Λ
1 x(1,1) x(1,en�0) (cid:20) (cid:104)
2 PC∗ PTPV (cid:104)X(j,t(cid:48)),X(i,T)+s+V(cid:105)≥nP
(cid:105) (cid:21)
x(i,t) No Error x(i,t00) + X(i,T),s+V forsomej=iandt(cid:48) Ke−nδ1
(cid:104) (cid:105) (cid:54) ≥
.
.. Error (cid:16) (cid:17)
exp (Klog2 10)exp((δ δ )n) .
0 1
≤ − − −
x(j,t0)
Proof: For the proof refer to the appendix.
enR x(enR,1) x(enR,en�0)
Lemma 2 (Quantizing Adversarial Vector). For a fixed jam-
ming vector s, for sufficiently small ε > 0, and for every
Figure2. (a)Thecodebookisconstructedsuchthatforsendingamessage
δ > δ > 0, there exists a codebook = x(i,t) of rate
fir∈om{1th,e..it.h,eronwR}ofththeeeanbcoovdeertacbhleo.o(sbe)sAonsseuomfinthgetheantδt0hecocdoedweworodrsdrxan(id,otm)liys R0 11log(1+ P ) comprises of vecCtors{x(i,t)} Rn of
sent,inourmodelanerroroccursiftheMLdecoderdeclaresx(j,t(cid:48))forsome size≤√2nP with1Λ+σi2 enR and1 t enδ0 whichp∈erforms
j (cid:54)= i. Note that there is no error if the decoder declares another codeword ≤ ≤ ≤ ≤
fromtheithrow. well over the AVC defined in Section II for all s(cid:48) n(s,ε),
∈ B
i.e., it satisfies
(cid:104)
e(s,i)=P P x(j,t),x(i,T)+s+V
T V (cid:48)
of radius √nP as it is shown in Figure 2 (caption (a)). Then, (cid:104) (cid:105)
(cid:105)
the ith row of the codebook, i.e., x(i,1),...,x(i,enδ0) , is nP + x(i,T),s+V forsomej =iandt(cid:48)
{ } ≥ (cid:104) (cid:105) (cid:54)
assigned to the ith message. In order to transmit the message
<Kexp( nδ ) (8)
1
i, the encoder randomly picks a codeword from the ith row of −
the codebook and sends it over the channel. for all s (s,ε).
(cid:48) n
∈B
Now, given that the message i has been transmitted, the Proof:Foraparticulars,insteadof(8),letusassumethat
error probability e(s,i) of an stochastic code used over a the code satisfies a stronger condition
quadratic-constrained AVC under the use of the minimum C
(cid:104)
distance decoder (defined by (6)) equals P P x(j,t),x(i,T)+s+V nP
T V (cid:48)
(cid:104) (cid:105)≥
(cid:105)
2ε√nP + x(i,T),s+V forsomej =iandt
(cid:48)
− (cid:104) (cid:105) (cid:54)
e(s,i)=P [φ(Ψ(i)+s+V)=i]
Ψ,V (cid:104) (cid:54) <Kexp(−nδ1). (9)
=P P x(i,T)+s+V x(j,t) 2
T V (cid:107) − (cid:48) (cid:107) Then it can be verified that for all s (s,ε) the code
(cid:105) (cid:48) ∈ Bn
≤(cid:107)s+V(cid:107)2 for some i(cid:54)=j and t(cid:48) Cs s=atsis+fieρsu(w8)hewreheureissanisarrbeiptrlaarcyedunbityvesc(cid:48)t.oTroansdhρow [thiεs,lεe]t.
(cid:104) (cid:48)
=PTPV (cid:104)x(j,t(cid:48)),x(i,T)+s+V(cid:105)≥nP Hence for all s(cid:48) ∈Bn(s,ε) we can write ∈ −
(cid:105) (cid:104)
+(cid:104)x(i,T),s+V(cid:105)for some j (cid:54)=i and t(cid:48) . (7) e(s(cid:48),i)=PTPV (cid:104)x(j,t(cid:48)),x(i,T)+s(cid:48)+V(cid:105)
(cid:105)
nP + x(i,T),s +V forsomej =iandt
(cid:48) (cid:48)
≥ (cid:104) (cid:105) (cid:54)
where T is a uniformly distributed random variable defined (cid:104)
over the set 1,...,enδ0 . Figure 2 (caption (b)) pictorially =PTPV (cid:104)x(j,t(cid:48)),x(i,T)+s+V(cid:105)+ρ(cid:104)x(j,t(cid:48)),u(cid:105)
{ }
demonstrates the decoding errors at the decoder. nP + x(i,T),s+V
≥ (cid:104) (cid:105)
(cid:105)
+ρ x(i,T),u forsomej =iandt
(cid:48)
(cid:104) (cid:105) (cid:54)
(cid:104)
P P x(j,t),x(i,T)+s+V +ε√nP
A. Achievability proof of Theorem 1 ≤ T V (cid:104) (cid:48) (cid:105)
nP + x(i,T),s+V
≥ (cid:104) (cid:105)
(cid:105)
ThemainstepinprovingtheachievabilitypartofTheorem1 ε√nP forsomej =iandt
(cid:48)
consistsinassertingthedoublyexponentialprobabilitybounds − (cid:54)
(a)
which is stated in Lemma 1. Kexp( nδ ),
1
≤ −
Lemma 1. Let = X(i,t) in which 1 i exp(nR) where (a) follows from (9).
∗
C { } ≤ ≤
and 1 t exp(nδ ) be a random codebook comprises of Now,inLemma1wecanusethestrongererrorrequirement
0
≤ ≤
independentrandomvectorsX(i,t)eachuniformlydistributed (9) to show that there exists a code which satisfies (9). This
onthen-dimensionalsphereofradius√nP.First,fixavector strongerrequirementresultsinarateloss,butasεgoestozero
s (0,√nΛ). Then for every δ > δ > 0 and for the rate loss due to that vanishes. By the above argument, we
n 0 1
∈ B
know that this code satisfies (8) for all s (s,ε) and we
(cid:48) n
∈ B
are done.
Finally, Lemma 3 shows the existence of a good codebook
for the quadratic constrained AVC problem with stochastic
encoderwhichhavebeenintroducedinSectionII-Bandhence
completes the proof of Theorem 1. " pn⇤
Lemma 3 (Codebook Existence). For every δ >δ >0 and
0 1
n n (δ ,δ ) there exist a codebook = x(i,t) of rate
0 0 1
R≥ 1log(1+ P ) comprises of vecCtors{x(i,t)} Rn of
≤ 2 σ2+Λ ∈
size √nP with 1 i enR and 1 t enδ0 such that for
≤ ≤ ≤ ≤
every vector s and every transmitted message i we have
(cid:104)
e(s,i)=P P x(j,t),x(i,T)+s+V
T V (cid:48)
(cid:104) (cid:105) √
(cid:105) Figure 3. This figure shows that how the whole sphere B(0, Λ) can be
≥nP +(cid:104)x(i,T),s+V(cid:105)forsomej (cid:54)=iandt(cid:48) coveredbyε-densesubsetsχn.Herethesetχn comprisesofpointsfroma
hyper-cubicallattice.
<Kexp( nδ ). (10)
1
−
Proof: For any fixed codebook = x(i,t) , let us the following upper bound on the capacity of an AVC with
C { }
explicitly mention to the dependency of the error probability stochastic encoder
on by defining e (s,i)(cid:44)e(s,i). Then in order to prove the (cid:18) (cid:19)
1 P
asseCrtion of lemmaCwe can equivalently show that C log 1+ .
≤ 2 Λ+σ2
liminfP ∗(cid:2) s, i e ∗(s,i)<Ke−nδ1(cid:3)>0. Now, it only remains to show that C =0 for P Λ where
n C ∀ ∀ C ≤
→∞ we use a similar argument to [7] (also see [1]). To this end,
However, by using Lemma 2, it is not necessary to check for
we show that the adversary can fool the decoder and make it
all s but only for those belonging to an ε-net7 χn that covers confused. Because P Λ, the adversary can use a stochastic
Bn(0,√nΛ). encoder Ψ(cid:48) with the≤same probabilistic characteristic of Ψ
Hence, we can write where we assume that Ψ and Ψ are independent9. Then for
(cid:48)
P ∗(cid:2) s χn, i e ∗(s,i)<Ke−nδ1(cid:3) any decoder φ and for any i(cid:54)=j we can write
C ∀ =(∈a)11−∀P(cid:88)C∗(cid:2)C(cid:88)∃ensR∈Pχ∗n(cid:2),e∃i∗(esC,∗i)(s,iK)≥e−Knδe1−(cid:3)n,δ1(cid:3) P[φ(Ψ(i)+Ψ(cid:48)(j)==+P1V[)φ(cid:54)=P(Ψ[iφ(]j()Ψ+(jΨ)+(cid:48)(iΨ)+(cid:48)(iV)+)(cid:54)=Vi)]=i]
≥ − C C ≥ −
s∈χni=1 1 P[φ(Ψ(j)+Ψ(cid:48)(i)+V)=j].
≥ − (cid:54)
where (a) follows from the union bound.
Hence we have
Now, note that to bound χ one might cover (0,√nΛ)
by a hypercube of edge size| 2n√| nΛ; see Figure 3B. nSo we can 1 (cid:88)M 1 (cid:88)M
write χ (cid:16)2√nΛ(cid:17)n. Then, by using Lemma 1 we have M emax(Ψ(cid:48)(j))≥ M2 e(Ψ(cid:48)(j),i)
| n|≤ ε j=1 i,j=1
M
PC∗(cid:2)∀s∈χn,∀(cid:32)i eC∗(s(cid:33),in)<Ke−nδ1(cid:3) = M12 (cid:88) P[φ(Ψ(i)+Ψ(cid:48)(j)+V)(cid:54)=i]
2√nΛ (cid:16) (cid:17) i,j=1
≥1− ε ×enR×exp −K(cid:48)en(δ0−δ1) , 1 (cid:88)M (cid:104)
P[φ(Ψ(i)+Ψ(j)+V)=i]
(cid:48)
≥ M2 (cid:54)
where, assuming δ > δ , the right hand side goes to 1 as n i,j=1
0 1
(cid:105)
goes to infinity and this completes the proof of lemma. +P[φ(Ψ(j)+Ψ(i)+V)=j]
(cid:48)
(cid:54)
B. Converse proof of Theorem 1 1 M(M 1)
−
The converse of Theorem 1 follows by combining two ≥ M2 2
1
differentupperboundsonthecapacity.Thefirstboundfollows ,
by observing that if the randomness of the stochastic encoder ≥ 4
is also shared with the decoder we can achieve higher rates. where M =enR. This shows that
So by using result of [2] for randomized codes8, we have M
1 (cid:88) 1
E[e (Ψ(j))] ,
max (cid:48)
M ≥ 4
7An ε-net is a set of points in a metric space such that each point of the j=1
spaceiswithindistanceεofsomepointintheset.
8Similartoourwork,[2]alsoconsidersthemaximumprobabilityoferror 9 Such a jamming strategy is equivalent to the notion of symmetrizability
criterion. conditionintheAVCliterature(see,forinstance[3],[4],and[5]).
which means there exists at least a k such that proceed as follows
E[e (Ψ(k))] 1 and this completes the proof.
max (cid:48) ≥ 4 E ∗[ft(X(i,1),...,X(i,t))X(i,1),...,X(i,t 1)]
APPENDIX C=E ∗(cid:20)PV(cid:104) X(j,t(cid:48)),X(i,|t)+s+V −
C (cid:104) (cid:105)
Fact 1. For two events and we can write
A B P + X(i,t),s + X(i,t),V
P[ ]=P[ ( ¯)] P[ ]+P[ ¯]. ≥ (cid:104) (cid:105) (cid:105)(cid:104)(cid:12)(cid:12) (cid:105) (cid:21)
A A∩ B∪B ≤ B A∩B for some j =i and t(cid:48) (cid:12)X(i,1),...,X(i,t 1)
(cid:54) (cid:12) −
lemOmura”prporoofverneqiunir[e1s,tLheemfomllaowAi1n]g. “martingale concentration =(a)PV(cid:20)P ∗(cid:104) (cid:91) (cid:8) X(j,t(cid:48)),X(i,t)+s+V
C (cid:104) (cid:105)
aLnedmfm(aX4(,[.1.,.L,eXmm)abAe1a]r)b.iLtreatrXy 1fu,n..c.ti,oXnLwbiteha0rbitrafryr.v.1’s, (j,t(cid:48)):j(cid:54)=i (cid:9)(cid:105)(cid:21)
i 1 L i P + X(i,t),s+V
≤ ≤
i=1,...,L. Then the condition ≥ (cid:104) (cid:105)
E[f (X ,...,X )X ,...,X ] a a.s., i=1,...,L, (b)PVP ∗(cid:2) X(i,t),s+V δ2(cid:3)
impliies t1hat L | 1 i−1 ≤ ≤+PVC(cid:20)P(cid:104)∗(cid:104) (cid:91) (cid:8) X(cid:105)≤(j,−t(cid:48)),X(i,t)+s+V
C (cid:104) (cid:105)
(j,t(cid:48)):j=i
P(cid:34)1 (cid:88)L f (X ,...,X )>τ(cid:35) exp( L(τlog2 a)). P + X(i,t),s(cid:54)+V (cid:9), X(i,t),s+V > δ (cid:105)(cid:21),
i 1 i 2
L ≤ − − ≥ (cid:104) (cid:105) (cid:104) (cid:105) −
i=1
(12)
Lemma 5 ([1, Lemma 2]). Let the random vector U be
where (a) follows because X(i,t) are independent random
uniformly distributed on the n-dimensional unit sphere. Then
variables so the conditioning can be removed and also using
for every vector u on this sphere and any 1 <α<1, we
have √2πn tahnedf(abc)tftohlaltowfosrfarnomevFeancttA1.we have EC∗PV[A] = PC∗PV[A]
P[|(cid:104)U,u(cid:105)|≥α]≤2(1−α2)(n−21). Now, for δ2 > 0, by using Fact 1 we can bound the first
term of (12) as follows
Proof of Lemma 1: For notational convenience let us
normalizeallvectorss,V,andX(i,t)by1/√ninthisproof. PVP ∗(cid:2) X(i,t),s+V δ2(cid:3)
Toderivethedoublyexponentialboundstatedinthelemma, PC(cid:2)(cid:104)s+V 2 s(cid:105)2≤+−σ2+δ (cid:3)
1we≤utse≤Leenmδ0maas4f.oTlloowthsisendletusdefinethefunctionsft for ≤+VPV(cid:107)PC∗(cid:2)(cid:104)X(cid:107)s(+i≥,tV)(cid:107),s(cid:107)2+<V(cid:105)s≤2+−2δσ22, +δ (cid:3)
2
ft(X(i,1),...,X(i,t)) P (cid:2) s+V(cid:107)2 s(cid:107) 2+(cid:107)σ2(cid:107)+δ (cid:3)
(cid:44)PV(cid:104)(cid:104)X(j,t(cid:48)),X(i,t)+s+V(cid:105) ≤+VPV(cid:107)P ∗(cid:2) X(cid:107) (≥i,t(cid:107)),(cid:107)s+V δ22,
≥P +(cid:104)X(i,t),s+V(cid:105)for some j (cid:54)=i and t(cid:48)(cid:105). C |(cid:104)(cid:107)s+V(cid:107)2 <(cid:107)s(cid:105)(cid:107)|2≥+σ2+δ2(cid:3). (13)
Firstnotethat s+V 2 = s 2+ V 2+2 s,V .Then,since
Now, by using the functions ft, the probability expression in (cid:107) (cid:107) (cid:107) (cid:107) (cid:107) (cid:107) (cid:104) (cid:105)
V = (V ,...,V ) is a sequence of i.i.d. Gaussian random
the statement of lemma can be written as follows 1 n
variables N(0,σ2), the first term of (13) can be bounded as
(cid:20) (cid:104) n
P ∗ PTPV X(j,t(cid:48)),X(i,T)+s+V P follows
C (cid:104) (cid:105)≥
(cid:105) (cid:21) PV(cid:2) s+V 2 > s 2+σ2+δ2(cid:3)
+(cid:104)X(i,T),s+V(cid:105)forsomej(cid:54)=iandt(cid:48) ≥Ke−nδ1 (cid:107)=P(cid:2) V(cid:107) 2+(cid:107)2(cid:107)s,V >σ2+δ2(cid:3)
(cid:107) (cid:107) (cid:104) (cid:105)
=PC∗(cid:20)en1δ0 e(cid:88)tn=δ10PV(cid:104)(cid:104)X(j,t(cid:48)),X(i,t)+s+V(cid:105)≥P ≤((=ba))PP[(cid:20)(cid:104)su,V,V(cid:105)≥η]+η P(cid:21)[(cid:107)+VP(cid:107)(cid:2)2+V2η2 >>σσ22++δδ2] 2η(cid:3),
(cid:105) (cid:21) (cid:104) (cid:105)≥ s (cid:107) (cid:107) 2−
+ X(i,t),s+V forsomej=iandt(cid:48) Ke−nδ1 (cid:107) (cid:107) (14)
(cid:104) (cid:105) (cid:54) ≥
(cid:20) enδ0 (cid:21) where (a) follows from Fact 1 for η >0 and in (b) we define
1 (cid:88)
=PC∗ enδ0 t=1 ft(X(i,1),...,X(i,t))≥Ke−nδ1 . uto=shsow/(cid:107)sth(cid:107)a.tBeuc,aVuse uNis(a0,unσi2t)a.ryHveenccteorthiteifisrssttrateigrmhtfoinrw(1a4rd)
(11) (cid:104) (cid:105)∼ n
can be bounded as follows
In order to bound (11) we use Lemma 4. To this end, we (cid:18)√nη(cid:19) 1 (cid:18) η2n (cid:19)
P [ s,V η]=Q exp , (15)
have to bound the expected values of the functions ft. So we V (cid:104) (cid:105)≥ σ s ≤ 2 −2σ2Λ
(cid:107) (cid:107)
where in the above equation we have used the approximation (13), (14), (15), (16), and (17) we can bound the first term in
Q(y)≤ 12e−y22.Inordertoboundthesecondtermin(14)note (12) as follows
that n V 2 has the Chi-squared distribution with n degree
of freσe2d(cid:107)om(cid:107). Then by using [24, Lemma 1] we can bound the PVP ∗[ X(i,t),s+V δ2]
C (cid:104) (cid:105)≤− ≤
secPoVnd(cid:20)tσenr2m(cid:107)Vof(cid:107)2(1>4)(cid:18)a1s+folδlo2wσ−s22η(cid:19)n(cid:21) 2exp(cid:18)−n−2 1(cid:107)s(cid:107)2δ+22σ/P2+δ2(cid:19)+e−nξ+ 12e−2nση22Λ(.18)
(cid:32) (cid:34) (cid:114) (cid:35) (cid:33)
1 δ 2η δ 2η
exp 1+ 2− 1+2 2− n Now we bound the second term in (12) as follows. Suppose
≤ −2 σ2 − σ2 denotes for the event X(i,t),s+V > δ and let
2
A {(cid:104) (cid:105) − }
=exp( ξn), (16) φ = X(i,t),s+V . Then for the second term of (12), we
− note t(cid:104)hat (cid:105)
(cid:20) (cid:113) (cid:21)
twithyerifeδξ2=>122η.1+ δ2σ−22η − 1+2δ2σ−22η is a positive quan- PVC∗(cid:20) (cid:91) (cid:8)(cid:104)X(j,t(cid:48)),X(i,t)+s+V(cid:105)≥P +φ(cid:9),A(cid:21)
(j,t(cid:48)):j=i
Remark2. Notethatbecause√1+2x 1+x x2 forallx (cid:54)
[0,1] then by choosing x = δ2σ−22η we≤have ξ−≥8116(δ2σ−22η)∈2 (≤a)PV (cid:2)(cid:20)(cid:107)s+V(cid:107)2 ≥(cid:107)s(cid:107)2+σ2+δ2(cid:3) (cid:21)
and exp(−nξ)≤exp(−116(δ2σ−22η)2). +PVC∗ (cid:91) (cid:8)(cid:104)X(j,t(cid:48)),X(i,t)+s+V(cid:105)≥P +φ(cid:9),A,B
Now it remains to bound the second term of (13). To this (j,t(cid:48)):
j=i
end let us write (cid:54)
PVP ∗(cid:2) X(i,t),s+V δ2, s+V 2 < s 2+σ2+δ2(cid:3) (≤b)e−nξ+ 21e−2nση22Λ
ˆC |(cid:104) (cid:105)|≥ (cid:107) (cid:107) (cid:107) (cid:107) (cid:88)
= 0(cid:107)s(cid:107)2+σ2+δ2PVPC∗(cid:104)|(cid:104)X(i,t),s+V(cid:105)|>δ2(cid:12)(cid:12)(cid:12) +(j,t(cid:48)):j(cid:54)=iPVC∗[(cid:104)X(j,t(cid:48)),X(i,t)+s+V(cid:105)≥P +φ,A,B],
(cid:105)
s+V 2 =r dF(r) where (a) follows from Fact 1 and we use to denote the
(cid:107) (cid:107) B
event s + V 2 < s 2 + σ2 + δ . The first two terms
where F(r)=P(cid:2) s+V 2 r(cid:3). Then we can write {(cid:107) (cid:107) (cid:107) (cid:107) 2}
in (b) follow from (14), (15), and (16) while the third term
(cid:107) (cid:107) ≤
PVPˆC∗(cid:2)|(cid:104)X(i,t),s+V(cid:105)|≥δ2,(cid:107)s+V(cid:107)2 <(cid:107)s(cid:107)2+σ2+δ2(cid:3) iX˘s (aj,rets)ul=t ofXth(ej,tu(cid:48))nioanndboUund=. LeXt(ui,st)+des+finVe .thTehuennitwveecntoortes
= 0(cid:107)s(cid:107)2+σ2+δ3PVPC∗(cid:104)(cid:104)X(i,t),(cid:107)ss++(cid:107)sVV+(cid:107)(cid:105)V>(cid:107)2√δ=2r(cid:12)(cid:12)(cid:12)r(cid:105)dF(r) thPaVtC∗(cid:20)(cid:48) (cid:91)(cid:107)X(j(cid:8),t(cid:48)(cid:104))X(cid:107) (j,t(cid:48)),X(i,(cid:107)tX)(+i,ts)++s+VV(cid:105)(cid:107)≥P +φ(cid:9),A(cid:21)
(cid:34) (cid:35) (j,t(cid:48)):j=i
(a) δ /√P (cid:54)
≤PUPC˘∗ (cid:104)X˘(i,t),U(cid:105)> (cid:112)(cid:107)s(cid:107)22+σ2+δ2 ≤(a)e−nξ+ 21e−2nση22Λ
(cid:20)
whereU = s+V ,X˘(i,t)= X(i,t) ,and(a)istruebecause + (cid:88) P X˘(j,t),U
evaluating th(cid:107)es+tVer(cid:107)m inside the(cid:107)iXnt(ei,gt)r(cid:107)ation for the point r = UC˘∗ (cid:104) (cid:48) (cid:105)
(j,t(cid:48)):j=i
s 2+σ2+δ2 can only increase the probability term. Next, (cid:54) P +φ (cid:12) (cid:21)
i(cid:107)t f(cid:107)ollows that ≥ √P(cid:112)P + s+V 2+2φ(cid:12)(cid:12)A,B
PVPˆC∗(cid:2)|(cid:104)X(cid:34) (i,t),s+V(cid:105)|≥δ2,(cid:107)s+V(cid:107)2 <(cid:107)s(cid:107)2+(cid:35)σ2+δ2(cid:3) (b)e−nξ+ 1e−2nση22Λ (cid:107) (cid:107)
δ /√P (cid:12) ≤ 2
= PC˘∗ (cid:104)X˘(i,t),u(cid:105)> (cid:112)(cid:107)s(cid:107)22+σ2+δ2(cid:12)(cid:12)U =u fU(u)du + (cid:88) PU˘∗(cid:20)(cid:104)X˘(j,t(cid:48)),U(cid:105)
ˆ C
≤(a) 2(cid:18)1− s 2δ+22σ/P2+δ (cid:19)n−21 fU(u)du (j,t(cid:48)):j(cid:54)=i P −δ2 (cid:21)
(cid:107) (cid:107) 2 ≥ √P√P +Λ+σ2+δ 2δ
(cid:18) δ 2/P (cid:19)n−21 2− 2
2
=2 1− s 2+σ2+δ where in (a) we use the fact that P[ , , ] P[ , ] and
2 E A B ≤ E|A B
(b)2exp(cid:18)(cid:107)n(cid:107)−1 δ22/P (cid:19) (17) a(bn)dfφoll=owsδb2ectahueseprboybasbuiblistytitutetrinmgi(cid:107)nsf+ronVt(cid:107)o2f=theΛs+umσm2a+tioδn2
≤ − 2 s 2+σ2+δ −
2 in (a) can only increase; this implies that we can remove the
(cid:107) (cid:107)
where (a) follows from Lemma 5 and (b) follows from the conditioningwithrespecttoevents and .Now,byapplying
A B
inequality 1 x e x for 0<x<1. Finally, by combining Lemma 5, we can further bound the second term of (12) as
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+ X(i,t),s+V forsomej=iandt(cid:48)(cid:3)(cid:105) Ke−nδ1(cid:21)
(cid:104) (cid:105) (cid:54) ≥
(cid:16) (cid:16) (cid:17)(cid:17)
exp exp(nδ ) Klog2exp( nδ ) 10exp( nδ )
0 1 1
≤ − − − −
(cid:16) (cid:17)
=exp (Klog2 10)exp(n(δ δ )) .
0 1
− − −
By assuming δ > δ > 0 we obtain the desired doubly
0 1
exponential bound, hence we are done.
