1

Optimal Performance for DS-CDMA Systems withHard Decision Parallel Interference Cancellation

Remco van der Hofstad, Marten J. Klok1

Abstract—We study a multiuser detection system using codedivision multiple access (CDMA). We show that applyingmultistage hard decision parallel interference cancellation(HD-PIC) significantly improves performance compared tothe matched filter system. In (multistage) HD-PIC, esti-mates of the interfering signals due to other users are usediteratively to improve knowledge of the desired signal. Weuse large deviation theory to show that the bit-error proba-bility (BEP) is exponentially small and investigate the expo-nential rate of the BEP after several stages of HD-PIC. Wepropose to use the exponential rate of the BEP as a mea-sure of performance, rather than taking the signal-to-noiseratio, which is not reliable in multiuser detection models.

We show that the exponential rate of the BEP remains fixedafter a finite number of stages, resulting in an optimal harddecision system. When the number of users becomes large,the exponential rate of the BEP converges to (log 2)/2−1/4.We provide intuition concerning the number of stages neces-sary to obtain this asymptotic exponential rate. Finally, wegive Chernoff bounds on the BEP’s. These estimates showthat the BEP’s are quite small as long as k = o(n/ logn)when the number of stages of HD-PIC is fixed, and even ex-ponentially small when k = O(n) for the optimal HD-PICsystem.

Keywords—Code division multiple access, hard-decision par-allel interference cancellation, large deviations, exponentialrate, performance measures, optimal hard-decision system.

I. Introduction

CURRENTLY, the third generation (3G) mobile com-munication system is being introduced. This system is

based on code division multiple access (CDMA). The per-formance of CDMA is mainly limited by interference fromother users: the multiple access interference. Particularly,the susceptibility to the near-far situation, which can sig-nificantly reduce capacity in the absence of good powercontrol, is a problem. Therefore, there exists a great in-terest in techniques which improve the capacity of CDMAreceivers (see [15], [2], [16] and the references therein). Ini-tial research on multi-user receivers for CDMA has demon-strated the potential improvements in capacity and near-far resistance. The best known technique is a maximumlikelihood sequence estimator [19], which obtains jointlyoptimum decisions for all users using maximum likelihooddetection. Unfortunately, this technique is of high com-plexity. A straightforward technique is called interferencecancellation, see [16], Chapter 4, [14], [9], [18] and thereferences therein. The idea is that we try to cancel the

Faculty of Information Technology and Systems, Delft Univer-sity of Technology, Mekelweg 4, 2628 CD Delft, the Nether-lands, tel : +31-15-2789215, fax: +31-15-2787255, email:r.w.vanderhofstad@its.tudelft.nl, m.j.klok@its.tudelft.nl

interference due to the other users. Interference cancel-lation, especially hard-decision parallel interference can-cellation (HD-PIC), is the most promising and the mostpractical technique for uplink receivers (see [16]). This isbecause in WCDMA proposals all user signals are demod-ulated coherently, which makes implementation efficient.Apart from that, as we will see in this paper, HD-PICsignificantly improves performance.

In the literature, not much attention has been given to ob-tain rigorous analytical results for (HD-)PIC systems. Thepapers that exist on this subject focus on approximatingthe signal-to-noise ratio (SNR), for example [1] and [20].However, using the SNR as a measure of performance bysubstituting the SNR in the Gaussian error function im-plicitly relies on Gaussian assumptions. For PIC systems,this assumption is false and it leads to incorrect results aswe will argue in Section II-E.

We use large deviation theory [3], [8] to obtain analyticalresults for the bit-error probability (BEP). More precisely,rather than calculating the SNR, we calculate the expo-nential rate of the BEP. We will first explain the essenceof this rate. Suppose pn is a sequence of probabilities andpn → 0 as n→∞. Often, this decay is exponentially in n,so that we investigate the exponential rate defined as

I = − limn→∞

1

nlog pn.

For finite, but not too small n, we can read this as

pn ∼ e−nI . (1)

Thus, the probability pn is mainly characterized by its ex-ponential rate. We stress that the exponential rate of theBEP is not a probability.

It turns out that the BEP for matched filter (MF) and HD-PIC systems have the same structure as pn. This makes itpossible to use the exponential rate as a measure for theperformance.

For the MF model, the SNR and the exponential rate areasymptotically equivalent, meaning that when the numberof users is sufficiently large, both substituting the SNR intothe Gaussian distribution and the use of the exponentialrate imply that the BEP has the same asymptotic form.This is shown in Section II-E. In [17], a MF model is in-vestigated using large deviation techniques. However, theemphasis lies on sampling techniques. For a one-stage soft-decision PIC model, results have been obtained in [6]. Inthat paper, large deviation techniques are used to proveproperties of the exponential rate and the BEP. In [7] and

2

[11], various results have been proven for the HD-PIC sys-tem.

In this paper, we start by repeating results of one-stageand multi-stage HD-PIC of [7] and [11]. We extend theseresults to values of the processing gain and the number ofusers that are comparable using Chernoff bounds.

The main novelty of this paper is that we obtain resultsfor systems with infinite stages of HD-PIC, the so-calledoptimal HD-PIC system. One of the striking results is thatafter finitely many iterations, the exponential rate does notincrease any further. In other words, applying more stagesof interference cancellation does not improve performanceanymore. Another striking result is that the obtained ex-ponential rate is strictly positive, regardless of the numberof users. We will give bounds on this exponential rate.Furthermore, we give intuition on the number of stagesnecessary to obtain this exponential rate.

II. System and background

In this section, we explain the system, and give previousresults concerning HD-PIC.

A. System model

We first introduce a mathematical model for CDMA sys-tems. We define the data signal bm(t) of the mth useras bm(t) = bm bt/Tc, for 1 ≤ m ≤ k, where bm =

(. . . , bm,−1, bm0, bm1, . . .) ∈ {−1,+1}Z and where for x ∈R, bxc denotes the largest integer smaller than or equalto x. For each user m, 1 ≤ m ≤ k, we have a coding se-quence am = (. . . , am,−1, am0, am1, . . .) ∈ {−1,+1}Z andwe put am(t), the rectangular spreading signal, generatedby (am)∞m=−∞, i.e., am(t) = am bt/Tcc, where Tc = T/n,for some integer n. The variable n is often called the pro-cessing gain.

The transmitted coded signal of the mth user is then

sm(t) =√

2Pm bm(t)am(t) cos(ωct), 1 ≤ m ≤ k, (2)

where Pm is the power of the mth user and ωc the carrierfrequency. The total transmitted signal is given by

r(t) =k∑

j=1

sj(t). (3)

In practice, the signals do not need to be synchronized, i.e.,it is not necessary that all users transmit using the sametime grid. However, for technical reasons we do assumeso. In [17], an asynchronous MF system is investigated,using large deviations techniques. One important resultin that paper is that the asynchronous system has muchsimilarities, compared to the synchronous system. In fact,the exponential rate for the synchronous and asynchronoussystem are the same. For the more advanced HD-PIC sys-tem, this idea carries over, so that the results investigatedin this paper turn out not to differ from the results for an

asynchronous system. Indeed, we expect that the asyn-chronous model has the same exponential rate of the BEPas the synchronized model. Therefore, the synchronicityassumption is without loss of generality.

We assume that there is no additive white Gaussian noise(AWGN). Thus, we are dealing with a perfect channel.However, we believe that many of our results remain truewhen there is little AWGN.

To retrieve the data bit bm0, the signal r(t) is multiplied byam(t) cosωct and then averaged over [0, T ]. For simplicity,we pick ωcTc = πfc, where fc ∈ N to get (cf. [6], Eqn.(3)-(4))

1

T

∫ T

0

r(t)am(t) cos(ωct) dt (4)

=

√

Pm2bm0 +

k∑

j=1j 6=m

√

Pj2bj0

1

n

n−1∑

i=0

ajiami.

As is seen from (4), the decoded signal consists of thedesired bit and interference due to the other users. Inthe ideal situation the vectors (am0, . . . , am,n−1) and(aj0, . . . , aj,n−1), j 6= m, would be orthogonal, so that∑n

i=1 ajiami = 0. In practice, the a-sequences are gen-erated by a random number generator. To model thepseudo-random sequence a, let Ami, m = 1, 2, . . . , k,i = 1, 2, . . . , n, be an array of independent and identicallydistributed (i.i.d.) random variables with distribution

P(A11 = +1) = P(A11 = −1) = 1/2. (5)

Assuming the coding sequences to be random, we modelthe signal of (4) as

√

Pm2bm0 +

k∑

j=1j 6=m

√

Pj2bj0

1

n

n∑

i=1

AjiAmi,

where we have replaced i = 0, . . . , n − 1 by i = 1, . . . , n,for notational convenience.

In practice, the a-sequences are not chosen as i.i.d. se-quences. Rather, they are carefully chosen to have goodcorrelation properties. Examples are Gold sequences [4] orKasami sequences [10]. However, it is common in the liter-ature to use random sequences, so that a detailed analysisis possible. Better performance can be achieved for well-chosen deterministic codes.

We let b(1)m0 be the estimator for bm0 given by

sgnrm

{

√

Pm2bm0 +

k∑

j=1j 6=m

√

Pj2bj0

1

n

n∑

i=1

AjiAmi

}

,

where, for x ∈ R, the randomized sign-function is definedas

sgnrm(x) =

+1, x > 0,Um, x = 0,−1, x < 0.

(6)

3

with P(Um = −1) = P(Um = +1) = 1/2. The random vari-ables Um are independent of all other random variables inthe system. Note that the mth user always makes the samedecision when its signal produces a zero. There are otherways to define the sign-function, such as the choice whereevery time when a zero is detected, a new independent ran-dom U is chosen. Another option is to let the sign of 0 beequal to 0. We choose the definition in (6) for technicalreasons only. We will comment more on other choices ofsign functions in Section III below. We use the notation(1) in b(1)m0 to indicate that this is a tentative decision only.

We now describe the hard-decision procedure. The powersPm are assumed to be known. We estimate the data signalsj(t) for t ∈ [0, T ] by (recall (2))

s(1)

j (t) =√

2Pj b(1)

j0 aj(t) cos(ωct).

Then we estimate the total interference for the mth user inr(t) due to the other users by (recall (3))

r(1)

m (t) =

k∑

j=1j 6=m

s(1)

j (t)

We use the above to find a better estimate of the data bitbm0, denoted by b(2)m0, which is the sgnrm of

1

T

∫ T

0

(r(t)− r(1)

m (t))am(t) cos(ωct)dt (7)

=

√

Pm2bm0 +

k∑

j=1j 6=m

(

1

n

n∑

i=1

AjiAmi

)

√

Pj2

(

bj0 − b(1)j0

)

.

We are now interested in P(b(2)m0 6= bm0), which is the prob-ability of a bit error after one stage of interference cancel-lation. We will see that this probability is indeed smallerthan P(b(1)m0 6= bm0), the probability of a bit error withoutcancellation. This motivates a repetition of the previousprocedure. We obtain, similar to (7), the estimates b(s)m0,which are the sgnr of√

Pm2bm0 +

k∑

j=1j 6=m

(

1

n

n∑

i=1

AjiAmi

)

√

Pj2

(

bj0 − b(s−1)

j0

)

.

This is called multistage HD-PIC. When we have applieds steps of interference cancellation we speak of s-stageHD-PIC and the corresponding bit error probability isP(b(s+1)

m0 6= bm0).

B. Reformulation of the problem

We can write the probability of a bit error in a more con-venient way. Namely, because b2mi = 1, we have

√

Pm2bm0 +

k∑

j=1j 6=m

√

Pj2bj0

1

n

n∑

i=1

AjiAmi

= bm0

(

√

Pm2

+

k∑

j=1j 6=m

√

Pj2

1

n

n∑

i=1

bj0Ajibm0Ami

)

.

Since Ajid= bj0Aji we have

P(b(1)m0 6= bm0) = P

(

b(1)m0

bm06= 1

)

= P(sgnrm(Z(1)

m ) 6= 1) = P(Z(1)

m < 0) +1

2P(Z(1)

m = 0),

where Z(1)m , for 1 ≤ m ≤ k, is defined as

Z(1)

m =√

Pm +

k∑

j=1j 6=m

√

Pj1

n

n∑

i=1

AjiAmi.

We can easily bound

P(Z(1)

m < 0) ≤ P(b(1)m0 6= bm0) ≤ P(Z(1)

m ≤ 0). (8)

Note that the above bound is completely independent ofthe choice of the sign-function in (6). The upper and lowerbound in (8) can be shown to be almost equal when k ≥ 3and n is large.

Similarly, we define for s ≥ 2 and 1 ≤ m ≤ k,

Z(s)m =

√

Pm +

k∑

j=1j 6=m

√

Pj

(

1

n

n∑

i=1

AjiAmi

)

(9)

×[1− sgnrj(Z(s−1)

j )],

to obtain as in (8)

P(Z(s)m < 0) ≤ P(b(s)m0 6= bm0) ≤ P(Z(s)

m ≤ 0). (10)

We will now investigate the performance of HD-PIC. Wewill use as a measure of performance the exponential rateof a bit error, defined by

H (s)k;m = − lim

n→∞1

nlogP(b(s)m 6= bm). (11)

For systems with equal powers, the users are exchangeable,so that the above quantity does not depend on m. In thatcase, we define H (s)

k = H (s)k;m for all m.

We will often use the system with equal powers as a ref-erence model. We will denote this model by the simplesystem. Without loss of generality, we then take Pm = 2,so that the factors

√

Pm/2 disappear.

C. Previous results for HD-PIC

In [6], it is shown that the exponential rate without in-terference cancellation H (1)

k , (denoted there by Ik), for thesimple system for k ≥ 3, is given by 1

H (1)

k =k − 2

2log

(

k − 2

k − 1

)

+k

2log

(

k

k − 1

)

. (12)

In [7], we have obtained an analytic formula for the rateafter one stage of HD-PIC.

1A different definition of the sign function is used there, but thisdoes not influence the results.

4

Theorem II.1: When all powers are equal

H (2)

k = min1≤r≤k−1

H (2)

k,r, (13)

whereH (2)

k,r = sups,t{− log h(s, t)}, (14)

with h(s, t) = hk,r(s, t) equal to

2−rr∑

j=−r

(

rr+j2

)

es+2sj+tj+tj2(cosh tj)k−r−1.

In the statement of the theorem, one should not confuseH (2)

k,r with H (2)

k;m in (11).

We see that the problem is split into two optimizationproblems. First, r represents the number of bits that havebeen estimated wrongly in the first stage. For every r, wesolve the large deviation minimization problem. Then weminimize over r to obtain the rate H (2)

k .

We can give an analytical expression for H (2)

k,1, similar tothe rate for the MF system, that reads

H (2)

k,1 =3

4log 3− log 2 (15)

+2k − 5

4log(2k − 5

2k − 4

)

+2k − 3

4log(2k − 3

2k − 4

)

For general r, we cannot obtain a closed form expressionforH (2)

k,r. However, standard numerical packages allow us to

compute H (2)

k,r for all k and r. In Figure 1, H (2)

k,r is shown forr = 1, . . . , 5. Observe that the optimal r equals 1 for k ≤ 9.In probability language it means that the BEP, caused by2 or more bit errors in the first stage is negligible comparedto the BEP caused by 1 bit error in the first stage. Fork = 10, . . . , 25, r = 2 will give the minimal rate, meaningthat “typically” a bit error in the second stage (after HD-PIC) is caused by 2 bit errors in the first stage. This isfurther illustrated in Table I, where the optimal rk is givenfor k = 3, . . . , 250.

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

number of users k

Fig. 1. Exponential rates H(2)

k,rfor r = 1, . . . , 5 indicated with

◦, 4, ¦,×, ? respectively.

For 1/k → 0 a Taylor expansion of (12) yields

H (1)

k =1

2k+O

(

1

k2

)

.

k rk{3,. . . ,9} 1{10,. . . ,26} 2{27,. . . ,51} 3{52,. . . ,84} 4{85,. . . ,125} 5{126,. . . ,174} 6{175,. . . ,231} 7{232,. . . ,250} 8

TABLE I

Optimal value rk for k = 3, . . . , 250.

In fact,1

2k≤ H (1)

k ≤1

2k+O

(

1

k2

)

. (16)

In [7], it is shown that for the simple system, we have thatthe rate has an asymptotic scaling:

Theorem II.2: Fix 1 ≤ s <∞ and assume that the powersare all equal. Then, as k →∞,

H (s)k =

s

8s

√

4

k

(

1 +O(

1s√k

))

. (17)

However, these results are asymptotic only. In Figure 2,we see that the approximation becomes worse when s in-creases.

5 10 15 20 25 30 35 40 45 500

0.05

0.1

0.15

0.2

0.25

0.3

number of users k

Fig. 2. Exponential rates H(1)

k, H

(2)

k, respectively, H

(3)

kand the

asymptotic behaviour 12k, 1

2√

k, respectively, 3

83√

4k. Here ◦, 4, ¦

represent the exact values, and +, ∗, • the asymptotic values.

We next describe the extension of the above results forthe simple system with equal powers to the case where thepowers are unequal, shown in [11]. For the MF system, thefollowing result is proven for the exponential rate H (1)

k .

Theorem II.3: For P1/∑k

m=1 Pm → 0,

H (1)

k;1 =P1

2∑k

m=1 Pm+O

((

P1∑k

m=1 Pm

)2)

.

Theorem II.3 is based on a Taylor expansion of the rate.For the HD-PIC system, it is more involved to find anasymptotic expansion, but it can still be done.

5

Theorem II.4: For P1/∑k

m=1 Pm → 0,

H (2)

k;1 ≥1

2

√

P1∑k

m=1 Pm+O

(

P1∑k

m=1 Pm

)

.

When there exists a C > 0, such that, uniformly in∑k

m=1 Pm/P1,

minR

∣

∣

∣

∣

∑

m∈R PmP1

− 1

2

√

∑km=1 PmP1

∣

∣

∣

∣

≤ C

(∑km=1 PmP1

)1/4

,

where the minimum ranges over all R ⊂ {2, . . . , k}, equal-ity is attained:

H (2)

k;1 =1

2

√

P1∑k

m=1 Pm+O

(

P1∑k

m=1 Pm

)

.

A simple example where the above condition is satisfied iswhen maxj,m Pj/Pm ≤ C for some C. We refer to [11] formore examples.

To see that the results above imply that HD-PIC reallyimproves performance, it is useful to investigate the simplesystem. For this case, H (1)

k and H (2)

k reduce to

H (1)

k ≈1

2kand H (2)

k ≈1

2√k.

It is clear that HD-PIC gives a significant increase in per-formance compared to the MF receiver, since 1/(2

√k) is

much large than 1/(2k), so that exp(−n/(2√k)) is much

smaller than exp(−n/(2k)). Note the shift from 1/(2·) to1/(2√·). In order to preserve the desired quality level, it

is thus possible to consider for example a decrease of theprocessing gain by a factor

√k.

For the system with unequal powers, we see that

H (1)

k;1 ≈1

2

P1∑k

m=1 Pmand H (2)

k;1 ≈1

2

√

P1∑k

m=1 Pm.

The same conclusion conclusion holds: interference cancel-lation significantly improves performance. The differenceis that k is replaced by

∑km=1 Pm/P1.

We will now give a heuristic explanation of Theorem II.2We will start by explaining the result for s = 2, and thencomment on s > 2. We will argue that

H (2)

k,r ≈r

2k+

1

8r. (18)

Indeed, using (13), we then see that the minimum over ris obtained when r ≈ 2

√k, yielding H (2)

k ≈ 1/(2√k).

To see (18), note that H (2)

k,r is the rate of the event that

Z(1)

2 < 0, . . . , Z(1)

r+1 < 0, Z(2)

1 < 0, and all the other Z(1)

j > 0for j = r + 2, . . . , k. When we know all the signs of theZ(1)’s, we can substitute these signs in the formula for Z (2)

1 .This yields that

Z(2)

1 = 1 +2

n

r+1∑

j=2

n∑

l=1

A1lAjl. (19)

Note that E (Z(1)

j ) = 1 for all j. Therefore, the event

{Z(1)

j ≥ 0} is not a large deviation. This explains that

we can show that the events Z (1)

j > 0 for j = r + 2, . . . , kdo not contribute to the rate and we can leave these eventsout in the definition of H (2)

k,r. Finally, independence of the

sets {Z(1)

2 < 0}, . . . , {Z(1)

r+1 < 0}, and {Z(2)

1 < 0} is clearlyfalse for all finite k. However, it turns out to be asymptot-ically true for k large, as shown in [7]. This yields that

H (2)

k,r ≈ limn→∞

r+1∑

m=2

− 1

nlogP(Z(1)

m < 0)

− 1

nlogP

(

1 +2

n

r+1∑

j=2

n∑

l=1

A1lAjl < 0

)

.

The first term produces rH (1)

k ≈ r/(2k), the second termcan be computed to give 1/(8r) when r is large. Thisgives an informal explanation of the asymptotic result in(19), and explains that to have one bit error at the firststage of HD-PIC, we need roughly

√k/2 bit errors at level

one. This explains that one-stage of HD-PIC significantlydecreases the BEP.

We finally give an informal explanation of Theorem II.2for s > 2. For R = (R(s)

σ )sσ=1, let H(s)k,R denote the rate of

the event that {Z(σ)m < 0} for all m ∈ R(s)

σ , and {Z(σ)m > 0}

for all m /∈ R(s)σ . Then we have that

H (s)k = min

RH (s)k,R. (20)

Let r(s)σ = |R(s)

σ |. In [7], we show that we have

H (s)k,R ≈

r(s)1

2k+

r(s)2

8r(s)1

+ . . .+r(s)s−2

8r(s)s−1

+1

8r(s)s−1

. (21)

Note that the right hand side does not depend on the pre-cise structure of (R(s)

σ )sσ=1, but only on the sizes. The rea-son is that all dependence between Z (σ) is different levels isdictated by the signs of these quantities. After these signsare substituted, the random variables are asymptoticallyindependent for large k, like in the case for s = 2.

Minimization of the right hand side of (21) overr(s)1 , . . . , r(s)

s−1 yields

r(s)σ ≈

(k

4

)(s−σ)/s. (22)

Substitution yields (17).

D. Chernoff bounds and large k depending on n

The Chernoff bound can be used to bound probabilitiesfrom above by an exponential, and is true regardless of k, n.For the MF system, the Chernoff bound is straightforwardto calculate. The Chernoff bound is given by

P(b(1)10 6= b10) ≤ e−nH(1)k (23)

and this holds for any n and k. Recall from (16) that

H (1)

k ≥ 1/(2k), so that P(b(1)10 6= b10) ≤ e−n/(2k) for any n

6

and k. For the HD-PIC system, a similar statement canbe derived. When we investigate for example the case ofequal powers, we are able to prove

P(b(2)10 6= b10) ≤k−1∑

r=1

(

k − 1

r

)

e−nH(2)k,r . (24)

The Chernoff bound is not tight in the sense that it ap-proximates the BEP with any desired precision. Instead itgives an upper bound in a very simple form. Note that ifn is large, the Chernoff bound is dominated by the termwhich has the smallest rate. Take for example k = 12and n = 63. We can show that (see Table I). H (2)

k,2 mini-

mizes H (2)

k,r. Therefore we expect that most of the sum iscontributed by the second term. Substitution leads to

P(b(2)10 6= b10) ≤ 1.32 · 10−3 + 1.76 · 10−2 + 5.28 · 10−3

+3.44 · 10−4 + . . . = 2.45 · 10−2.

The main contribution clearly comes from r = 2 (72%),the contributions from r = 3 and r = 1 are also relevant(22% and 5%), but the remainder of the sum is negligible(1%).

In Figure 3, simulated BEP’s are compared with the Cher-noff bound in (24) as a function of n. It is seen that theChernoff bound indeed gives upper bounds. The exponen-tial rate appears in this figure as the slope of the BEPand the Chernoff bound. The difference in slopes is verysmall, indicating that the large deviation approach is in-deed promising for performance evaluation. These exactBEP’s are obtained by an importance sampling procedure(see [12] and [13] for the details).

20 40 60 80 10010

−10

10−8

10−6

10−4

10−2

100

processing gain n

Fig. 3. BEP and Chernoff bound. The BEP for k = 3, 6, 9 is markedwith ◦, 4, ¦, respectively. The Chernoff bound for k = 3, 6, 9 ismarked with +, ∗, •, respectively.

We next use the Chernoff bound in order to take k largewith n. From a practical point of view, one always wishesto have as many users as possible, so that the situationwhere k is fixed and n→∞ may be violated. Instead, wenow take k = kn →∞. We prove the following result:

Theorem II.5: When kn →∞ such that kn = o( nlog n ),

limn→∞

−s√knn

logP(b(s)10 6= b10) ≥s

8s√4. (25)

The proof is given in Section V-B.

The above result implies that when kn → ∞ such thatkn = o( n

logn ), we have that P(b(2)10 6= b10) is to leading

asymptotics equal to exp(− n2√kn

). This should be con-

trasted with the behaviour that is expected when k is ofthe order n. This is the central limit regime where we ex-pect that the bit error probability converges as n → ∞to a non-zero constant. It pays to take k slightly smallerthan n instead of a multiple of n, at least when consid-ering a finite number of HD-PIC stages! In Section III-Bwe will give a more precise explanation of the central limitbehaviour and its relation to the HD-PIC results.

We remark that the Chernoff bound for s = 2 can easilybe extended to certain cases with different powers. Indeed,when we assume that maxj,m

PjPm≤ C, we can easily show

that

limn→∞

−√∑

m Pm

nlogP(b(s)10 6= b10) ≥

1

2

√

P1. (26)

E. Measures of performance: signal-to-noise ratio vs. ex-ponential rate

We next investigate the relation of the above result to thesignal-to-noise ratio (SNR), which is defined as

SNR =E (Z)

√

Var(Z),

where Z denotes the decision statistic in the model underconsideration. We consider the case nÀ k À 1 and Pm =2. We will show that in this case the SNR is not a goodmeasure of performance, and therefore, we propose to usethe exponential rate instead.

The asymptotic result for H (1)

k in (16) yields that the ex-ponential rate of the BEP is approximately 1/(2k). Hence,the BEP is approximately

BEP ∼ e−n2k . (27)

For the MF system, we know that SNR=√

n/(k − 1),since

E (Z(1)

m ) = 1 and Var(Z(1)

m ) = (k − 1)/n.

To get an approximation for the BEP, one often substitutesthe SNR in the well-known Q-function. Since (see [5])

e−x2/2

√2πx2

(1− 1

x2) < Q(x) <

e−x2/2

√2πx2

,

substituting the SNR yields the approximation

BEP ∼ e−SNR2/2

This approach gives the approximation

BEP ∼ e−n

2(k−1) ∼ e−n2k ,

which agrees with the large deviation result in (27).

7

For the simple system with one-stage of HD-PIC, TheoremII.2 that H (2)

k ≈ 1/(2√k), yielding

BEP ∼ e− n

2√k . (28)

However, for a system with one stage of HD-PIC, usingthe SNR results in

BEP ≤ exp

{

− en2k

2(k − 1)2

}

, (29)

which is shown below. The latter value is far too smallcompared to the true asymptotics of the BEP in (28),which clearly indicates that the Gaussian approximationusing the SNR is no good. The reasoning above can beadapted for other models, such as multistage HD-PIC orthe optimal HD-PIC system described below. Therefore,using the SNR together with a Gaussian approximation forsuch multiuser detection systems leads to faulty approxi-mations for the BEP.

To prove the upper bound (29), observe that

E (b(2)1 ) = 1 + 2(k − 1)P(b(1)2 6= b2)E (A21A11|b(1)2 6= b2)

≈ 1, (30)

when n is large compared to k since P(b(1)2 6= b2) ≈ 0.Moreover,

Var(Z(2)

1 ) ≤ E(

(Z(2)

1 − 1)2)

≤ (k − 1)2P(b(1)2 6= b2),

where for the latter bound we use that |Aij | = 1. Fromthe Chernoff bound (see (23)), and the fact that H (1)

k ≥ 12k

(see (16)), we end up with

Var(Z(2)

1 ) ≤ (k − 1)2e−n2k .

This yields that

SNR ≥ en4k

(k − 1),

so that

e−SNR2/2 ≤ exp

{

− en2k

2(k − 1)2

}

.

III. The optimal system

In this section, we show that after a finite number of stagesof HD-PIC, the exponential rate of the BEP remains fixed.Then we study properties of this optimal hard decision sys-tem, such as its exponential rate, and Chernoff bounds forthe BEP’s. The results show that the BEP is exponentiallysmall whenever k ≤ δn for some δ > 0 sufficiently small,which is quite remarkable.

A. Optimal hard decisions

We first list some easy consequences for the system withunequal powers. First of all, we define the worst case rateof a bit error to be

H (s)k =

kminm=1

H (s)k;m. (31)

Hence, H (s)k is the exponential rate of the bit error proba-

bility of that user that has largest bit error probability.

Theorem III.1: (a) H (s)k is monotone in s.

(b) There exists a sk ≤ 2k + 1, such that H (s)k = H (sk)

k forall s ≥ sk.

The above shows that we can speak of the system for s = skas the optimal system. For equal powers, we know that therate for all users does not improve for s ≥ sk. For unequalpowers, we can think of this system as having the optimalworse case rate, in the sense that the rate of the bit errorminimized over all users is optimal.

We will use the definition of the sign-function in (6). How-ever, in the proof, we will comment on other definitions ofthe sign-function where a similar result can be shown.

We will now prove the above theorem.

Proof: First of all, s 7→ H (s)k is non-decreasing. In-

deed, to have a bit error, one of the interfering users hasto have a bit error in the previous stage.

For the simple system, we have exchangeability of users,so that the probability of a bit error at stage s is smallerthan that of stage s− 1 and the desired statement follows.The above statement even proves that the probability of abit error is non-decreasing.

This simple argument fails to hold when the powers areunequal. However, in this case, we can show in preciselythe same way that the maximal bit error is non-decreasing.Then we take m such that the bit error in stage s is max-imal for that user m. We can just repeat the argumentgiven above, and see that we need to have a bit error fromanother user in the previous stage. However, the latterprobability is at most the maximal probability of a bit er-ror.

By the monotonicity above H (∞)

k = lims→∞H (s)k exists.

To see why H (∞)

k with k fixed is reached in a finite numberof stages sk and sk ≤ 2k + 1, define

R(s)σ = {m : sgnrm(Z(σ)

m ) 6= 1}. (32)

Then R(s)σ is the set of bit errors at stage s and level σ,

i.e., the set of indices j for which sgnrj(Z(σ)

j ) < 0 when weenforce that sgnrm(Z(s)

m ) < 0. We observe that necessarily

R(s)σ′ = R(s)

σ′′ , for some σ′ < σ′′ ≤ 2k + 1. Then Z(σ′′+1)m ≡

Z(σ′+1)m for all 1 ≤ m ≤ k and thus R(s)

σ′+1 = R(s)σ′′+1. It

follows that R(s)σ′+1 = R(s)

σ′′+1 and thus R(s)σ+i(σ′′−σ) = R(s)

σ

for all σ′ ≤ σ < σ′′, for all i ∈ N ∪ {0}. By (6), all stagesbeyond σ′′ are determined by the stages 1, . . . , σ′′ and donot contribute to the rate. Clearly some users have biterrors for a stage s′ > σ′′. Thus, the scenario describedabove is one scenario to force a bit error for a user at stage

s′. However, H (s′)k is the worst-case rate over all scenarios

and all users. Therefore, H (s′)k ≤ H (s)

k . Since H (s)k is non-

decreasing, necessarily H (s′)k = H (s)

k for all s′ ≥ σ′′. Sinces ≥ σ′′, the desired statement follows. 2

The statement of Theorem III.1 can be extended to otherdefinitions of the sign-function.

8

For example, when we let the sign of 0 to be 0, then wecan copy the original proof, apart from the fact that welet R(s)

σ;0 be the set of values m where Z (σ)m = 0, and R(s)

σ;1

the set where Z(σ)m < 0. Then, when s ≥ 3k + 1, we must

R(s)σ′;i = R(s)

σ′′;i, for i = 0, 1 and some σ′ < σ′′ ≤ 3k+1. Thisagain implies a periodic scenario.

For the system with equal powers, we are also allowed touse a sign-function that assumes values ±1 independentlyevery time it is used, rather than fixing it per user. In-deed, the limit determining H (∞)

k exists, so we can reachthe limit just by using the odd n. In this case, we can-not have Z(σ)

m = 0, except when σ = 1. Therefore, everyuser draws at most one time a random sign-function, whichmakes the statement equivalent to the statement with ouroriginal definition of sgnr. For unequal powers this argu-ment unfortunately does not hold. However, Q is densein R, so that for almost all powers (P1, . . . , Pk) and for allintegers n, Z(σ)

m 6= 0. This explains that the problems aris-ing from the definition of the sign of zero are somewhatacademic.

We will next investigate this optimal system. Recall thatbecause of the hard decisions, the decisions at stage σ onlydepend on the decisions at stage σ − 1. For illustrationpurposes, we will first investigate the system with 2 users.We will use that when user 1 has no bit error at stageσ, user 2 necessarily has no bit error at stage σ + 1. Wedenote a bit error by 2 and no bit error by • in Figure4. All possible five scenarios (except the trivial scenariowhere no errors at all are made) are shown.

@@stage

user

1

2

3

4

2

•2

•

1

•2

•2

2

•2

•2

1

2

•2

•

2

2

2

••

1

2

•••

2

2

•••

1

2

2

••

2

2

2

2

2

1

2

2

2

2

2

Fig. 4. Possible scenarios for the HD-PIC model with 2 users.

It is clear that within 3 stages, the set where bit errorsare made becomes periodic. We will call this a periodicscenario. This is somewhat smaller than 22+1. A periodicscenario has a big “advantage” over non-periodic scenarios.As long as the scenario is not yet periodic, specifying whichusers have bit errors at a certain stage results in a decreaseof the BEP. Indeed, users do not tend to have a bit error;it is more likely to estimate a bit correctly. However, in aperiodic scenario, this is not true anymore. For example,to get a bit error for user 1 at stage s = 1001, it is sufficientto specify the positions of bit errors in stages 1,2 and 3 (thefirst and last scenario in Figure 4 will do). From that pointonwards, the bit errors are determined by the bit errors inthe first three stages.

Two essentially different scenarios are characterizing thebehaviour of the optimal system for k = 2. The first oneis the so-called disjoint scenario, where at every stage user

1 has a bit error and user 2 not, or vice versa. The otherscenario, which we will call the overlapping scenario is thescenario where at every stage both user 1 and user 2 havea bit error. Note that for both the disjoint as the overlap-ping scenario, the periodic behaviour kicks in at stage 1.For both scenarios, we can calculate the exponential rate.The minimum of the two exponential rates indicates whichscenario typically is observed.

When k ≥ 3, we extend those scenarios in the followingway. For every r, at stage 1, 3, 5, . . . bit errors are made forusers in some set R1, with |R1| = r. At stage 2, 4, 6, . . ., biterrors are made for users in the set R2 with |R2| = r. WhenR1 ∩R2 = ∅, we speak of the disjoint scenario. WheneverR1 = R2, we will call it the overlapping scenario. All otherscenarios are called partly overlapping.

The statement of Theorem III.1(b) is that after at most2k+1 stages, the set where bit-errors are made is periodicafter a certain stage. We expect that the scenarios that wetypically observe are periodic already at stage 1. Indeed,specifying the bit errors at the initial non-periodic stagesmake the BEP smaller, while for the periodic stages wedo not need to specify the positions of bit errors, as theseare determined by the first few stages. The longer thenumber of stages where the behaviour is not yet periodic,the smaller the BEP and thus this behaviour is less typi-cal. This suggests that the scenarios that we typically ob-serve are either the disjoint, the partly overlapping, or theoverlapping scenario. The overlapping scenario has period1, while the other scenarios have period 2, so stretchingthe above heuristic even further, it is tempting to assumethat the overlapping scenario gives the smallest exponen-tial rate, and is therefore typical. However, this is not truewhen k is small (see also 2) below).

We observe the following phenomena.

1) The partly overlapping scenario is never optimal. Itseems that both extremes (the overlapping and disjointscenario) do a better job.2) For small k, the disjoint scenario is optimal. The reasonis quite simple. For r fixed, in the disjoint scenario, a userat stage 2 has contribution from r noise terms, whereas forthe overlapping scenario the user has contribution fromonly r − 1 terms (the user does not interfere with its ownsignal). For higher k, however, the overlapping scenariois optimal. For the case k → ∞, it is implicitly proven inthe proof of Theorem III.2(a) that the overlapping scenariois indeed optimal, and the partly overlapping scenario isnever optimal.3) For k →∞, also r →∞, but much slower than k. Thenumerical results indicate that r ≈

√k/2.

To illustrate 2) and 3), the optimal r is shown for k =1 − 1000 in Table II. Also, it is indicated whether thedisjoint or the overlapping scenario is optimal.

In Figure 5, the exponential rates H (sk)

k are given, togetherH (s)k for s = 1, 2, 3. The results for s = 3 are obtained

using similar techniques as in Theorem II.1. However, wehave not stated the result for s = 3 in this paper. The

9

k r k r1-30 1(d)1 337-414 10(o)31-73 2(d) 415-499 11(o)74-83 3(d) 500-592 12(o)84-107 5(o)2 593-694 13(o)108-153 6(o) 695-803 14(o)154-206 7(o) 804-920 15(o)207-267 8(o) 920-1000 16(o)268-336 9(o)

1(d) means disjoint scenario is optimal,2(o) means overlapping scenario is optimal.

TABLE II

Expected optimal scenario for optimal HD-PIC model.

rate H (sk)

k is in fact the rate corresponding to the disjointscenario for r = 1 or r = 2. For k = 3, it is seen thatH (1)

3 = H (2)

3 = H (3)

3 , so that s3 = 1. For 4 ≤ k ≤ 9, wesee H (1)

k < H (2)

k = H (3)

k , so that sk = 2. We see that onestage of HD-PIC gives an improvement in exponential rate.However, adding one more stage does not result in anyimprovement. For 10 ≤ k ≤ 22, 2-stage HD-PIC gives animprovement over one stage HD-PIC. The numerical resultshow that there is a block scenario (in fact the disjointscenario with r = 1) with the same rate. Therefore sk = 3in this case. For 23 ≤ k ≤ 50, we expect sk = 4, eventhough we cannot calculate H (4)

k .

5 10 15 20 25 30 35 40 45 500

0.05

0.1

0.15

0.2

0.25

0.3

sk=1

sk=2

sk=3 s

k=4

number of users k

Fig. 5. The exponential rates H(sk)

kand H

(s)

kfor s = 1, 2, 3.

We now turn to a lower bound of the exponential ratefor the optimal system. Clearly, (17) has no meaningfor s → ∞, but assuming (17) and the monotonicity ofs 7→ H (s)

k it follows for the simple system that for all ε > 0,kεH (∞)

k →∞ when k →∞. Thus, if H (∞)

k converges to 0as k → ∞, it does so slower than any power of 1/k. Thetheorem below states that the exponential rate of the op-timal system remains strictly positive as k → ∞, and weidentify the limiting rate under a certain assumption onthe powers P1, . . . , Pm that we will define now. We will as-sume that (P1 −P4) hold, where the conditions (P1 −P4)are defined by

(P1) There exists a δ > 0 s. t. ]{j : Pj ∈ [δ, 1/δ]} → ∞,(P2) limk→∞ kP−1 <∞,

(P3) k−1Pmax → 0,

(P4) kPmin →∞.

Here Pmax = maxm Pm, Pmin = minm Pm. The main re-

sult in this section in the following theorem.

Theorem III.2: (a) For the general system, for all s > 2k,

H (s)k ≥

1

2log 2− 1

4≈ 0.09657. (33)

For the system with equal powers the bound is obtained fors > k.(b) If the powers are such that (P1 − P4) hold, then

limk→∞

H (sk)

k =1

2log 2− 1

4. (34)

We stress that the case of equal powers is covered in The-orem III.2(b).

In Figure 6, an upper and lower bound is given of theexponential rate H (sk)

k for equal powers. We expect thatthe upper bound is in fact equal to H (sk)

k , but we lack aproof.

200 400 600 800 10000

0.05

0.1

0.15

0.2

0.25

0.3

number of users k

Fig. 6. Upper and lower bounds for the exponential rates H(sk)

k.

The proof of the above result will be given in the nextsections. We will start with bounds on moment generatingfunctions in Section IV. These bounds will be used in theproof of Theorem III.2 in Section V.

B. Chernoff bounds and large k depending on n

In Section V-B, we prove the following Chernoff bound forthe optimal system:

Theorem III.3: When kn →∞ and for s > 2kn

P(b(s)10 6= b10) ≤ 8kne−In, (35)

where I = 12 log 2− 1

4 .When the powers are equal, the same result holds whens > kn.

The above result shows that the limit of k and n can betaken simultaneously, instead of the order common in largedeviation theory (first n→∞, and subsequently k →∞).Moreover, Theorem III.3 shows that when k = δn withδ < I/(3 log 2), that then the BEP is exponentially small.

10

Since for k = O(n), we have central limit behaviour, theabove Chernoff bound implies that applying HD-PIC suf-ficiently often enables us to reduce the BEP from O(1) forthe MF system, to exponentially small for the optimal HDsystem. We now give a more extensive heuristic explana-tion for this effect.

We take k = δn for some constant δ > 0. We as-sume that the powers are equal, and for simplicity takePm = 2. In this case, we have that by the centrallimit theorem, Z(1)

m ≈ N (1, δ). Hence, the BEP is ap-proximately ε(1) = Q(1/

√δ). Therefore, we expect ε(1)n

bit errors in the first stage (s = 1). Now, given thatthere are ε(1)n bit errors in the first stage, we have thatZ(2)m ≈ N (1, 4ε(1)). Therefore, the BEP is approximately

ε(2) = Q(1/(2ε(1))), and we expect ε(2)n bit errors in the sec-ond stage (s = 2). Repeating the above argument, we findthat, with ε(s) = Q(1/(2ε(s−1))), we expect approximatelyε(s)n bit errors in the sth stage. The above is probablyquite accurate when s is finite and n → ∞, but certainlynot when s tends to infinity with n. However, from The-orem II.2 combined with Theorem III.2, we do know thatsk → ∞ when k → ∞. Therefore, the above argumentfails to hold for the optimal system, and we see that theoptimal HD-PIC system has a much smaller BEP than theBEP after any finite number of stages.

C. On the number of stages to optimality

In this section, we will investigate the number of stagesnecessary to reach the asymptotic optimal rate 1

2 log 2− 14 .

We will prove the following theorem. In its statement, wewill recall that P =

∑

m Pm, and we will use the notationp = minm Pm.

Theorem III.4: For the general system, for all 0 < ε <12 log 2− 1

4 and for all s such that s ≥ ε−1 log Pp ,

H (s)k ≥

1

2log 2− 1

4− ε. (36)

We know that I = 12 log 2 − 1

4 is the asymptotically opti-mal rate for the system where all the powers are equal.Note that when all powers are comparable (e.g., when

maxj,mPjPm

< C for some constant C < ∞ uniformly in

k), that then ε−1 log Pp ≈ ε−1 log k. Theorem III.4 states

that when we apply at least that many stages of HD-PIC,that then the limiting rate will be asymptotically boundedfrom below by I. This suggests that sk grows roughlylike log k. It is an interesting problem to determine howsk grows more precisely. It follows from Theorem III.2 to-gether with Theorem II.2 that sk →∞. However, it wouldbe of practical importance to know the precise rate, or evenan upper bound on sk. We have the following conjecture:

Conjecture III.5: For the general system,

lim supk→∞

log sklog log k

= 1. (37)

Conjecture III.5 says that sklog k cannot grow or decrease

faster than any small power of log k.

Theorem III.4, together with Theorem III.2, suggests thatsk ≤ ε−1

k log k for every εk ↓ 0. We believe that in facta logarithmic number of stages is required to obtain theoptimal HD-PIC system. However, we have no proof forthis belief. This belief stems from the fact that we expectthe proof of Theorem II.2 to hold for some s that tendto infinity with k sufficiently slowly. More precisely, weexpect that the strategy described in Section II-C remaintrue for as long as sk ≤ (log k)1−ε for any ε > 0. Thereason is that in (22), there is an essential change whens¿ log k compared to sk = O(log k), in the sense that forthe former

r(s)σ

r(s)σ+1

→∞,

whereas for the latter, the above converges to a constant.When this convergence is towards a constant, we cannotexpect (21) to be a good approximation. Moreover, notethat when s = (log k)1−ε, then substitution of s into theright hand-side of (17) gives

s

8e− log k/s =

(log k)1−ε

8e−(log k)ε → 0

for all positive ε. This is clearly far away from the rate ofthe optimal HD-PIC system in Theorem III.2. Therefore,we believe sk ≥ (log k)1−ε for all ε > 0, which explainsConjecture III.5.

IV. Bounds on moment generating functions

In this section, we will give sharp bounds on certain mo-ment generating functions that will prove to be essentialin the analysis of the optimal HD-PIC system.

We define

Sn =

n∑

i=1

√

PiXi, (38)

where

P(Xi = ±1) =1

2.

We will also use the notation P =∑n

i=1 Pi. The mainresult of this section is

Proposition IV.1:(a) For all n ∈ N, 0 ≤ t < 1

P and all s ∈ R,

E [esSn+ t2S

2n ] ≤ e

Ps2

2(1−Pt)√1− Pt

, (39)

(b) For all − 1P ≤ t ≤ 1

P

E [et2S

2n ] ≤ 1√

1− Pt. (40)

Note that if n is large, then Sn ≈√PZ, where Z has a

standard normal distribution. The bounds in (39-40) showthat the moment generating function of Sn and S2

n, for theappropriate ranges of the variables t and s are at leastbounded from above by the moment generating functionsof√PZ and PZ2, ePs

2/2 and 1/√1− Pt, respectively.

11

Proof of Proposition IV.1(a). This bound is easiest. LetZ have a standard normal distribution, then we know thatE (etZ) = et

2/2. Hence, we get that

E [esSn+ t2S

2n ] = E

[

e(s+√tZ)Sn

]

= E[ n∏

i=1

cosh(√

Pi(s+√tZ))

]

.

We use that cosh(t) ≤ et2/2, to arrive at

E [esSn+ t2S

2n ] ≤ E

[ n∏

i=1

ePi2 (s+

√tZ)2

]

= ePs2

2 E [ePs√tZ+tPZ2

].

We complete the proof by noting that, for t ≤ 1P ,

E [ePs√tZ+tPZ2

] =e

P2s2t2(1−Pt)√1− Pt

,

and by rearranging terms. 2

Proof of Proposition IV.1(b). The claim for 0 ≤ t ≤ 1P

follows from Proposition IV.1(a) proved above. The claimfor t < 0 is more difficult, and we will use induction on n.Define

fn(t) = E [et2S

2n ]. (41)

The induction hypothesis is that

fn(t) ≤1√

1− Ptfor all − 1

P≤ t ≤ 0. (42)

Clearly, for n = 0, the above is trivial, as both the leftand right hand side are equal to 1. We next advance theinduction. We write

fn(t) = E [et2S

2n ] = e

tPn2 E [e

t2S

2n−1+

√PntSn−1Xn ]

= etPn2 E [e

t2S

2n−1 cosh(

√

PntSn−1)].

We again use that cosh(t) ≤ et2/2 to arrive at

fn(t) = E [et2S

2n ] ≤ e

tPn2 E [e

t+Pnt2

2 S2n−1 ] = e

tPn2 fn−1(t+Pnt

2).

To prove the claim, we first show that for − 1P ≤ t ≤ 0,

− 1

P − Pn≤ t+ Pnt

2 ≤ 0

Indeed, since − 1P ≤ t ≤ 0, we have 0 ≤ 1 + Pnt ≤ 1, so

that

−1 ≤ (−1− Pnt)(1 + Pnt) ≤ (Pt− Pnt)(1 + Pnt)

= (P − Pn)(t+ Pnt2) = (P − Pn)t(1 + Pnt) ≤ 0,

where the last inequality follows from P−Pn ≥ 0 and t ≤ 0.We therefore can substitute the induction hypothesis (42)for n− 1, so that it remains to show that

etPn/2√

1− (P − Pn)(t+ Pnt2)≤ 1√

1− Pt.

Since ex ≥ 1 + x+ x2/2 for all x ≥ 0,

etPn/2 =1√e−tPn

≤ 1√

1− Pnt+ P 2nt

2/2.

Multiplying and rearranging terms gives

(1− Pnt+ P 2nt

2/2)(1− (P − Pn)(t+ Pnt2))

= 1 + t[−Pn − (P − Pn)]

+t2[−Pn(P − Pn) + Pn(P − Pn) + P 2n/2]

+t3[−(P − Pn)P2n/2 + P 2

n(P − Pn)]

+t4[−P 3n(P − Pn)/2]

= 1− Pt+ P 2nt

2/2 + P 2n(P − 4Pn/3)t

3/2

−P 3n(P − Pn)t

4/3

= 1− Pt+P 2nt

2

2

(

1 + Pt− Pnt(1 + (P − Pn)t)

)

≥ 1− Pt,

since 1 + Pt ≥ 0, t ≤ 0 and 1 + (P − Pn)t ≥ 0. Thiscompletes the proof. 2

V. Proofs

In this section we will first prove Theorem III.2(a) andIII.4, where we will use Proposition IV.1. Then, we willprove the Chernoff bounds in Theorem II.5 and III.3 in Sec-tion V-B. This proof makes use of the proof of TheoremnTheorem III.2(a). Finally, we prove Theorem III.2(b) forequal powers. The proof of Theorem III.2(b) for unequalpowers satisfying the power conditions (P1 − P4) will begiven in the appendix.

A. Proof of Theorem III.2(a) and Theorem III.4

In this section, we will prove Theorem III.2(a) and Theo-rem III.4 simultaneously. For completeness, we will repeatthe statements. For s = 2k + 1,

H (s)k ≥

1

2log 2− 1

4. (43)

For every 0 < ε < I = 12 log 2− 1

4 . For s = ε−1 log(P/p),

H (s)k ≥

1

2log 2− 1

4− ε. (44)

In Section III we have shown that there is an optimal HD-PIC system, and that for s ≥ 2k+1, the set of bit errors isperiodic. For any set A ⊂ {1, . . . , k}, we let PA =

∑

i∈A Pi.When s = 2k + 1, there must be a σ ≤ 2k + 1 such thatPR

(s)σ≥ P

R(s)σ−1

. In fact, when the powers are equal, we

have that PR

(s)σ

= P |R(s)σ |, and the above must happen

already when s ≥ k. We focus on that level σ and are onlyinterested in the event {P

R(s)σ≥ P

R(s)σ−1

}.

Furthermore, when s = ε−1 log(P/p) + 1, there must bea σ ≤ ε−1 log(P/p) + 1 such that P

R(s)σ≥ (1 − ε)P

R(s)σ−1

.

12

Indeed, when this should not be the case after σ − 1 =ε−1 log(P/p) stages, we have

PR

(s)σ−1

≤ (1− ε)σ−1PR

(s)0

= exp

(

log(1− ε)

εlog

P

p

)

P.

Since 1− ε ≤ e−ε, also log(1− ε)/ε ≤ −1, so that

PR

(s)σ−1

≤ exp

(

− logP

p

)

P = p.

However, always PR

(s)σ≥ p, so that

PR

(s)σ≥ p ≥ (1− ε)P

R(s)σ−1

,

so certainly after at most s = log(P/p)/ε + 1 stages thedesired event has occurred. We focus on the level σ atwhich the desired event occurs and are only interested inthat occurrence.

At this point we remark that once we obtain for PR

(s)σ≥

(1− ε)PR

(s)σ−1

that

− limn→∞

1

nlogP

(

PR

(s)σ≥ (1− ε)P

R(s)σ−1

)

≥ 1

2log 2− 1

4− ε,

we immediately obtain the result for ε = 0, which is thestatement (43).

We use that when A = R(s)σ−1 and B = R(s)

σ , that then forall m ∈ B, we have that

1

2

√

2PmZ(σ)

m = Pm +2

n

n∑

l=1

∑

j∈A\{m}

√

PjPmAjlAml.

As the number of configuration (R(s)σ )sσ=1 is just finite, we

obtain that H (s)k is bounded from below by the minimum

over A and B such that PB ≥ (1− ε)PA of

− limn→∞

1

nlogP

(

∑

j∈A

n∑

l=1

√

PjPmAjlAml +nPm2≤ 0∀m ∈ B

)

≥ − limn→∞

1

nlogP

(

∑

j∈Am∈B\{j}

n∑

l=1

√

PjPmAjlAml +nPB2≤ 0

)

≥ − log(

ePB−2PA∩B

4 tE(

et2

∑

(j,m)∈A×B√PjPmAj1Am1

)

)

,

where the last inequality is the exponential Chebycheff’sinequality for t ≤ 0. We write SA =

∑

j∈A√

PjAj1 to endup with

H (s)k ≥ min

A,B

PB≥(1−ε)PA

− log(

E (et2SASB )

)

+PB − 2PA∩B

4t. (45)

We will now bound the moment generating function frombelow using Proposition IV.1.

We first write SA = SA∩B + SA\B , and we use the factthat SA\B is independent from (SB , SB∩A) to get that

E (et2SBSA) equals

E (et2SBSB∩Ae

t2SBSA\B ) (46)

= E (et2SBSB∩A

∏

j∈A\Bcosh(

t

2

√

PjSB))

≤ E (et2SBSB∩Ae

t2

8 PA\BS2B )

We write the right hand side of (46) as

E(

e(t2+ t2

8 PA\B)S2A∩Be(

t2+ t2

4 PA\B)SB∩ASB\A+ t2

8 PA\BS2B\A)

and use Proposition IV.1(a) with s = ( t2 + t2

4 PA\B)SB∩A

and t = t2

4 PA\B ≥ 0 to bound the expectation over SA\Bas

1√

1− PA\BPB\At2/4

×E

(

e(t2+ t2

8 PA\B)S2A∩Be

PA\B(t/2+PB\At

2/4)2

2(1−PA\BPB\A2t2/4)S2B∩A

)

=1

√

1− PA\BPB\At2/4E(

etS2B∩A/2

)

≤ 1√

1− PA\BPB\At2/4

1√

1− PB∩At,

where the last inequality is valid as long as |t| ≤ 1PB∩A

andwhere

t = t+t2PA\B

4+ PB\A

(t/2 + PA\Bt2/4)2

1− PA\BPB\At2/4

=t+ (PB\A + PA\B)t

2/4

1− PA\BPB\At2/4.

From the first expression for t it is clear that t ≥ t, so thatwe restrict −1/PA∩B ≤ t ≤ 0. From the second expressionfor t above it is straightforward to prove that t ≤ 0. Weproceed by multiplying the two square roots to obtain

E (et2SASB ) ≤ 1

√

1− tPB∩A − t2

4 (PAPB − P 2B∩A)

, (47)

since

PA\BPB\A + PA∩B(PA\B + PB\A) = PAPB − P 2B∩A.

Substituting (47) into (45) yields that for all t ≥ − 1PA∩B

we have that H (s)k is bounded from below by

minA,B

PB≥(1−ε)PA

1

2log

(

1− tPA∩B −t2

4(PAPB − P 2

B∩A)

)

−PB − 2PB∩A4

t. (48)

Since − 1PB∩A

≤ − 1PB≤ 0, substituting t = − 1

PBresults in

H (s)k ≥ min

A,B

PB≥(1−ε)PA

1

2log

(

1− PA4PB

+PB∩APB

+P 2B∩A4P 2

B

)

−1

4

(

2PB∩APB

− 1

)

. (49)

13

It is clear that this lower bound of H (s)k is decreasing in

PA and that PA ≤ 11−εPB . Therefore substituting PA =

11−εPB still gives a lower bound. Clearly, PA∩B

PB∈ [0, 1].

The above lower bound is therefore attained at

min0≤α≤1

1

2log

(

1− 1

4(1− ε)+ α+

α2

4

)

− 1

4(2α− 1)

= min0≤α≤1

f(α). (50)

Differentiating f(α) w.r.t. α gives that f ′(α) equals

−1

2+

1

2

1 + α2

1− 14(1−ε) + α+ α2

4

=

14(1−ε) − α

2 − α2

4

1− 14(1−ε) + α+ α2

4

.

Hence, f ′(α) > 0 for α <√

1 + 1/(1− ε)−1 and f ′(α) < 0

for α >√

1 + 1/(1− ε)− 1. Therefore, the minimum of fis attained at either α = 0 or at α = 1. Substitution yieldsthat

f(0) =1

2log

(

1− 1

4(1− ε)

)

+1

4

=1

2log

3

4+

1

4+

1

2log

(

1− 1

3

( 1

1− ε− 1)

)

and

f(1) =1

2log

(

9

4− 1

4(1− ε)

)

− 1

4

=1

2log 2− 1

4+

1

2log

(

1− 1

8

( 1

1− ε− 1)

)

.

Finally observe that for 0 ≤ ε ≤ 2/3,

e−2ε ≤ 1− 2ε+ 2ε2 ≤ 1− 2ε+ 4ε/3 = 1− 2ε/3,

so that −2ε ≥ log(1−2ε/3). Furthermore, for 0 ≤ ε ≤ 1/2,1/(1− ε) ≤ 1/(2ε). Substituting this yields

log

(

1− 1

3

(

1− 1

1− ε

)

)

≥ −2ε,

log

(

1− 1

8

(

1− 1

1− ε

)

)

≥ −3

4ε ≥ −2ε.

Therefore, since 12 log

34 + 1/4 > 1

2 log 2 + 1/4,

H (s)k ≥ min{f(0), f(1)} = 1

2log 2− 1

4− ε.

This completes the proof of Theorem III.4. Substitutingε = 0 gives Theorem III.2(a). 2

B. Proof of the Chernoff bounds

We will start by proving Theorem II.5 for s = 2. The ex-tension to s > 2 will follow later, and is a small adaptationof the proof for s = 2.

We have that

P(sgnr1(Z(2)

1 ) < 0) ≤ P(Z(2)

1 ≤ 0)

≤kn−1∑

r=1

(

kn − 1

r

)

P(

Z(2)

1 ≤ 0,r+1maxm=2

Z(1)

m ≤ 0,knmin

m=r+2Z(1)

m ≥ 0)

.

We split the sum over r in two parts: 1 ≤ r ≤ 4√kn and

r > 4√kn. We start with the first term. The Chernoff

bound gives

P(

Z(2)

1 ≤ 0,r+1maxm=2

Z(1)

m ≤ 0,knmin

m=r+2Z(1)

m ≥ 0)

≤ e−nH(2)kn,r .

We bound, using that(

kn−1r

)

≤ krn and e−nH(2)kn,r ≤

e−nH(2)kn ,

4√kn

∑

r=1

(

kn − 1

r

)

e−nH(2)kn,r ≤ k4

√kn+1

n e−nH(2)kn

= e4 kn log kn√

kn+

√kn log kn√

kn e−nH(2)kn .

The first term on the right hand-side is eo( n√

kn), since kn =

o( nlogn ) implies kn log kn = o(n). Therefore, this term is

bounded from above by

eo( n√

kn)e−nH

(2)kn = e

o( n√kn

)e− n

2√kn

(1+o(1)).

This proves that the first term has the right order, and itremains to show that the other term is an error term. Inorder to do this, we will first prove the following lemma.

Lemma V.1: For every k, n

P(

Z(2)

1 ≤ 0,r+1maxm=2

Z(1)

m ≤ 0,k

minm=r+2

Z(1)

m ≥ 0)

≤ e−r4kn.

Proof: We bound, using P(A ∩B) ≤ P(A),

P(

Z(2)

1 ≤ 0,r+1maxm=2

Z(1)

m ≤ 0,k

minm=r+2

Z(1)

m ≥ 0)

≤ P(

rmaxm=1

Z(1)

m ≤ 0)

≤ P(

r∑

m=1

Z(1)

m ≤ 0)

.

We can compute that

r∑

m=1

Z(1)

m =1

n

n∑

l=1

r∑

m=1

Aml

k∑

i=1

Ail, (51)

Therefore, by the exponential Chebycheff inequality, forevery t ≤ 0, the probability of interest is bounded by

(

mint≤0

E (et2SrSk)

)n

≤(

mint≤0

E (et2S

2r e

t2Sr(Sk−Sr))

)n

,

where we write Sm =∑m

l=1 A1l. We next bound the mo-ment generating function from above. We first use theindependence of Sr and Sk − Sr to obtain

E(

et2S

2r e

t2Sr(Sk−Sr)

)

= E(

et2S

2r cosh

( t

2Sr

)k−r). (52)

We next use the bound 1 ≤ cosh(t) ≤ et2/2 to bound the

expression above as

E(

et2S

2r cosh

( t

2Sr

)k)

≤ E(

e(t+kt2

4 )S2r/2)

. (53)

14

As long as t + kt2

4 ≥ − 1r , we can use Proposition IV.1(b)

to obtain

E et2S

2r e

t2Sr(Sk−Sr) ≤ 1

√

1− rt− rkt2

4

. (54)

Therefore, we arrive at

P(

Z(2)

1 ≤ 0,r+1maxm=2

Z(1)

m ≤ 0,k

minm=r+2

Z(1)

m ≥ 0)1/n

≤ mint≤0

exp

(

− 1

2log(

1− rt− rkt2

4

)

)

.

The optimal t is attained at t = −2/k. For this choice,we have that t + kt2/4 = −1/k ≥ −1/r. This justifies(54). We further observe that for t = −2/k, we have that

log(

1− rt− rkt2

4

)

= log(

1 + rk

)

. Finally, observe that12 log(1 + x) ≥ x/4 for all 0 ≤ x ≤ 1, which completesthe proof of the lemma. 2

We now complete the proof of Theorem II.5 for s = 2using Lemma V.1. Since kne

− n4kn = o(1), the sum over r

satisfying r > 4√kn is bounded by

∑

r>4√kn

krne− nr

4kn =(

elog kn−n

4kn

)4√kn∑

r>0

(

elog kn−n

4kn

)r

≤ 2 exp

(

4√

kn log kn −n√kn

)

.

This satisfies the required bound since√kn log kn =

o( n√kn

), so that we have

P(sgnr1(Z(2)

1 ) < 0)

≤ e− n

2√kn

(1+o(1))+ 2e

− n√kn

(1+o(1))= e

− n

2√kn

(1+o(1)).

The proof for s > 2 is similar, and we point out the dif-ferences only. We can use the proof of Lemma V.1 to

show that the probability that there are at least c1k(s−1)/sn

bit errors in the first stage is an error term if c1 > 0 islarge enough. Therefore, we only have to deal with the

case where there are at most c1k(s−1)/sn bit errors in the

first stage. In this case, an easy extension of the proof ofLemma V.1 shows that the probability that there are at

least c2k(s−2)/sn bit errors in the second stage is an error

term if c2 > 0 is large enough. Therefore, we may also as-

sume that there are at most c2k(s−2)/sn bit errors in the first

stage. We can repeat this argument, so that we only haveto deal with the probability that user 0 has a bit error

in stage s, intersected by the events |R(s)σ | ≤ cσk

(s−σ)/sn .

We now can use the Chernoff bound and show that thebinomial factors are of lower order.

2

We next turn to the proof of Theorem III.3, which is quiteeasy when we use results of the proof of Theorem III.2 inSection V-A. In the proof, we have used the fact that for

s ≥ 2kn + 1, there must be a stage σ such that PR

(s)σ≥

PR

(s)σ−1

. The rate of this event is proven to be bounded from

below by I = 12 log 2 − 1

4 . The number of possible stages

at which this can happen is 2kn + 1, and the number ofpossibilities for R(s)

σ and R(s)σ−1 are 2kn − 1 each, since R(s)

σ

and R(s)σ−1 cannot be empty. The above argument leads

to an overall factor of (2kn + 1)(2kn − 1)2 ≤ 8kn . Thiscompletes the proof, since

P(b(s)10 6= b10) ≤ P(

⋃

σ≤2kn+1

{PR

(s)σ≥ P

R(s)σ−1

})

≤ 8kn minA,B:PB≥PA

P(R(s)σ = A,R(s)

σ−1 = B)

≤ 8kne−In.

We note that the above proof also holds if we choose thesign(0) to be equal to ± independently every time. 2

C. Proof of Theorem III.2(b)

In order to prove Theorem III.2(b), we have to find a strat-egy that has asymptotic rate 1

2 log 2 − 14 . For simplicity,

we will assume that all powers are equal. The proof forunequal powers is more technical and is therefore deferredto Appendix A. For simplicity, we assume that the powersare all equal to 2. We note that when

R2 = R1,

that necessarily for all σ ≥ 1

Rσ = R1.

Hence, we have now found a strategy that implies bit errorsat all stages, so that the rate of this event is an upperbound for the rate of the optimal system. We fix r = |R1|and for technical reasons we assume r to be odd. We willfirst investigate the rate

− limn→∞

1

nlogP(R2 = R1, |R1| = r).

Due to the fact that all users are exchangeable, and therate function of the vector ((Z (1)

m , Z(2)m ))km=1 is convex, we

have that

− limn→∞

1

nlogP(R2 = R1, |R1| = r)

= − limn→∞

1

nlogP

( r∑

m=1

Z(1)

m ≤ 0,k∑

m=r+1

Z(1)

m ≥ 0,

r∑

m=1

Z(2)

m ≤ 0,

k∑

m=r+1

Z(2)

m ≥ 0

)

,

where Z(2)m denotes Z(2)

m where the signs of the Z (1)m are

substituted. This statement will be proven in more detailin Appendix A.

15

We next note that

k∑

m=1

Z(1)

m =1

n

n∑

i=1

( k∑

m=1

(

1 +k∑

j=1j 6=m

AjiAmi

)

)

=1

n

n∑

i=1

( k∑

m,j=1

AjiAmi

)

=1

n

n∑

i=1

( k∑

m=1

Ami

)2

≥ 0.

Moreover, we have that∑r

m=1 Z(1)m ≤ 0. Hence,

∑km=r+1 Z

(1)m ≥ 0, so that we can remove the event

{∑km=r+1 Z

(1)m ≥ 0}. Since P(X ≤ 0, Y ≤ 0) ≥ P(X −

Y/2 ≤ 0, Y ≤ 0), we can bound the exponential rate fromabove by

− limn→∞

1

nlogP

( r∑

m=1

Z(1)

m −1

2Z(2)

m ≤ 0,

r∑

m=1

Z(2)

m ≤ 0,

k∑

m=r+1

Z(2)

m ≥ 0

)

,

where for 1 ≤ m ≤ r,

Z(1)

m −1

2Z(2)

m =1

n

n∑

i=1

(

1

2+

k∑

j=r+1

AjiAmi

)

,

Z(2)

m =1

n

n∑

i=1

(

1 + 2

r∑

j=1j 6=m

AjiAmi

)

=1

n

n∑

i=1

(

−1 + 2

r∑

j=1

AjiAmi

)

,

while for r + 1 ≤ m ≤ k,

Z(2)

m =1

n

n∑

i=1

(

1 + 2

r∑

j=1

AmiAji

)

.

We abbreviate

E1 =

{

−r + 2r∑

m,j=1

1

n

n∑

i=1

AmiAji ≤ 0

}

,

E2 =

{

r

2+

k∑

m=r+1

r∑

j=1

1

n

n∑

i=1

AmiAji ≤ 0

}

,

E3 =

{ k∑

m=r+1

r∑

j=1

[

1 +1

n

n∑

i=1

AmiAji

]

≥ 0

}

.

Clearly,

P(E1 ∩ E2) = P(E1 ∩ E2 ∩ E3) + P(E1 ∩ E2 ∩ Ec3),

so that, according to the largest-exponent-wins principle,

−limn→∞

1

nlogP(E1∩E2) =min

{

−limn→∞

1

nlogP(E1∩E2∩E3),

−limn→∞

1

nlogP(E1∩E2∩Ec

3)

}

.

We wish to show that for k →∞ and r large,

− limn→∞

1

nlogP(E1 ∩ E2) = − lim

n→∞1

nlogP(E1 ∩ E2 ∩ E3).

In order to do so, we show that for all r

lim infk→∞

− limn→∞

1

nlogP(E1 ∩ E2 ∩ Ec

3) ≥1

2log

3

4+

1

4. (55)

Furthermore, we will show that

− limk→∞

limn→∞

1

nlogP(E1 ∩ E2) ≤ − lim

n→∞1

nlogP(E1) (56)

and as r →∞,

− limn→∞

1

nlogP(E1) =

1

2log 2− 1

4+ o(1). (57)

This implies directly when r is large

− limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩E3)

= − limk→∞

limn→∞

1

nlogP(E1 ∩ E2)

≤ − limn→∞

1

nlogP(E1)

=1

2log 2− 1

4+ o(1) = 0.096573 . . .+ o(1).

Indeed, when x ≤ min{y, z} and x < z, then necessarilyx = y. Equations (55), (56) and (57) also imply the state-ment in Theorem III.2(b) as we will show now. Takingr →∞ gives

− limr→∞

limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3) ≤

1

2log 2− 1

4.

The remainder of this proof therefore focuses on proving(55), (56) and (57). We prove (55) in the following lemma,(56) in Lemma V.3 and (57) in Lemma V.4.

Lemma V.2: As k →∞,

lim infk→∞

− limn→∞

1

nlogP(E1 ∩ E2 ∩ Ec

3) ≥1

2log

3

4+

1

4. (58)

Proof: We bound the rate of interest from below by

−limn→∞

1

nlogP(Ec

3) = −limn→∞

1

nlogP

( k∑

m=r+1

Z(2)

m ≤ 0

)

= −limn→∞

1

nlogP

(

1

n

n∑

i=1

k∑

m=r+1

[

1 + 2

r−1∑

j=0

AjiAmi

]

< 0

)

.

We can follow the proof of Section V-A with A ={0, . . . , r − 1} and B = {r, . . . , k − 1} (see e.g. (49) withA ∩B = ∅). This results in

− limn→∞

1

nlogP(Ec

3) ≥1

2log(

1− r

4(k − r)

)

+1

4≥ 1

2log

3

4+1

4,

when r ≤ k/2. 2

16

The strategy of the proof is first to characterize the be-haviour for k → ∞ and then showing that r → ∞ givesthe desired result.

Lemma V.3: For r fixed,

limk→∞

− limn→∞

1

nlogP(E1 ∩E2) ≤ − lim

n→∞1

nlogP(E1). (59)

Proof: We focus on

− limn→∞

1

nlogP

(

r

2+

k−1∑

m=r

r−1∑

j=0

1

n

n∑

i=1

AmiAji ≤ 0,

−r + 2

r∑

m,j=0

1

n

n∑

i=1

AmiAji ≤ 0

)

.

Using Cramer’s Theorem and invoking the notation R ={0, . . . , r − 1} and R+

0 = {r, . . . , k − 1} gives that the rateabove is given by

− inft1,t2≤0

logE(

et1(r/2+SRSR+

0)t2(2S

2R−r)

)

.

Since SR and SR+0are independent, we can bound the rate

of interest from above by

− inft1,t2≤0

log

(

et1r/2 cosh(t1SR)k−rE et2(2S

2R−r)

)

. (60)

It is sufficient to prove that for all δ > 0, |t1| ≤ δ as k →∞.Indeed, since coshx ≥ 1, the rate on the right hand-side isbounded of (60) from above by

− inf|t1|≤δ,t2≤0

log

(

et1r/2(cosh(t1SR))k−rE et(2S

2R−r)

)

≤ − inft2≤0

log

(

e−δr/2E et(2S2R−r)

)

=δr

2− inf

t2logE et2(2S

2R−r) =

δr

2−limn→∞

1

nlogP(E1).

Since δ is arbitrary small and r is fixed, the desired state-ment then follows. It is hence sufficient to prove that under|t1| ≥ δ, the infimum can not be attained for k sufficientlylarge. We prove this statement by contradiction. Assumethat the infimum is attained for a |t1| ≥ δ. Since r is odd,cosh(t1j) ≥ cosh(t1) ≥ e|t1|/2, so that

et1r/2 cosh(t1j)r ≥ 2−re−|t1|r/2er|t1| ≥ 2−r.

Substituting the result above gives

− inf|t1|≥δ,t2≤0

log

(

et1r/2(cosh(t1j))k−rE et(2S

2R−r)

)

≤ − inf|t1|≥δ,t2≤0

log

(

e−r log 2(cosh(t1j))k−2rE et(2S

2R−r)

)

≤ − inft2≤0

log

(

e−r log 2(cosh δ)k−2rE et(2S2R−r)

)

= − inft2≤0

logE et2(2S2R−r) + r log 2− (k − 2r) log cosh δ.

Since cosh(δ) > 1, the last term tends to −∞ for k →∞.Thus, assuming t1 ≥ δ leads to − limn→∞

1n logP(E1) < 0,

which is a contradiction.

2

To complete the proof, we let r →∞.

Lemma V.4: For r →∞,

− limn→∞

1

nlogP(E1) ≤

1

2log 2− 1

4+ o(1).

Proof: We show

limr→∞

− inft2≤0

logE et2(2S2r−r) =

1

2log 2− 1

4.

The first step is to show that whenever t2 ≤ −1/r, h(t2) =E et2(2S

2r−r) > 1. We then can conclude that the infimum is

not attained, since h(0) = 1 < h(t2) for t2 ≤ −1/r. Sinceh(t2) is a moment-generating function, it is log-convex, sothat it suffices to show that h(−1/r) > 1. Indeed, for allt2 ≤ −1/r, there exists an α ∈ (0, 1], such that αt2 =−1/r. It now follows that

1 < h(αt2) = h(

αt2 + (1− α) · 0)

≤ αh(t2) + (1− α)h(0) = αh(t2) + (1− α),

so that h(t2) > 1. In order to prove that h(−1/r) > 1, weobserve that

e(−1r )(2S2

r−r) = e1−2( Sr√

r)2 D→ e1−2Z2

,

where Zd= N (0, 1). Furthermore, since E e(−

αr )(2S2

r−r) ≤eα < ∞ for all α > 1, it then follows from [5] Example

7.10.15, that as r →∞,

E e1−2( Sr√

r)2 → E e1−2Z2

=e√5≈ 1.21565 > 1.

Indeed, the infimum is not attained. Finally, using thatt2 = −β/r for some β ∈ [0, 1), we can again use the argu-ment above to conclude that

E eβ−2β( Sr√

r)2 → E eβ−2βZ2

=eβ√

1 + 4β.

Minimizing over β gives β∗ = 1/4, resulting in

limr→∞

− inft2≤0

logE et2(2S2r−r) = − log

e1/4√2

=1

2log 2− 1

4,

which is the desired result. 2

This completes the proof of Theorem III.2(b) in the caseof equal powers. 2

VI. Conclusions

Investigated is a DS-CDMA system with HD-PIC. Themodel incorporates interfering users with unequal pow-ers. The processing gain is denoted by n and the num-ber of users by k. Large deviation theory is used to ob-tain qualitative statements concerning the performance of

17

the system. The exponential rate has been proven to be agood measure of quality. For the MF system, the exponen-tial rate is asymptotically equivalent with the well-knownsignal-to-noise ratio. We note that for the HD-PIC systemthe signal-to-noise ratio is not a good measure, since thereis no Gaussian behaviour for large k. However, the rate isnot based on (Gaussian) assumptions and this allows us todraw qualitative conclusions concerning the behaviour ofHD-PIC systems by using the rate of the BEP.

We have also investigated the asymptotic exponential rateof the BEP for multistage HD-PIC as the number of userstends to infinity. The asymptotic exponential rate of theBEP shows that HD-PIC keeps increasing performance asthe number of users increases!

We have shown that the exponential rate of the BEP re-mains unchanged after a finite number of HD-PIC stages.The number of stages of HD-PIC after which this happensis 1 for 3 up to 9 users, and 2 for 10 up to 22 users when allpowers are equal. This is the first analytical proof of thiseffect often observed in practice. We have shown that theexponential rate of the BEP remains uniformly positive asthe number of users increases. Under mild conditions onthe powers, we have identified the limiting exponential rateof the BEP after applying sufficiently stages of HD-PIC as12 log 2− 1

4 when the number of users increases.

The exponential rate have been used to give Chernoffbounds. The bound is not tight in the sense that it ap-proximates the BEP with any desired precision. Instead,it gives an upper bound that is valid for all n and k, makingit a robust upper bound. For s fixed, the Chernoff boundis valid as long as k = o( logn

n ), whereas for the optimalHD-PIC system, it is valid as long as k = o(n).

The analysis in this paper answers many questions. Nev-ertheless, many other questions remain open. We try tosummarize the most important ones:1. What happens to the results when AWGN and unequalpowers are considered in the case where we apply morethan one stage of HD-PIC? We expect that the result willbe generalized as in Theorem II.4. The proof of TheoremII.4 has been extended to include AWGN in [11].2. Can we include AWGN in the proof for the optimal HD-PIC system? We believe that we can prove upper and lowerbounds on the asymptotic exponential rate of the BEP us-ing the methods in this paper, but that these bounds arenot sharp. In order to prove asymptotics, new ideas willneed to be developed.3. It is well known that e−nI is not a good approximationto pn when − 1

n log pn → I. Can we compute the second or-der asymptotics, e.g. in the case when we apply one-stageon HD-PIC?4. Can we say more about how many iterations of HD-PICare necessary to obtain the optimal system? For instance,can we prove Conjecture III.5?

Appendix

A. Proof of Theorem III.2(b) for unequal powers

We recall that we have the following condition on the pow-ers (P1, . . . , Pk):

(P1) There exists a δ > 0 s. t. ]{j : Pj ∈ [δ, 1/δ]} → ∞,(P2) limk→∞ kP−1 <∞,(P3) k

−1Pmax → 0,(P4) kP

min →∞,

where Pmax = maxm Pm, Pmin = minm Pm.

The result of the statement is that under these power con-ditions, we have that

limk→∞

H (sk)

k =1

2log 2− 1

4.

Theorem III.2(a) states that H (sk)

k ≥ 12 log 2 − 1

4 , so thatit is sufficient to prove

limk→∞

H (sk)

k ≤ 1

2log 2− 1

4.

We will prove the theorem similar to the way we haveproven the result for equal powers. However, the deriva-tions will be more technical.

Similarly to the case in which all powers are equal, we willfocus on the event {R2 = R1 = R}. This specifies Rσ forall σ. We will show the theorem, using a symmetrizingargument. More precisely, we replace different powers bythe same value. In this case, we can use exchangeability,together with convexity arguments in order to simplify theanalysis. We intend to follow the strategy in Section V-Cas closely as possible.

We take R ⊂ {j : Pj ∈ [δ, 1/δ]} and we assume that |R| isfixed and odd and k is large.

Our starting point is the probability

P(R2 = R1 = R) = P(

maxm∈R

Z(1)

m ≤ 0, minm∈Rc

Z(1)

m ≥ 0,

maxm∈R

Z(2)

m ≤ 0, minm∈Rc

Z(2)

m ≥ 0

)

,

where

Z(1)

m =1

n

n∑

i=1

(

P 1/2m +

k∑

j=1j 6=m

P1/2j AjiAmi

)

, (61)

Z(2)

m =1

n

n∑

i=1

(

P 1/2m + 2

∑

j∈Rj 6=m

P1/2j AjiAmi

)

. (62)

Since P(A ≤ 0, B ≤ 0) ≥ P(A − B/2 ≤ 0, B ≤ 0), we canbound the probability from below by

P(

maxm∈R

Z(1)

m −1

2Z(2)

m ≤ 0, minm∈Rc

Z(1)

m −1

2Z(2)

m ≥ 0,

maxm∈R

Z(2)

m ≤ 0, minm∈Rc

Z(2)

m ≥ 0

)

,

18

where for m ∈ R,

Z(1)

m −1

2Z(2)

m =1

n

n∑

i=1

(

1

2P 1/2m +

∑

j∈RcP

1/2j AjiAmi

)

,

Z(2)

m =1

n

n∑

i=1

(

P 1/2m + 2

∑

j∈Rj 6=m

P1/2j AjiAmi

)

,

while for m ∈ Rc

Z(1)

m −1

2Z(2)

m =1

n

n∑

i=1

(

1

2P 1/2m +

∑

j∈Rcj 6=m

P1/2j AjiAmi

)

Z(2)

m =1

n

n∑

i=1

(

P 1/2m + 2

∑

j∈RP

1/2j AjiAmi

)

.

We can further bound the probability of interest from be-low by the probability of the intersection of the events

maxm∈R,j∈Rc

1

n

n∑

i=1

AmiAji ≤ −(Pmax

Rc Pmax

R )1/2

2PRc,

maxm,j∈R, j 6=m

1

n

n∑

i=1

AmiAji ≤ −(Pmax

R )1/2

2(|R| − 1)(Pmin

R )1/2, (63)

minm,j∈Rc,j 6=m

1

n

n∑

i=1

AmiAji ≥ 0,

minm∈R,j∈Rc

1

n

n∑

i=1

AmiAji ≥ −(Pmin

Rc )1/2

2|R|(Pmax

R )1/2,

where Pmax

R = maxm∈R Pm, Pmin

R = minm∈R Pm.

Indeed, using∑

m∈Rc P1/2m ≥ PRc

(PmaxRc

)1/2, we have for m ∈ R

Z(1)

m −Z(2)m

2=

1

n

n∑

i=1

(

1

2P 1/2m +

∑

j∈RcP

1/2j AjiAmi

)

≤ 1

2(Pmax

R )1/2 −∑

j∈RcP

1/2j

(Pmax

Rc Pmax

R )1/2

2PRc

≤ 1

2(Pmax

R )1/2 − (Pmax

R )1/2

2= 0.

Furthermore,

Z(2)

m =1

n

n∑

i=1

(

P 1/2m + 2

∑

j∈Rj 6=m

P1/2j AjiAmi

)

≤ (Pmax

R )1/2 − 2∑

j∈Rj 6=m

(Pmin

R )1/2(Pmax

R )1/2

2(|R| − 1)(Pmin

R )1/2

≤ (Pmax

R )1/2 − (|R| − 1)(Pmax

R Pmin

R )1/2

(|R| − 1)(Pmin

R )1/2= 0.

For m ∈ Rc, similar arguments give the desired result:

Z(1)

m −Z(2)m

2=

1

n

n∑

i=1

(

1

2P 1/2m +

∑

j∈RcP

1/2j AjiAmi

)

≥ 1

2P 1/2m ≥ 0,

Z(2)

m =1

n

n∑

i=1

(

P 1/2m + 2

∑

j∈RP

1/2j AjiAmi

)

≥ (Pmin

Rc )1/2− 2

∑

j∈R(Pmax

R )1/2(Pmin

Rc )1/2

2|R|(Pmax

R )1/2= 0.

In order to obtain a positive probability, we must assurethat

− (Pmax

Rc Pmax

R )1/2

2PRc≥ − (Pmin

Rc )1/2

2|R|(Pmax

R )1/2, (64)

or equivalently,

(Pmax

R )2Pmax

Rc |R|2Pmin

Rc P2Rc

≤ 1.

We use Pmax

Rc ≤ Pmax, together with Pmin

Rc ≥ Pmin andPmax

R ≤ 1/δ, to obtain from (P2), (P3) and (P4) that

(Pmax

R )2Pmax

Rc |R|2Pmin

Rc P2Rc

≤ δ−2|R|2 Pmax

P 2Pmin

= δ−2|R|2 k2

P 2

Pmax

k

1

kPmin,

which tends to zero for k → ∞. Hence, the inequalityholds, since |R| is fixed and k is sufficiently large.

We have now symmetrized the problem. The next step is toobserve that the events are exchangeable form, j. Togetherwith convexity of the rate function, we can replace the maxand min by

∑

, as we will now show. This is a standardlarge deviation argument.

We write ~a 7→ Jk(~a) to be the rate function of the cross

correlations ~W =(

Wjm

)

1≤j<m≤k, where

Wjm =1

n

n∑

i=1

AijAim. (65)

This means that, for any set A, we let

Jk(A) = − limn→∞

1

nlogP( ~W ∈ A), (66)

and we can compute Jk(A) using

Jk(A) = min~a∈A

Jk(~a). (67)

This rate function ~a 7→ Jk(~a) exists by Cramer’s Theorem,is convex by [8], Thm. III.27, and is minimal for ~a = ~0,where it takes the value Jk(~0) = 0. We note that

H (sk)

k ≤ min~a∈A

Jk(~a), (68)

where A is defined to be the set of vectors such that thecoordinates satisfy the inequalities in (63). We now let

E1 =

{

∑

m,j∈Rm6=j

1

n

n∑

i=1

AmiAji ≤ −|R|(Pmax

R )1/2

2(Pmin

R )1/2

}

,

19

E2 =

{

∑

m∈Rj∈Rc

1

n

n∑

i=1

AmiAji ≤ −|R||Rc| (Pmax

Rc Pmax

R )1/2

2PRc

}

,

E3 =

{

∑

m,j∈Rcm6=j

1

n

n∑

i=1

AmiAji ≥ 0

}

, (69)

E4 =

{

∑

m∈Rj∈Rc

1

n

n∑

i=1

AmiAji ≥ −|R||Rc| (Pmin

Rc )1/2

2|R|(Pmax

R )1/2

}

.

Then we see that

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ E4) = min

~b∈BJk(~b), (70)

where B is the set such that the inequalities in (69) hold.Note that A ⊂ B, and we wish to show that the minimumin (70) is attained for a ~b∗ ∈ A. That proves that the rateon the r.h.s. of (68) equals the rate of P(E1∩E2∩E3∩E4).

Let ~b∗ be the minimizer of (70). Note that by exchange-ability, if we interchange the elements in R and in Rc, weobtain the same rate. We write PR for the set of permuta-tions of the elements in R and PRc the set of permutationsof the elements in Rc, and we define P = PR×PRc . Hence,for a p ∈ P, we have p = (p1, p2) where p1 ∈ PR, p2 ∈ PRc .Then we obtain that

min~b∈B

Jk(~b) =1

|P|∑

p∈PJk(p(~b

∗)). (71)

By convexity, we obtain that

min~b∈B

Jk(~b) ≥ Jk

( 1

|P|∑

p∈Pp(~b∗)

)

. (72)

We have ended up with the vector ~a∗ = 1|P|∑

p∈P p(~b∗).

This vector takes only three different values, as we willexplain now. Indeed, let j 6= j ′ be any two indices in R,and m 6= m′ any two indices in Rc. Then a∗nl takes thevalue a∗jj′ for n, l ∈ R, the value a∗jm for n ∈ R, l ∈ Rc anda∗mm′ for any n, l ∈ Rc. Hence, we must have that eachof the elements satisfies the inequalities in (63), so that~a∗ ∈ A. This completes the proof.

We continue to determine the rate of P(E1∩E2∩E3∩E4).We write

P(E1∩E2∩E3) = P(E1∩E2∩E3∩E4)+P(E1∩E2∩E3∩Ec4).

By the largest-exponent-wins principle, we have

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3)

= min

{

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ E4),

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ Ec

4)

}

.

We wish to show that for k →∞ and |R| large,

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3) (73)

= − limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ E4).

In order to do so, we show that for all R

lim infk→∞

− limn→∞

1

nlogP(E1∩E2∩E3∩Ec

4) ≥1

2log

3

4+

1

4. (74)

Furthermore, we will show that

− limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3) = − lim

n→∞1

nlogP(E1)

(75)and, when |R| → ∞

− limn→∞

1

nlogP(E1) ≤

1

2log 2− 1

4+ o(1). (76)

This implies directly that when |R| is large

− limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ E4)

= − limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3)

≤ − limn→∞

1

nlogP(E1) ≤

1

2log 2− 1

4+ o(1).

Indeed, we note that 12 log

34 + 1

4 > 12 log 2 − 1

4 . Whenx ≤ min{y, z} and x < z, then necessarily x = y. We willapply this principle with

x = − lim|R|→∞

limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3),

y = − lim|R|→∞

limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ E4),

z = − lim|R|→∞

limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ Ec

4).

Indeed, (75) and (76) show that

x = − lim|R|→∞

limk→∞

limn→∞

1

nlogP(E1∩E2∩E3) ≤

1

2log 2− 1

4.

Equation (73) shows that z > x, so that necessarily y = x.Moreover, (76) shows that x ≤ 1

2 log 2− 14 . This proves the

claim, since it gives that

y = − lim|R|→∞

limk→∞

limn→∞

1

nlogP(E1∩E2∩E3) ≤

1

2log 2− 1

4.

(77)

The remainder of this proof therefore focuses on proving(74), (75) and (76). We prove (74) in the following lemma,(75) in Lemma A.2 and (76) in Lemma A.3.

Lemma A.1: For k sufficiently large, under power condi-tions (P1) and (P4),

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ Ec

4) ≥1

2log

3

4+

1

4.

20

Proof: We have

− limn→∞

1

nlogP(E1 ∩E2 ∩ E3 ∩ Ec

4)

≥ − limn→∞

1

nlogP(Ec

4)

= supt4≤0− log

{

E et4|Rc|

(PminRc

)1/2

2(PmaxR

)1/2+t4

∑

m∈R,j∈Rc Am1Aj1}

.

Recall the definition of SR, which is SR =∑

m∈RAm1.Since Rc and R are by definition disjoint, and cosh(t) ≤et

2/2, it follows that

E et4SRcSR = E (cosh(t4SR))|Rc| ≤ E e|R

c|t24S2R/2.

Furthermore, by (40), this is further bounded from aboveby (1− t24|R||Rc|)−1/2, as long as t24|R||Rc| ≤ 1. Note thatconditions (P1) and (P4) imply that Pmin

Rc |Rc| ≥ Pmax

R |R|when k is sufficiently large, because Pmax

R |R| is boundedand |Rc| = k(1− o(1)). Substituting this yields

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ Ec

4)

≥ supt4≤0−t4|Rc|(Pmin

Rc )1/2

2(Pmax

R )1/2+

1

2log(1− t24|R||Rc|)

≥ supt4≤0−t4|R|1/2|Rc|1/2

2+

1

2log(1− t24|R||Rc|).

We substitute t4 = − 12|R|1/2|Rc|1/2 to obtain a lower bound.

Observe that the substituted t4 indeed fulfills t24|R||Rc| ≤1. Substitution of the above result yields

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3 ∩ Ec

4) ≥1

2log

3

4+

1

4.

2

The second lemma is more involved than the one above.

Lemma A.2: For |R| fixed and odd, under condition (P1−P4),

− limk→∞

limn→∞

1

nlogP(E1 ∩ E2 ∩ E3) ≤ − lim

n→∞1

nlogP(E1).

Proof: Cramer’s Theorem gives that

− limn→∞

1

nlogP(E1 ∩ E2 ∩ E3) = sup

t∈D− logE eY1+Y2+Y3 ,

where t = (t1, t2, t3), D = {t : t1, t3 ≤ 0, t2 ≥ 0} and

Y1 = t1

( |R|(Pmax

R )1/2

2(Pmin

R )1/2+∑

m,j∈Rm6=j

Am1Aj1

)

,

Y2 = t2

(

∑

m,j∈Rcm6=j

Am1Aj1

)

,

Y3 = t3

(

−|R||Rc| (Pmax

Rc Pmax

R )1/2

2PRc+

∑

m∈R,j∈RcAm1Aj1

)

.

We rewrite this, invoking SA =∑

j∈AAj1 as

Y1 = t1

(

|R|(

(Pmax

R )1/2

2(Pmin

R )1/2− 1

)

+ S2R

)

,

Y2 = t2

(

−|Rc|+ S2Rc

)

, (78)

Y3 = t3

(

−|R||Rc| (Pmax

Rc Pmax

R )1/2

2PRc+ SRSRc

)

.

It is sufficient to prove that for k →∞, for all t1, t2

E eY1+Y2+Y3 ≥ E eY1(1− o(1)).

Indeed, then

− limn→∞

1

nlogP(E1) = sup

t1≤0− logE eY1

≤ supt1≤0{− logE eY1+Y2+Y3}+ o(1)

≤ supt∈D{− logE eY1+Y2+Y3}+ o(1),

so that for k → ∞, the statement of the lemma follows.We abbreviate

h(t1, t2, t3) = E eY1+Y2+Y3

We will prove that

h(t1, t2, t3) ≥ E eY1(1− o(1)), k →∞. (79)

Instead of maximizing − log h(t1, t2, t3), we will minimizeh(t1, t2, t3). To do this, we will define an appropriate ellipseE with elements t2, t3 with (0, 0) ∈ E0, the interior of E .In order to show that the supremum of − log h(t1, t2, t3) isattained in E0, it is sufficient to show that on the boundaryof the ellipse h(t1, t2, t3) > E eY1 . Since h(t1, 0, 0) = E eY1

and h is log-convex (since it is a moment-generating func-tion), we can then conclude that h(t1, t2, t3) > E eY1 out-side the ellipse. Indeed, whenever (t2, t3) 6∈ E , there existsa unique 0 < α < 1 such that α(t2, t3) ∈ ∂E . From con-vexity of h and h

(

t1, α(t2, t3))

> E eY1 it follows that

E eY1 < h(α(t1, t2), t3) = h(α(t1, t2) + (1− α)(0, 0), t3)

≤ αh(t1, t2, t3) + (1− α)h(0, 0, t3)

= αh(t1, t2, t3) + (1− α)E eY1 .

It immediately follows that h(t1, t2, t3) > E eY1 , so that(t1, t2, t3) cannot be the minimizer of h(t1, t2, t3) over D.When (t2, t3) ∈ E0, we can prove h(t1, t2, t3) ≥ E eY1(1 −o(1)) for k →∞.

Since ex ≥ 1 + x+ x2/2 + x3/6, we have

E eY1+Y2+Y3 ≥E(

eY1(1+Y2+Y3+1

2(Y2+Y3)

2+1

6(Y2+Y3)

3)

)

.

We will first calculate the required moments. Since |R| isodd, we have that S2

R ≥ 1, so that

E eY1Y2 = 0, E eY1Y2Y3 = 0,

21

E eY1Y3 = E eY1

(

−t3|R||Rc| (Pmax

Rc Pmax

R )1/2

2PRc

)

,

E eY1Y 22 = E eY1t22(3|Rc|2 + 2|Rc|) ≥ E eY33t22|Rc|2,

E eY1Y 23 = E eY1

(

t23|R|2|Rc|2Pmax

Rc Pmax

R

4P 2Rc

+ t23S2R|Rc|

)

≥ E eY1t23|Rc|. (80)

Concerning E eY1(Y2 + Y3)3,

E eY1(Y2 + Y3)3 ≥ −4E eY1(|Y 3

2 |+ |Y 33 |).

Using (78) and |SR| ≤ |R| results in

|Y 33 | ≤ 4|t3|3

(

|R|3|Rc|3 (Pmax

Rc )3/2(Pmax

R )3/2

8P 3Rc

+ |R|3|S3Rc |)

,

and

|Y 32 | ≤ 4t32

(

|Rc|3 + |S6Rc |)

,

so that (using E |SmRc | ≤ C|Rc|m/2 for some C <∞)

−E eY1(|Y 32 |+ |Y 3

3 |)

≥ −E eY1

(

C|t3|3|R|3|Rc|3 (Pmax

Rc )3/2(Pmax

R )3/2

P 3Rc

+C|t3|3|R|3|Rc|3/2 + Ct32|Rc|3)

.

We next use the moments of (80) to obtain

h(t1, t2, t3)

≥ E eY1

(

1 + 2|Rc|2t22 − t3|R||Rc| (Pmax

Rc Pmax

R )1/2

2PRc

+t23|Rc|

2+ e(t2, t3)

)

(81)

= E eY1

(

1−H+3|Rc|2

2t22 +

|Rc|2

(t3 − t∗3)2 + e(t2, t3)

)

,

where

H =|R|2|Rc|Pmax

Rc Pmax

R

8P 2Rc

,

t∗3 = −|R|(Pmax

Rc Pmax

R )1/2

2PRc,

e(t1, t2) = −Ct32|Rc|3 − C|t1|3|R|3|Rc|3/2

−C|t1|3|R|3|Rc|3 (Pmax

Rc )3/2(Pmax

R )3/2

P 3Rc

.

According to (P1), we have that PR ≤ Pmax

R |R| ≤ δ−1|R|is bounded, so that P = PRc + PR = PRc(1 − o(1)). Thisimplies, using (P1 − P4), that as k →∞,

H ≤ |R|2kPmaxδ−1

8P 2(1− o(1))=|R|2δ−1

8

Pmax

k

k2

P 2(1 + o(1))→ 0.

(82)We define the ellipse E as

E =

{

(t2, t3) :3|Rc|2

2t22 +

|Rc|2

(t3 − t∗3)2 ≤ 2H

}

.

To show that on ∂E , h(t1, t2, t3) > E eY1 , it is sufficient toshow that for every |R| fixed

|e(t2, t3)| = o(H). (83)

Indeed, on ∂E , according to (82)

h(t1, t2, t3) = E eY1

(

1−H+ 2H+ e(t2, t3)

)

> E eY1 .

Therefore, we focus on e(t2, t3) for (t2, t3) ∈ ∂E . Note thatfor (t2, t3) ∈ E by the triangle inequality,

|t2| ≤1√6

|R|(Pmax

Rc Pmax

R )1/2

|Rc|1/2PRc, (84)

|t3| ≤|R|(Pmax

Rc Pmax

R )1/2

PRc. (85)

Substituting this into the formula for e(t2, t3) gives

|e(t2, t3)| ≤ C|R|3(Pmax

Rc Pmax

R )3/2

P 3Rc

×(

|R|3|Rc|3 (Pmax

Rc Pmax

R )3/2

P 3Rc

+ |R|3|Rc|3/2)

+C|R|3(Pmax

Rc Pmax

R )3/2

|Rc|3/2PRc|Rc|3

≤ C(H3 + |R|3H3/2) + CH3/2.

Since H = o(1) according to (82) and |R| is fixed, indeed|e(t2, t3)| = o(H). This proves that

h(t1, t2, t3) > E eY1 , (t2, t3) 6∈ E . (86)

Finally, for (t2, t3) ∈ E0, we use the results above to obtain

h(t1, t2, t3) ≥ E eY1

(

1−H+3|Rc|22

t22+|Rc|2

(t3−t∗3)2−o(H))

.

Minimization over t2 and t3 of the right hand side clearlygives t2 = 0 and t3 = t∗3, so that for all (t2, t3) ∈ E

h(t1, t2, t3) ≥ E eY3

(

1−H− o(H))

.

Finally, since H = o(1), 1−H−o(H) = 1−o(1). Togetherwith (86), this proves the claim in (79). 2

Lemma A.3: For |R| → ∞, under condition (P1),

− limn→∞

1

nlogP(E1) ≤

1

2log 2− 1

4+ o(1).

Proof: Clearly,

− limn→∞

1

nlogP(E1) = sup

t1≤0

{

− logE eY1

}

,

where

Y1 = t1

(

|R|(

(Pmax

R )1/2

2(Pmin

R )1/2− 1

)

+ S2R

)

.

22

When we prove

lim|R|→∞

supt1≤0

{

− logE eY1

}

≤ 1

2log 2− 1

4, (87)

we conclude that for |R| sufficiently large,

− limn→∞

1

nlogP(E1) ≤

1

2log 2− 1

4+ o(1).

Thus, it remains to show (87). We denote

jR =1

2

(Pmax

R )1/2

(Pmin

R )1/2− 1/2.

The crucial step in this part of the proof is that we are stillallowed to choose R. So far, we have only assumed thatR ⊂ {j : Pj ∈ [δ, 1/δ]} and that |R| is odd. By condition(P1), we can choose R such that whenever |R| = r, thereexists a P with

maxm∈R

|Pm − P | ≤ P

r.

Thus, we know

Pmax

R ≤ P (1 + 1/|R|)P and Pmin

R ≥ P (1− 1/|R|)P ,

so that for |R| → ∞,

0 ≤ jR ≤1

2

√

1 + 1/|R|(1− 1/|R|) −

1

2=

√

1 +2

|R| − 1− 1→ 0.

We next write

− inft1≤0

logE exp

(

t1

(

Pmax

R

2Pmin

R

− 1

2

)

+ t1S2R

)

= − inft1≤0

logE exp

(

t1jR + t1(S2R − |R|/2)

)

.

We can now follow the argument in the proof of Lemma

V.4, together with the observation that whenever XnD→ X

and cn → c that also cnXnD→ cX. Indeed, in all weak

convergence arguments −1/|R| ≤ t1 ≤ 0, so that we areallowed to replace et1jR by its limit 1. This proves thatindeed for |R| large

− inft1≤0

logE eY1 ≤ 1

2log 2− 1

4+ o(1).

2

This completes the proof of Theorem III.2(b) in the caseof unequal powers. 2

Acknowledgement

The work of M.J.K. was partly supported by a grant of TheNetherlands Organization for Scientific Research (NWO),and by a grant of the Stieltjes Institute for Mathemat-ics. The work of R.v.d.H. was supported in part by theUniversity of British Columbia during a visit in October-December 2001.

References

[1] Buehrer, R.M. and Woerner, B.D. (1996). Analysis of adap-tive multistage interference cancellation for CDMA using an im-proved Gaussian approximation, IEEE Trans. Comm., 44, 10,1308-1329.

[2] Buehrer, R.M., Nicoloso, S.P. and Gollamudi, S. (1999).Linear versus non-linear interference cancellation, Journal of Com-munications and networks, 1, 2.

[3] Dembo, A. and Zeitouni, O. (1992). Large Deviations Tech-niques and Applications, Jones and Barlett Publishers, England.

[4] Gold, R. (1967). Optimum binary sequences for spread spectrummultiplexing, IEEE Trans. Inform. Theor., IT-13, 619-621.

[5] Grimmett, G.R. and Stirzaker, D.R. (1992). Probability andRandom Processes, second edition, Oxford University Press, Ox-ford, New York.

[6] van der Hofstad, R., Hooghiemstra G. and Klok M.J.(1999). Large deviations for code division multiple access systems,accepted for publication, SIAM J. on Appl. Math.

[7] van der Hofstad, R. and Klok M.J. (2001). Improving theperformance of third generation wireless communication systems,Preprint.

[8] den Hollander, F. (2000). Large Deviations, Fields InstituteMonographs, AMS, USA.

[9] Juntti, M. (1998). Multiuser Demodulation for DS-CDMA Sys-tems in Fading Channels, Ph.D. thesis, University of Oulu, Fin-land.

[10] Kasami, T. (1966). Weight distribution formula for some classof cyclic codes, Coordinated Science Lab, Univ. Illinois, Urbana,Tech.Rep. R-285.

[11] Klok M.J. (2001). On the asymptotic rate of error probabilitiesin third generation wireless systems, Preprint.

[12] Klok M.J. (2001). Importance sampling for DS-CDMA sytemeswith 1-stage HD-PIC, WPMC 2001, Aalborg, Denmark, Septem-ber 2001.

[13] Klok M.J. (2001). Performance Analysis of Advanced ThirdGeneration Receivers, Ph.D. thesis Delft University of Technology.

[14] Latva-aho, M. (1999). Advanced Receivers for WidebandCDMA Systems, Ph.D. thesis, University of Oulu, Finland.

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[19] Verdu, S. (1986). Minimum probability of error for asyn-chronous Gaussian multiple access IEEE Trans. Inform.Theory,IT-32, No. 1, 85-96.

[20] Xue, G., Weng, J., Le-Ngoc, T. and Tahar, S. (1999). Ananalytical model for performance evaluation of multistage parallelinterference cancellation detectors in CDMA systems, proceedingsof VTC’99 Fall, Amsterdam, the Netherlands, 1144-1149.

Remco van der Hofstad was born in Eind-hoven, the Netherlands, on May 3, 1971. Hereceived his M.Sc. degree with honors at theDepartment of Mathematics of the Universityof Utrecht in 1993, and subsequently did hisPh.D. at the same university. He receivedhis Ph.D. degree in 1997 on research in Sta-tistical Physics. In the next year, he was apost-doc at McMaster University in Hamilton,Canada, and worked at Microsoft Research inRedmond, U.S.A. In 1998, he obtained an as-

sistant professor position at Delft University of Technology. His re-search is in applications of probability, mostly in Statistical Physicsand in Electrical Engineering. In Delft, he started with researchon the modelling of the Internet as a graph, and on improving theperformance of CDMA systems. In all his research in Electrical En-gineering, the training in Statistical Physics, and Probability in gen-eral, turned out to be extremely relevant. Starting February 2002,Remco will move to the Technical University in Eindhoven, where heobtained an associate professorship.

23

Marten J. Klok was born in Utrecht, theNetherlands, on September 16, 1975. He re-ceived the M.Sc. degree (with honors) in tech-nical mathematics from Delft University ofTechnology in 1998. Since May 1998, he ob-tained a research position at the IRCTR, DelftUniversity of Technology. Currently, he worksat the mathematics department at Delft Uni-versity of Technology, where he is working to-wards his Ph.D. His scientific interests are inthe field of CDMA, interference cancellation

and probability theory, particularly large deviation theory.