Interference Channel with Binary Fading: Effect of ... - CiteSeerX

Interference Channel with Binary Fading: Effect of Delayed Network State Information Alireza Vahid† , Mohammad Ali Maddah-Ali∗ , and A. Salman Avestimehr† † Cornell University, Ithaca, NY, USA ∗ Bell Labs, Alcatel-Lucent, Holmdel, NJ, USA Abstract—We investigate the fundamental limits of communication over binary interference channels with finite states and with delayed network state information at the transmitters. Our results include a novel achievability strategy that systematically utilizes the stale network state information available at the nodes. We also derive new bounds on the capacity region of such networks.

I. I NTRODUCTION One of the practical constraints that limits the access of the transmitters to up-to-date Network State Information (NSI) is the inherent delay in the process of channel estimation and acquisition through feedback channel. Recently, there has been a growing interest in studying this issue in wireless networks. In practical settings, network state information is provided to the transmitters through the feedback channel which is based on previous observations at the receivers. The feedback channel can be of one of the following forms: 1) Output Feedback: In this case, each receiver based on its previous observations, can create a feedback message and send it to a subset of transmitters [1–4]. 2) Delayed-NSI: In this case, only the previous channel gain realizations is provided to the transmitters [5–8]. In this paper, we focus on the latter scenario. A conventional approach to address the feedback delay is to predict the current NSI exploiting the time correlation in channel parameters. However, if the delay becomes comparable to the coherence time of the channel, the current NSI becomes uncorrelated with the delay and therefore prediction schemes fail. In [5], it is shown that even in these scenarios, the completely stale NSI is still very useful and can change the scale of the capacity, measured by degrees of freedom (DoF). The main idea behind the algorithm is that instead of predicting future network state information, we should focus on sideinformation provided in past signaling stages. The point is that the side-information is evaluated based on outdated channel information. This idea has been initially proposed for channels with ARQ feedback [9–13]. The idea is that if the transmitter sends data to one receiver, but is informed through ACK/NACK feedback that the message is received by the unintended receiver, then this information can be exploited in later stages of communication to reduce the time needed for communication. This idea has been used to develop achievable schemes for wireless interference and X channels [6, 7, 14– 16]. However, except some particular examples, the achievable schemes are not optimal in general. This motivates us to

look at interference networks with delayed-NSI available at the transmitters. As a stepping stone, we consider a two-user binary interference channel with binary fading (finite states) to gain a deeper understanding of such scenarios. Our contribution in this paper is two-fold: 1) We first derive a new achievable rate region for the two-user binary interference channel with delayed-NSI available at the transmitters. This achievable region is strictly larger than the capacity region of the case where transmitters only have the knowledge of the distribution from which the channel gains are drawn, but not the actual value of the channel gains. This result is interesting since it shows that although channel gains are distributed independently from each other and over time, still the outdated channel gain information at the transmitters could be used to improve the achievable rate region. 2) Second, we derive a new outer-bound on the capacity region that is strictly smaller than that of the case where transmitters have instantaneous knowledge of the network state information. We incorporate a novel approach in order to obtain the outer-bound. This outerbound shows that outdated network state information is strictly suboptimal compared to the full network state information in terms of the capacity region. The rest of the paper is organized as follows. In Section II, we formulate our problem and give a brief overview of the channel model that we consider in this paper. In Section III, we present our main results. In Section IV, we derive the capacity region for two extreme cases of NSI knowledge at the transmitters and compare it with our results. We will provide our achievability scheme in Section V. Next in Section VI, we describe our new outer-bound on the capacity region of the two-user binary IC with delayed-NSI. II. P ROBLEM S ETTING We consider an interference channel (IC) with two transmitters and two receivers as shown in Figure 1. Each transmitter Txi wishes to communicate an independent message Wi to its corresponding receiver Rxi , i = 1, 2. The channel gain from transmitter Txi to receiver Rxj at time instant t is denoted by Cij [t], i, j ∈ {1, 2}. We assume that the channel gains are either 0 or 1 (i.e., Cij [t] ∈ {0, 1}), and they are distributed as i.i.d. Bernoulli 0.5 (B(0.5)) random variables (independent from each other and over time).

At each time instant t, the transmit signal at transmitter Txi is denoted by Xi [t] ∈ {0, 1}, i = 1, 2, and the received signal at receiver Rxi is given by Yi [t] = Cii [t]Xi [t] ⊕ C¯ii [t]X¯i [t],

i ∈ {1, 2},

(1)

where ¯i = 3 − i, and the summation is in F2 . Furthermore, the network state information (NSI) at time instant t is denoted by the quadruple C[t] = (C11 [t], C12 [t], C21 [t], C22 [t]). n

Tx1

n

W1

X1

C11 C

Rx1

(2)

n

ˆ1 W

n

ˆ1 W

Y1

n

12

n

1

C2 n

W2

X2

Tx2 Fig. 1.

Y2

n

C22

Rx2

Two-user binary interference channel.

We use the following notations in this paper. We use capital letters to denote random variables (RVs), e.g. Cij [t] is a random variable at time instant t. We use small letters to denote a particular realization of the corresponding random variable, e.g. cij [t] is a particular realization of random variable Cij [t]. Suppose S ⊆ Z+ is of size |S| > 0 and S = {s1 , s2 , . . . , s|S| } where s1 < s2 < . . . < s|S| . Then for random variable C, we define  0 C S = C[s1 ], C[s2 ], . . . , C[s|S| ] . (3) and cS is a particular realization of C S ,  0 cS = c[s1 ], c[s2 ], . . . , c[s|S| ] .

(4)

Moreover, we set ( i:j=

{i, i + 1, . . . , j} ∅

i ≤ j, i > j.

(5)

Also, for simplicity of notation, we represent C 1:i by C i , i ≥ 1. Finally, we set t Cij Xit

⊕

t C¯ij X¯it

(6) 0 = Cij [1]Xi [1] ⊕ C¯ij [1]X¯i [1], . . . , Cij [t]Xi [t] ⊕ C¯ij [t]X¯i [t] . 

Consider the scenario in which transmitter Txi wishes to reliably communicate message Wi ∈ {1, 2, . . . , 2nRi } to receiver Rxi during n uses of the channel, i = 1, 2. We assume that the messages and the channel gains are mutually independent and the messages are chosen uniformly. For each transmitter Txi , let message Wi be encoded as Xin using the encoding function fi (.), which depends on the available NSI at transmitter Txi . Receiver Rxi is only interested in decoding Wi , and it will decode the message using the decoding b i = gi (Yin , C n ). function W b i 6= Wi . The average probability An error occurs when W of error is given by b i 6= Wi ]], λi,n = E[P [W

i = 1, 2,

(7)

and the expectation is taken with respect to the random choice of the transmitted messages W1 and W2 . A rate tuple (R1 , R2 ) is said to be achievable, if there exists encoding and decoding functions at the transmitters and the receivers respectively, such that the decoding error probabilities λ1,n , λ2,n go to zero as n goes to infinity. In this paper, receivers will always have full knowledge of the NSI, and we consider three models for the available NSI at the transmitters: • Full-NSI: In this model, the network state information C t is available at each transmitter at any time instant t, t = 1, 2, . . . , n. • No-NSI: In this model, transmitters only know the distribution from which the channel gains are drawn, but not the actual realization of them. • Delayed-NSI: In this model, at time instant t, each transmitter has the knowledge of the network state information up to time instant t, i.e. C t−1 , and the distribution from which the future channel gains are drawn, t = 1, 2, . . . , n. III. M AIN R ESULTS Our contribution in this paper is two-fold, first we derive a new achievable rate region for the two-user interference channel described in Section II with delayed-NSI available at the transmitters. This achievable region is strictly larger than the capacity region of the no-NSI model. This result is interesting since it shows that although channel gains are distributed independently from each other and over time, still the outdated channel gain information at the transmitters could be used to improve the achievable rate region. Second, we derive a new outer-bound on the capacity region that is strictly smaller than that of the full-NSI model. We incorporate a novel approach in order to derive the outerbound. This immediately shows that outdated network state information can never be as good as full network state information in this problem. The following theorem presents our main results. Theorem 1. For the two-user binary interference channel described in Section II with delayed-NSI at the transmitters, the capacity region C delayed−NSI satisfies Rdelayed−NSI ⊆ C delayed−NSI ⊆ C¯delayed−NSI ,

(8)

where

Rdelayed−NSI

and C¯delayed−NSI

 R1 ≤ 21     R ≤1 2 2 = 29  R + 1  14 R2 ≤   29 14 R1 + R2 ≤  1   R1 ≤ 2 R2 ≤ 21 =   R1 + R2 ≤

9 7 9 7

(9)

(10) 11 12

The proof of Theorem 1 is presented in Sections V and VI.

R1

equal to  for sufficiently large n. It is straight forward to see that receiver Rx2 can also decode W2 with decoding error probability less than or equal to . Therefore, a sum-rate of 3 4 − 2δ is achievable with decoding error probability less than or equal to 3. In the following section, we describe our transmission strategy which results the achievable region of Theorem 1.

full-NSI

delayed-NSI outer-bound no-NSI delayed-NSI achievable

V. ACHIEVABILITY

R2 Fig. 2. Two-user IC with binary fading: capacity region with no-NSI and full-NSI, alongside the achievable region and outer-bound with delayed-NSI.

In the following section, we present the capacity region of the two-user interference channel described in Section II with no-NSI and full-NSI at the transmitters. IV. C APACITY R ESULTS FOR NO -NSI AND FULL -NSI For the two-user binary interference channel described in Section II with no-NSI and full-NSI, we have the following result. Lemma 2. The capacity region of the two-user binary interference channel described in Section II with no-NSI, C no−NSI , and with full-NSI, C full−NSI , are given by  1 (   R1 ≤ 2 R1 ≤ 12 full−NSI no−NSI 1 R2 ≤ 2 and C = C =  R2 ≤ 12  R1 + R2 ≤ 34 (11) Remark 3. Comparing the capacity region for the no-NSI and the full-NSI model with the achievable region and the outer-bound for the delayed-NSI model, we conclude that the capacity region of the two-user interference channel with delayed-NSI is strictly larger than that of the no-NSI model and strictly smaller than that of the full-NSI model, i.e. C no−NSI ⊂ C delayed−NSI ⊂ C full−NSI .

(12)

See Figure 2 for a depiction. The proof of Lemma 2 is straightforward and omitted due to space limitations. The idea to obtain the result for the fullNSI model is similar to that of [17]. We briefly describe the achievability scheme for the sum-rate with no-NSI, i.e. a sumrate of 3/4. The channel from transmitter Tx2 to receiver Rx1 can be modeled as a point-to-point erasure channel with erasure probability equal to 41 (any time C11 [t] = 1, we consider an erasure has happened). Hence from [18], we know that for any , δ > 0 and sufficiently large n, a rate of 14 − δ is achievable for transmitter Tx2 to receiver Rx1 with decoding error probability less than or equal to . Receiver Rx1 decodes W2 and removes X2n from its received signal. Now the channel from transmitter Tx1 to receiver Rx1 becomes a point-to-point erausre channel with erasure probability 21 (since C11 [t] = 0 with probability 12 ), and as a result, a rate of 21 −δ is achievable for tansmittre Tx1 with decoding error probability less than or


The achievability scheme for rate tuples (R1 , R2 ) = 12 , 14
and (R1 , R2 ) = 41 , 12 is the same as the one presented for Lemma 2. In the remaining of this section, we describe a transmission strategy that
achieves a rate tuple (R1 , R2 ) 18 , arbitrary close to 18 43 43 as n → ∞. Therefore with time sharing, we can achieve region Rdelayed−NSI as described in Theorem 1. Before going into details of our transmission strategy, we first provide a brief overview of it. A. Overview Our transmission strategy for the two-user IC with delayedNSI consists of two phases as follows. • Phase 1 (uncategorized transmission): During phase 1, each transmitter sends its bits without worrying about the decodability at the corresponding receiver. This phase will go on until a contribution of each bit is available at at least one of the receivers, i.e. for any bit, at least one of the receivers gets it with or without interference. • Phase 2 (categorizing the bits and retransmission): At the end of phase 1, the transmitters know the NSI of the first phase. Hence, each transmitter Txi , i = 1, 2, can categorize its bits into four states, as in Table I: (A) Qi→{1,2} : The bits that are required by both receivers. More precisely, a bit is in Qi→{1,2} if one of the following cases happen – both receivers get the bit with interference; – receiver Rx¯i gets the bit with interference and receiver Rxi does not get it interference-free. (B) Qi→i|¯i : The bits that are required by the intended receiver Rxi but are available at the unintended receiver Rx¯i . A bit is in Qi→i|¯i if receiver Rx¯i gets it without interference and receiver Rxi does not get it with or without interference. (C) Qi→¯i|i : The bits that are required by the unintended receiver Rx¯i but are available at the intended receiver Rxi . More precisely, a bit is in Qi→¯i|i if receiver Rxi gets the bit without interference and receiver Rx¯i gets it with interference. (D) Qi→F : The bits that are transmitted successfully and they do not fall in category (C). Based on this categorization, the bits in different states can be elevated to different sub-problems that can be solved more efficiently. In particular, the bits in (A) are of interest to both receivers, hence, they will be transferred to the sub-problem of two-source multicast, and as we will show in the following subsection a sum-rate of 3/4 is achievable in this sub-problem.

The bits in (B) and (C) will be transferred to the problem of the interference channel with side information, and as we will show in the following subsection a sum-rate of 1 is achievable in this sub-problem. B. Two Sub-problems Two-user IC with side-information: Consider the two-user IC described in Section II and assume that Wi is also available at receiver Rx¯i , i = 1, 2. We refer to this network as two-user IC with side-information. Lemma 4. For the two-user IC with side-information as described above, we have delayed−NSI no−NSI full−NSI Cside = Cside = Cside ,

(13)

and, we have ( full−NSI Cside

=

R1 ≤ R2 ≤

1 2 1 2

(14)

The proof of Lemma 4 is omitted due to space limitations. Here, we briefly describe the achievability scheme of Lemma 4. Since message Wi is available at receiver Rx¯i , receiver Rx¯i can create Xin and remove it from its received signal. After removing the interfering signal at each receiver, the channel between each transmitter and its corresponding receiver is an erasure channel with erasure probability 12 (since Cii [t] = 0 with probability 12 , i = 1, 2), and as a result, a rate of 12 − δ is achievable for tansmittre Tx1 with decoding error probability less than or equal to  for sufficiently large block length, see [18]. Two-source Multicast: Consider a network with two transmitters and two receivers where each transmitter Txi wishes to communicate an independent message Wi to both receivers, i = 1, 2. The channel gain model is the same as described in Section II. We refer to this network as the two-source multicast network. Lemma 5. For the two-source multicast network as described above, we have delayed−NSI no−NSI full−NSI Cmulticast = Cmulticast = Cmulticast ,

full−NSI Cmulticast

(16) 3 4

The proof of Lemma 5 is omitted due to space limitations. Here, we briefly describe the achievability scheme of Lemma 5. With no-NSI, it is easy to verify that I(X1n ; Y1n |C n ) = I(X1n ; Y2n |C n ).

We now formally describe the achievability strategy of Theorem 1. Define m = bnR1 c = bnR2 c, where R1 = R2 = 18 43 and n is such that m ∈ Z, and we assume W1 = a1 , a2 , . . . , am , and W2 = b1 , b2 , . . . , bm , where ai ’s and bi ’s are picked uniformly and independently from {0, 1}, i = 1, . . . , m. Note that we first transmit raw (uncoded) bits in our transmission strategy.  Phase 1 (uncategorized transmission): At the beginning of the communication block, we assume the bits at transmitter Txi are in state Qi→i (the initial state of the bits), i = 1, 2. Based on the channel gain realizations, a total of 16 possible configurations may occur at any time instant. In some cases, no change happens in the status of a transmitted bit of transmitter Txi (it has not been received with or without interference at either of the two receivers), and therefore such bits will remain in state Qi→i , i = 1, 2. In any other case, the transmitted bit is moved to a new state. Table I describes how transition from the initial state to other states takes place. In this table, a solid arrow from transmitter Txi to receiver Rxj represents Cij [t] = 1, i, j ∈ {1, 2}, t = 1, 2, . . . , n. Below, we describe how transitions from state Qi→i , i = 1, 2, to other states takes place for cases 1, 2, and 7, and the description for the remaining cases is removed due to space limitations. • Case 1 (C11 [t] = 1, C12 [t] = 1, C21 [t] = 1, and C22 [t] = 1): If at time instant t case 1 occurs, then each receiver gets a linear combination of the bits that were transmitted, t = 1, 2, . . . , n. Consider the two bits that each of the receivers has a linear combination of them. Now, if either of such bits is provided to both receivers then the receivers can decode both bits, see Figure 3(a). In other words, broadcasting one of the two bits is sufficient. To form the problem of twosource multicast with equal rate from each transmitter, we would like to have almost equal number of bits in Q1→{1,2} and Q2→{1,2} . To do so, if t is odd, then the transmitted bit of transmitter Tx1 joins Q1→{1,2} , and if t is even, then the transmitted bit of transmitter Tx2 joins Q2→{1,2} , t = 1, 2, . . . , n.

(15)

and, we have  1   R1 ≤ 2 R2 ≤ 12 =   R1 + R2 ≤

C. Transmission Strategy

(17)

In other words, for any scheme if receiver Rx1 can decode W1 , then so can receiver Rx2 and vice versa. Based on the argument presented above, the achievability scheme for Lemma 2 works for Lemma 5 as well.

Tx1

Rx1

Tx1

Rx1

Tx2

Rx2

Tx2

Rx2

(a)

(b)

Fig. 3. (a) Case 1, if either of the bits is broadcasted then the receivers can decode both bits; (b) Case 7, if the transmitted bit of transmitter Tx2 is broadcasted then the receivers can decode the intended bits.

• Case 2 (C11 [t] = 1, C12 [t] = 1, C21 [t] = 0, and C22 [t] = 1): In this case, receiver Rx1 has already received its corresponding bit while receiver Rx2 has a linear combination of the transmitted bits, see Table I. As a result, if the transmitted bit of transmitter Tx1 is provided to receiver Rx2 , it will be

TABLE I A LL POSSIBLE CHANNEL GAIN REALIZATIONS AND TRANSITIONS FROM THE INITIAL STATE TO OTHER STATES ; SOLID ARROW FROM TRANMSITTER Txi TO RECEIVER Rxj INDICATES THAT Cij [t] = 1, i, j ∈ {1, 2}, t = 1, 2, . . . , n. case ID

channel realization at time instant n Tx1

state transition

case ID

Rx1

Tx1

 1

channel realization at time instant n

a → Q1→{1,2} b → Q2→{1,2}

n odd n even

Rx1

Rx2

Tx2

Rx2

Tx1

Rx1

Tx1

Rx1

a → Q1→2|1 Rx2

Tx2

Rx2

Tx1

Rx1

Tx1

Rx1

b → Q2→1|2 Rx2

Tx2

Rx2

Rx1

Tx1

Rx1

4

a → Q1→{1,2}

12

Tx2

Rx2

Tx2

Rx2

Tx1

Rx1

Tx1

Rx1

 5

a → Q1→F b → Q2→2

Rx2

Tx2

Rx2

Tx1

Rx1

Tx1

Rx1

6

a → Q1→F b → Q2→2

(

a → Q1→1 b → Q2→2|1

(

a → Q1→1|2

13

Tx2



a → Q1→1 b → Q2→F

a → Q1→{1,2}

Tx1

a → Q1→F b → Q2→F



11

Tx2



a → Q1→1 b → Q2→F

10

Tx2

3

 9

Tx2

2

state transition

14

b → Q2→2

Tx2

Rx2

Tx2

Rx2

Tx1

Rx1

Tx1

Rx1

( b → Q2→{1,2}

7

15

b → Q2→2|1

Tx2

Rx2

Tx2

Rx2

Tx1

Rx1

Tx1

Rx1

 b → Q2→{1,2}

8 Tx2

Rx2

able to decode both. In other words, the transmitted bit from transmitter Tx1 is available at receiver Rx1 and is required by Rx2 . Therefore, transmitted bit from transmitter Tx1 joins Q1→2|1 . • Case 7 (C11 [t] = 1, C12 [t] = 0, C21 [t] = 1, and C22 [t] = 0): In this case, receiver Rx1 has a linear combination of the transmitted bits, while receiver Rx2 has not received anything, see Table I. Note that it is sufficient to provide the transmitted bit of transmitter Tx2 to both receivers, see Figure 3(b). Therefore, the transmitted bit of transmitter Tx2 joins Q2→{1,2} .  Phase 2 (categorizing the bits and retransmission): The aforementioned discussion provides the justification for categorizing the bits as mentioned before. To form the problem of the two-source multicast, the bits in states Q1→{1,2} and Q2→{1,2} will be transmitted together from the corresponding transmitters, i.e. Tx1 and Tx2 , respectively1 . In other words, we pair up bits from Q1→{1,2} and Q2→{1,2} to form the prob1 At any point in our description of the transmission strategy, if bits from different states are combined, add 0’s to the state of shorter length such that the states have equal length.

a → Q1→1|2

16 Tx2

a → Q1→1 b → Q2→2

Rx2

lem of the two-user multicast, see Figure 4. More precisely, the bits in Qi→{1,2} will be considered as the message of transmitter Txi and they will be encoded as in the achievability scheme of Lemma 5, i = 1, 2. 1→{1,2} 3

2

1

1

1

1

2

2

2→{1,2} 3

2

Fig. 4. We pair up bits from Q1→{1,2} and Q2→{1,2} to form the problem of the two-user multicast. Note that each receiver is interested in decoding the bits coming from both transmitters.

Similarly, we pair up bits in Q1→1|2 and Q2→2|1 to form the problem of the two-user IC with side information. We treat these bits as the message of the corresponding transmitter and they will be encoded as in the achievability scheme of Lemma 4. Interestingly the bits in Q1→2|1 and Q2→1|2 can be also

paired up to form another two-user IC with side information. More precisely, we swap the role of the transmitters, i.e. a bit in Qi→¯i|i is now of interest of receiver Rx¯i , i = 1, 2. In other words, the bits in Qi→¯i|i are already received by the intended receiver, however, they are used in order to help the unintended receiver to decode its corresponding bits. In the following subsection, we provide the analysis of the achievable sum-rate using our two-phase transmission strategy and its decoding error probability.

i = 1, 2. Upon completion of phase 1, since all transitions from the initial state to other states are equiprobable, we have

D. Analysis of the Transmission Strategy

and since transition of a bit to each state is distributed as i.i.d. Bernoulli RV, applying Chernoff bound, we have

In this subsection, we provide the analysis of our proposed transmission strategy. In particular, we show that with probability approaching 1 as the total number of bits, i.e. 2m, goes to infinity, the required time to successfully transmit all the bits is less than or equal to 13 2 43 m + m3 , 18 3

(18)

36 which results in a sum-rate of 43 bits per channel use. We break our analysis into 3 steps. In step 1, we evaluate the required time to finish phase 1. Step 2 is dedicated to evaluating the required time to finish phase 2. Finally, step 3 uses the results derived in the first two steps to calculate the achievable sum-rate. • Step 1 (required time to finish phase 1): The number of remaining bits in Qi→i during phase 1 is a Markov chain that at each time instant, the value will either decrease by 1 or remain unchanged, i = 1, 2. The total initial number of bits is equal to 2m. Let random variable T1 denote the number of time instants in which phase 1 will be over (hitting time for the aforementioned Markov chain). At the beginning of the communication block, all 2m bits are in Q1→1 and Q2→2 , and the other states are empty. As described in Table I, 12 out of 16 possible configurations result in a transition from state Qi→i for any bit at transmitter Txi , i = 1, 2. Since all the configurations occur with equal probability, the average number of time instants for a bit to leave the initial state is equal to 4/3. The transition from the initial state to other states is distributed as i.i.d. Bernoulli RV and hence, applying Chernoff bound, we get

E[T1 ] =

4 m, 3 4

m3 ! −   2 2 4 m+ 1 m 3 2 4 3 3 3 P r T1 ≥ m + m ≤ e . (19) 3 h i 2 We consider the event E = T1 ≥ 43 m + m 3 as error, and we halt the transmission strategy if E occurs at the end of phase 1. • Step 2 (required time to finish phase 2): In this step, we analyze the required time to transmit the bits in each of the remaining states. Let random variable Ni→{1,2} denote the number of bits in Qi→{1,2} ; Ni→i|¯i denote the number of bits in Qi→i|¯i ; and Ni→¯i|i denote the number of bits in Qi→¯i|i ,

E[N1→{1,2} ] = E[N2→{1,2} ] =

5 m, 12

1 m, 3 1 E[N1→2|1 ] = E[N2→1|2 ] = m, 6 E[N1→1|2 ] = E[N2→2|1 ] =

−

h

P r N ≥ E[N ] + m

i 2 3

≤e

(20)

4 m3 2 2 E[N ]+ 1 m 3 3

!

,

(21)

where N can be Ni→{1,2} , Ni→i|¯i , or Ni→¯i|i , i = 1, 2. If event h i 2 N ≥ E[N ] + m 3 occurs for any of the 6 states mentioned above, we consider it as error and we halt the transmission strategy. Upon completion of phase 1, we add 0’s (if necessary) in order to make all states of size equal to the corresponding 2 expected value plus m 3 (note that this can only increase the communication time). Therefore, as m → ∞, the number of bits in each state goes to infinity. We are now ready to caluculate the required time to transmit the remaining bits in phase 2. 1) Transmission time required for bits in Q1→{1,2} and Q2→{1,2} : We use the achievability scheme of the twosource multicast problem for the bits in states Q1→{1,2} and Q2→{1,2} . In other words, for fixed , δ ∈ R+ and for sufficiently large m, a sum-rate of 34 −δ is achievable with decoding error probability less than or equal to  using random block linear coding as in Lemma 5. Therefore for sufficiently large m, the required time (block length) to transmit the bits in states Q1→{1,2} 2

]+m 3

E[N

1→{1,2} and Q2→{1,2} is t2,1 = with decoding 3 4 −δ error probability less than or equal . 2) Transmission time required for bits in Q1→1|2 and Q2→2|1 : We use the achievability scheme of the IC with side-information for the bits in Q1→1|2 and Q2→2|1 . For sufficiently large m and for fixed , δ ∈ R+ , a sum-rate of 1 − δ is achievable with decoding error probability less than or equal to  as in Lemma 4. Therefore for sufficiently large m, the required time to transmit the 2

E[N

]+m 3

1→1|2 with bits in Q1→1|2 and Q2→2|1 is t2,2 = 1−δ decoding error probability less than or equal . 3) Evaluating the transmission time required for bits in Q1→2|1 and Q2→1|2 is similar to the previous case and 2

E[N

]+m 3

1→2|1 we have t2,3 ≤ , with decoding error 1−δ probability less than or equal .

As a result, for sufficiently large m, the bits in phase 2 can be communicated with decoding error probability less than or

equal to 3 in a total of t2 =

5 12 m + m 3 4 −

Then we have

2 3

2 3

1 3m

1 +m m+m + 6 1− 1−

+

2 3

(22)

Therefore, as m → ∞ and , δ → 0, we achieve a (symmetric) sum-rate of 2m 43 18 m

+

2 13 3 3 m

→

36 43

(24)

−

1 − P r [E] −

e

n n = H(Y1n |C n ) − H(C21 X2n |C n ) + H(C22 X2n |C n ) n X (c) 3 i−1 i−1 ≤ n+ H(C22 [i]X2 [i]|C22 X2 , C i−1 ) 4 i=1  i−1 i−1 −H(C21 [i]X2 [i]|C21 X2 , C i−1 ) n 1 X (d) 3 i−1 i−1 H(X2 [i]|C22 X2 , C i−1 ) = n+ 4 2 i=1  i−1 i−1 −H(X2 [i]|C21 X2 , C i−1 ) , (29)

where n → 0 as n → ∞; (a) follows from Fano’s and data processing inequalities, i.e. = I (X2n ; Y2n |X1n , C n ) + I (W2 ; Y2n |X1n , X2n , C n ) , (30)

4 m3 2 2 E[N ]+ 1 m 3 3

(b)

I (W2 ; Y2n |X1n , C n ) ≤ I (W2 , X2n ; Y2n |X1n , C n )

with probability greater than or equal to X

(a)

≤ I(X1n ; Y1n |C n ) + I(X2n ; Y2n |X1n , C n )

time instants. For any , δ > 0 and sufficiently large m, we have " # 2 2 1 5 3 3 2 4 12 m + m 2 m + 2m 3 P r T1 + t2 ≥ m + m + + 3 3 1− 4 −   2 4 = P r T1 ≥ m + m 3 = P r [E] . (23) 3

R1 + R2 =

n(R1 + R2 − 2n )

!

→ 1,

(25)

N

where N can be Ni→{1,2} , Ni→i|¯i , or Ni→¯i|i , i = 1, 2, and the decoding error probability is less than or euqal to 3 → 0. This completes the achievability proof of Theorem 1.

and the last term is 0; (b) holds since given C n , X1n and X2n are independent; (c) holds since H(Y1 [i]) ≤ 43 and the fact that X2 [i] is independent of C i:n ; and (d) is derived by conditioning on C22 [i] and C21 [i] and the fact that X2 [i] is independent of them. Pn i−1 i−1 Claim 6. 21 i=1 [H(X2 [i]|C22 X2 , C i−1 )

VI. C ONVERSE OF T HEOREM 1 Suppose there are enconding functions at the transmitters and decoding functions at the receivers such that rate tuple (R1 , R2 ) is achievable with decoding error probabilities going to zero, as the block length n goes to infinity. Suppose each message Wi is picked uniformly from {1, 2, . . . , 2nRi }, i = 1, 2, and that the messages are independent. To derive the outer-bound on R1 , we have (a)

nR1 ≤ I(X1n ; Y1n |X2n , C n ) + nn = H(Y1n |X2n , C n ) − H(Y1n |X1n , X2n , C n ) + nn 1 n = H(C11 X1n |X2n , C n ) + nn ≤ n + nn , (26) 2 where n → 0 as n → ∞; (a) follows from Fano’s and data processing inequalities and the fact that W1 , W2 and C n are mutually independent. Similar result holds for R2 . Dividing both sides by n and let n → ∞, we get Ri ≤ 1/2, i = 1, 2. To obtain the outer-bound on R1 + R2 , we have (a)

n(R1 + R2 ) = H(W1 ) + H(W2 ) = H(W1 |C n ) (b)

+ H(W2 |C n ) = H(W1 |C n ) + H(W2 |X1n , C n ),

(27)

where (a) holds since W1 , W2 and C n are mutually independent; and (b) holds since = I (X1n ; W2 |C n ) ≤ I (W1 , X1n ; W2 |C n ) = I (W1 ; W2 |C ) + I

(X1n ; W2 |W1 , C n )

n−1 1 1 n 1 X −i i4 . n− × n − 6 6 4 2 i=1

Using (29) and Claim 6, we have n(R1 + R2 − 2n ) ≤

n−1 3 1 1 n 1 X −i n+ n− × n − i4 , 4 6 6 4 2 i=1

(31)

dividing both sides by n and let n → ∞, we get R1 + R2 ≤ 11/12. In the remaining of this section, we provide the proof of Claim 6. However, before presenting the proof of Claim 6, we introduce a few notations that will be used in the rest of this i−1 i−1 section. From the channel gains only C22 and C21 appear, therefore, we can further simplify the notations. We define ˜ i−1 = [(C22 [1], C21 [1]) , . . . , (C22 [i − 1], C21 [i − 1])]0 . C

(32)

Moreover, for a given channel gain realization c˜k (a matrix of size k × 2) and a given set A where |A| = n − k and A ⊆ {2, 3, . . . , n−1}, we define ck,A = Π(˜ ck , A) as follows. Suppose {1, 2, 3, . . . , n} \ A = {i1 , i2 , . . . , ik−1 } where 1 = i1 < i2 < . . . < ik−1 = n, then ck,A is matrix of size n × 2, and  (0, 0) j ∈ A, th j row of ck,A = (33) `th row of c˜k j = i` . Definition 7. F(˜ cn ) =

0 ≤ H(W2 |C n ) − H(W2 |X1n , C n ) n

i−1 i−1 −H(X2 [i]|C21 X2 , C i−1 )] ≤

n h X

i−1 i−1 ˜ i−1 H(X2 [i]|C22 X2 , C = c˜i−1 )

i=1

= 0.

(28)

i i−1 i−1 ˜ i−1 −H(X2 [i]|C21 X2 , C = c˜i−1 ) .

(34)

Note that since X2i is independent of C˜ i:n , we can change the condition from C˜ i−1 = c˜i−1 to C˜ n = c˜n in (34). We prove Claim 6 in 2 steps. Step 1: Lemma 8. Fix a channel gain realization C˜ n = c˜n and assume @ i ∈ {1, 2, . . . , n − 1} : c˜[i] = (0, 0), we have

Step 2: Lemma 9. Fix a channel gain realization C˜ k = c˜k , and assume @ i ∈ {1, 2, . . . , k − 1} : c˜[i] = (0, 0), we have

F(˜ cn ) ≤ |{j : j < n, c˜[j] = (0, 1)}|.

where m is the number of indices for which the corresponding row of c˜k is equal to (0, 1) and n ≥ k. The proof of Lemma 9 is omitted. Using the results of the aforementioned two steps, we can prove Claim 6. This completes the proof of converse. VII. ACKNOWLEDGMENT The work of A. S. Avestimehr and A. Vahid is supported in part by the NSF CAREER award 0953117 and U.S. Air Force Young Investigator Program award FA9550-11-1-0064. R EFERENCES

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Proof: Suppose |{j < n : c˜[j] = (0, 1)}| = m, and {j < n : c˜[j] = (0, 1)} = {j1 , j2 , . . . , jm }, where j1 < j2 < . . . < jm , m ≥ 0. Set j0 = 0 and jm+1 = n. We have F(˜ cn ) =

n h X

i−1 i−1 ˜ n H(X2 [i]|C22 X2 , C = c˜n )

i=1

i i−1 i−1 ˜ n −H(X2 [i]|C21 X2 , C = c˜n ) jk+1 h m X X i−1 i−1 ˜ n H(X2 [i]|C22 X2 , C = c˜n )

=

k=0 i=jk +1

i i−1 i−1 ˜ n −H(X2 [i]|C21 X2 , C = c˜n ) jk+1 m X (a) X

≤

h

i−1 i−1 ˜ n H(X2 [i]|C22 X2 , C = c˜n )

k=0 i=jk +1

i {j:j