On the inner and outer bounds of 3-receiver broadcast channels with 2-degraded message sets

On 3-recei v er broadcast channels with 2-de graded message sets Chandra Nair Departmen t of I nform ation Engineering Chinese University of Hong K ong Sha T in, N.T ., Hong K ong Email: ch andra@ie.cu hk.edu .hk Zizhou V incent W a ng Departmen t of I nform ation Engineer ing Chinese University of Hong K ong Sha Tin, N.T ., Ho ng Kong Email: zzwang6@ie.cuhk .edu.h k Abstract — W e consider a broadcast ch annel with 3 receiv ers and 2 message s ( M 0 , M 1 ) where two of the three receiv ers need to decode messages ( M 0 , M 1 ) while t he r emainin g one just needs to decode the message M 0 . W e study the best k nown inner and outer bounds u nder this setting, in an attempt to find the deficiencies w ith the current techn iques of establi shing the bounds. W e produce a simple example where we ar e a ble to explicitly ev aluate the inn er b ound and show that it differs from the general outer bound. Fo r a class of channels where the general inner and outer bounds differ , we u se a new arg ument to show that the i nner bou nd is tight. I . I N T R O D U C T I O N The broad cast channe l with degrad ed message sets was initially studied by K ´ orner and M ¨ arton [1] for two r eceiv ers and more recently in [2], [3], [4] for three and more receivers. K ´ orner and M ¨ arton [1] established the capa city region for the degraded message sets with two receivers and some cap acity regions for thre e or mo re receivers were established [ 2], [3] b y showing that the straigh tforward extension of the inner bound in [1] was op timal. In [4], an idea called indirect decoding was introdu ced and the author s showed that this c ould be used to enhance (in some cases strictly) the straigh tforward extension of th e inn er bou nd by K ´ orner an d M ¨ arto n. Unfo rtunately , the new i nner boun ds [ 4] become quite messy and unwieldy due to the in troductio n of many aux iliary r andom variables. Howe ver there is still one class of broad cast chann els with degraded message sets wh ere the id ea of in direct dec oding does no t yield any r egion better than the straightforward extension of the K ´ orner and M ¨ arton inner b ound, an d this is the scena rio of in terest here. Consider a 3-r eceiv er broad cast ch annel with 2 messages ( M 0 , M 1 ) with the following decoding requ irement. Receiv ers Y 1 and Y 2 need to decode both m essages ( M 0 , M 1 ) while receiver Y 3 needs to decode o nly message M 0 . The tr aditional inner and outer bounds presented belo w remain the best kno wn inner and outer bo unds fo r this class of broad cast ch annels. In this pa per we look at the genera l in ner and outer bound s for this scenario in a gr eater detail. W e show th at these bo unds differ in gene ral, and th at there is a class of channels wh ere the inner bo und is tight and the outer bound is weak . There ar e two main contributions in this paper: the first one is the techn ique (same spirit as Mr s. Ge rber’ s lem ma [5]) used to e valuate the boundary of a particular inn er bou nd; the seco nd is the use of a ( 1 2 + ǫ )-co deboo k 1 rather than an ǫ -codeb ook to establish the cap acity region. Bound 1: T he un ion of th e fo llowing set of r ate pairs ( R 0 , R 1 ) satisfy ing R 0 ≤ I ( U ; Y 3 ) R 1 ≤ min { I ( X ; Y 1 | U ) , I ( X ; Y 2 | U ) } R 0 + R 1 ≤ min { I ( X ; Y 1 ) , I ( X ; Y 2 ) } over all pairs of r andom variables ( U, X ) such that U → X → ( Y 1 , Y 2 , Y 3 ) forms a Markov chain constitutes an inner bound to the cap acity r egion. Bound 2: T he union over the set of rate pairs ( R 0 , R 1 ) satisfying R 0 ≤ min { I ( U 1 ; Y 3 ) , I ( U 2 ; Y 3 ) } R 0 + R 1 ≤ min { I ( U 1 ; Y 3 ) + I ( X ; Y 1 | U 1 ) , I ( U 2 ; Y 3 ) + I ( X ; Y 2 | U 2 ) } R 0 + R 1 ≤ min { I ( X ; Y 1 ) , I ( X ; Y 2 ) } over all possible choices of ra ndom v ariables ( U 1 , U 2 , X ) such that ( U 1 , U 2 ) → X → ( Y 1 , Y 2 , Y 3 ) forms a Markov chain constitutes an outer bou nd fo r this chan nel. The a bove b ound s are trad itional, i.e. can be obtained using standard techniqu es. The inner bou nd is a straigh tforward extension of the achiev ab ility argum ent in [1] an d the outer bound can b e ded uced b y argum ents in [6], [7], etc. Remark 1 : It is also possible to in clude the co nstraint R 0 ≤ min { I ( U 1 ; Y 1 ) , I ( U 2 ; Y 2 ) } into the ou ter bou nd. Howe ver , it is quite straightforward to show that the region obtained b y adding this ine quality is identical to the bo und we presented. These bou nds are k nown to be tig ht in all of the fo llowing special cases, • Recei ver Y 1 is a less n oisy receiv er than Y 3 and Y 2 is a less no isy receiv er than Y 3 [2], [4], • Y 3 is a deterministic fu nction of X , • Y 1 is a mor e capa ble receiver th an Y 2 (or vice-versa), • Y 3 is a mor e capa ble receiver th an Y 2 (or Y 1 ), 1 An η -codeboo k is used to denote a codebook whose probabi lity of error is bounded abov e by η . The last two cases are very straig htforward and the p roof is omitted. When Y 3 is a determ inistic fun ction of X , note that it is not difficult to sh ow that by taking the c onv ex clo sure of the regions obtained by setting ( i ) U = Y 3 and ( ii ) U = ∅ in the inner bo und exha usts the following region, R 0 ≤ H ( Y 3 ) R 0 + R 1 ≤ min { I ( X ; Y 1 ) , I ( X ; Y 2 ) } and this clear ly f orms a n o uter b ound to the capa city region. One class o f chann els that doe s not fall into any of th e cases is the fo llowing channe l sho wn in Figure 1 b elow . The chann el X → ( Y 1 , Y 2 ) represents a binary skew-symmetric (BSSC) broadc ast c hannel [ 7], [8] and the chan nel X → Y 3 represents a b inary symmetric (BSC) with crossover probab ility p , with 0 ≤ p ≤ 1 2 . P S f r a g r e p l a c e m e n t s 0 1 0 1 0 1 0 1 X Y 1 Y 2 Y 3 1 2 1 2 p Fig. 1. 3-recei ver broadcast channel In the next section we evaluate Boun d 1 for this channel. Based on the symmetr y , it is very natu ral to b elieve that th e auxiliary channel U → X must be a BSC with s ome cross o ver probab ility s . In the next sectio n, we p rove th at this is indeed the case. T his uses a techniqu e similar in spirit to W yn er and Ziv’ s technique of using Mrs. Gerber’ s lemma [5]. W e will also show tha t the Bound 2 yield s a strictly larger region f or this chann el. Fin ally , we will show that the region represented by Bo und 1 co nstitutes the cap acity r e gion for this ch annel. I I . E V A L U A T I O N O F T H E I N N E R B O U N D In the ev aluation of the inn er bo und, we divide the range 0 ≤ p ≤ 1 2 into two r egions, 0 ≤ p ≤ p max and p max ≤ p ≤ 1 2 , where p max ∈ [0 , 1 2 ] is the u nique solu tion o f 1 − h ( p ) = h ( 1 4 ) − 1 2 ; i.e. the value of p at which capacity o f the BSC m atches the term ma x p ( x ) min { I ( X ; Y 1 ) , I ( X ; Y 2 ) } . The n umerical value of p max ≈ 0 . 18 4 . A. Evaluation of the in ner bound , 0 ≤ p ≤ p max In the region 0 ≤ p ≤ p max it is straigh tforward to see that the inner bou nd red uces to the following region (o btained via a time-division betwee n th e two au xiliary cha nnels: ( i ) U = ∅ and ( ii ) U = X , and in each case, setting P( X = 0) = 0 . 5 ), R 0 + R 1 ≤ h ( 1 4 ) − 1 2 , which clearly matches the ou ter b ound (Bound 2). Thu s for 0 ≤ p ≤ p max ≈ 0 . 184 , th e inner and outer boun ds are tight and giv e the cap acity re gion. B. Evaluation of the in ner bound , p max ≤ p ≤ 1 2 Let U = { 1 , 2 , ..., m } and let P( U = i ) = u i and P( X = 0 | U = i ) = s i . Further, let h ( x ) = − x log 2 x − (1 − x ) lo g 2 (1 − x ) denote th e binary en tropy fun ction. Using th ese n otations we have, I ( U ; Y 3 ) = h ( X i u i ( s i (1 − p ) + (1 − s i ) p )) − X i u i h ( s i (1 − p ) + (1 − s i ) p ) I ( X ; Y 1 | U ) = X i u i h ( s i 2 ) − X i u i s i I ( X ; Y 2 | U ) = X i u i h ( 1 − s i 2 ) − X i u i (1 − s i ) I ( X ; Y 1 ) = h ( X i u i s i 2 ) − X i u i s i I ( X ; Y 2 ) = h ( X i u i (1 − s i ) 2 ) − X i u i (1 − s i ) . Define ˜ U = { 1 , 2 , ..., m } × { 1 , 2 } , P ( ˜ U = ( i, 1)) = u i 2 , P( X = 0 | ˜ U = ( i, 1)) = s i , P( ˜ U = ( i, 2)) = u i 2 , and P( X = 0 | ˜ U = ( i, 2)) = 1 − s i . This indu ces an ˜ X with P ( ˜ X = 0) = 1 2 . It is straigh tforward to see the fo llowing: I ( ˜ U ; ˜ Y 3 ) ≥ I ( U ; Y 3 ) I ( ˜ X ; ˜ Y 1 | ˜ U ) = I ( ˜ X ; ˜ Y 2 | ˜ U ) = 1 2 ( I ( X ; Y 1 | U ) + I ( X ; Y 2 | U )) ≥ min { I ( X ; Y 1 | U ) , I ( X ; Y 2 | U ) } I ( ˜ X ; ˜ Y 1 ) = I ( ˜ X ; ˜ Y 2 ) ≥ 1 2 ( I ( X ; Y 1 ) + I ( X ; Y 2 )) From this it follows that for every U re placing U b y ˜ U leads to a larger achiev able region. Hence to evaluate Bou nd 1, it suffices to maximize over all auxiliary rando m variables of the form U defined by: U = { 1 , 2 , ..., m } × { 1 , 2 } , P( U = ( i, 1)) = u i 2 , P( X = 0 | U = ( i, 1)) = s i , P( U = ( i, 2)) = u i 2 , and P( X = 0 | U = ( i, 2)) = 1 − s i . Under this notation we have th e following expression for the rate region given in Boun d 1, R 0 ≤ I ( U ; Y 3 ) = h  1 2  − X i u i h ( s i (1 − p ) + (1 − s i ) p ) , R 1 ≤ min { I ( X ; Y 1 | U ) , I ( X ; Y 2 | U ) } = X i u i 2  h ( s i 2 ) + h ( 1 − s i 2 )  − 1 2 , R 0 + R 1 ≤ min { I ( X ; Y 1 ) , I ( X ; Y 2 ) } = h  1 4  − 1 2 . Using the symme try of the function h ( x ) = h (1 − x ) we note that h ( s i (1 − p ) + (1 − s i ) p ) = h ((1 − s i )(1 − p ) + s i p ) and thus the above r egion is con stant u nder the transfo rmation s i → 1 − s i , implyin g we can r estrict s i to take values on ly in 0 ≤ s i ≤ 1 2 . Before we pro ceed to determine the bou ndary of this region, we prove the fo llowing lemma. C. An inequ ality for a class of fun ctions Lemma 1 : Let f ( x ) and g ( x ) be two n on-negative and strictly increasing functions that are dif f erentiable in the region x ∈ [ x 1 , x 2 ] . Further assume that f (1) ( x ) g (1) ( x ) is a decreasin g function , where f (1) ( x ) and g (1) ( x ) denote the deriv atives o f the function . Given any u, 0 ≤ u ≤ 1 , let x int be uniq uely defined accord ing to f ( x int ) = uf ( x 1 ) + (1 − u ) f ( x 2 ) . Then the following h olds, g ( x int ) ≤ ug ( x 1 ) + (1 − u ) g ( x 2 ) . Pr oof: W e h ave u ( f ( x int ) − f ( x 1 )) = (1 − u )( f ( x 2 ) − f ( x int )) , and we wish to s how that u ( g ( x int ) − g ( x 1 )) ≤ (1 − u )( g ( x 2 ) − g ( x int )) . Since all th e terms are positive, this reduces to showing f ( x int ) − f ( x 1 ) g ( x int ) − g ( x 1 ) ≥ f ( x 2 ) − f ( x int ) g ( x 2 ) − g ( x int ) . Howe ver, this is immediate as shown b elow . From th e fact that f (1) ( x ) g (1) ( x ) is a decreasing fu nction, we hav e R x int x 1 f (1) ( x ) dx R x int x 1 g (1) ( x ) dx ≥ f (1) ( x int ) g (1) ( x int ) ≥ R x 2 x int f (1) ( x ) dx R x 2 x int g (1) ( x ) dx Repeated application s of Lemma 1 leads to the following corollary - po tentially o f in depend ent interest. Cor ollary 1: Let f ( x ) an d g ( x ) b e two non- negativ e and strictly increasing functions that are dif f erentiable in the region x ∈ [ x 1 , x 2 ] . Further assume that f (1) ( x ) g (1) ( x ) is a decreasin g function , wh ere as before f (1) ( x ) an d g (1) ( x ) d enote the deriv ativ es of the function. Gi ven any u i ≥ 0 , P i u i = 1 , and y i ∈ [ x 1 , x 2 ] , let x int be uniq uely d efined according to f ( x int ) = P i u i f ( y i ) . Then th e following holds g ( x int ) ≤ X i u i g ( y i ) . D. Determining the bound ary rate pairs W e use the Corollary 1 to determ ine the bou ndary of th e region. W e make the following iden tifications, let f ( x ) = h ( x 2 ) + h ( 1 − x 2 ) − 1 , and g ( x ) = h ( x (1 − p ) + (1 − x ) p ) . Ob serve that f ( x ) and g ( x ) a re increasing dif ferentiab le f unctions in the region [0 , 1 2 ] . Claim 1: F or 1 6 ≤ p ≤ 1 2 , the ratio of th e d eriv atives f (1) ( x ) g (1) ( x ) is a dec reasing fu nction. The p roof of this fact is f ound in th e App endix. (Numerical simulations ind icate that this is true for p min ≤ p ≤ 1 2 for p min ≈ 0 . 05 , but for the p urpo ses of establishing the inner bound clearly this r egion o f p suffices, as 1 6 ≤ p max ≈ 0 . 184 ) . Remark 2 : By combining Claim 1 and Corollary 1 note that h ( p ∗ f − 1 ( y )) is con vex in y , and this is very similar to Mrs. Gerber’ s Lemma [5 ]. Now let s int be d efined according to h ( s int 2 ) + h ( 1 − s int 2 ) = X i u i  h  s i 2  + h  1 − s i 2   . Then fr om Co rollary 1 , for p min ≤ p ≤ 1 2 we hav e h  1 2  − X i u i h ( s i (1 − p ) + (1 − s i ) p ) ≤ h  1 2  − h ( s int (1 − p ) + (1 − s int ) p ) . This im plies that the optimal auxiliary ch annel U → X is a BSC with a cro ss-over p robab ility s and P( U = 0) = 1 2 . Thus for p max ≤ p ≤ 1 2 , the bou ndary is ch aracterized b y the pair o f p oints o f the fo rm, R 0 = 1 − h ( s (1 − p ) + (1 − s ) p ) , R 1 = min  1 2  h  s 2  + h  1 − s 2  − 1  , (1) h  1 4  − 3 2 + h ( s (1 − p ) + (1 − s ) p )  , for 0 ≤ s ≤ 1 2 . The second ter m in R 1 comes from taking into ac count the sum rate constrain t, R 0 + R 1 ≤ h  1 4  − 1 2 . A simple calcu lation shows that for p o ≤ p ≤ 1 2 one can ignore the s um ra te constrain t, wh ere p o = √ 3 − 1 2 √ 3 ≈ 0 . 21 1 . This p o correspo nds to the smallest value of p where the conv ex region characterized by the pairs R 0 = 1 − h ( s (1 − p ) + (1 − s ) p ) , R 1 = 1 2  h  s 2  + h  1 − s 2  − 1  . has a slo pe o f − 1 at the point ( R 0 , R 1 ) =  0 , h  1 4  − 1 2  . Therefo re the inner bound has three different expressions: • 0 ≤ p ≤ p max : the inn er bou nd r educes to R 0 + R 1 ≤ h  1 4  − 1 2 , • p max ≤ p ≤ p o : the in ner bou nd is given by eq uation (1) where all in equalities ar e nec essary , • p o ≤ p ≤ 1 2 : the inn er boun d is ch aracterized by pair of points of the fo rm R 0 = 1 − h ( s (1 − p ) + (1 − s ) p ) , R 1 = 1 2  h  s 2  + h  1 − s 2  − 1  . E. Comparison with the outer bound T o sho w that the ou ter b ound gi ves a larger region, we produ ce a p articular ch oice o f the pair ( U 1 , U 2 , X ) . Consider a U 1 , U 2 defined as follows, P( U 1 = 1 ) = P ( U 2 = 1 ) = u , P( U 1 = 2 ) = P ( U 2 = 2 ) = 1 − u , P( X = 0 | U 1 = 1) = P( X = 1 | U 2 = 1 ) = 1 , P( X = 0 | U 1 = 2) = P( X = 1 | U 2 = 2 ) = s, where s = 0 . 5 − u 1 − u for 0 ≤ u ≤ 0 . 5 . Ex istence of the triple ( U 1 , U 2 , X ) is guar anteed by the con sistent distribution on X . Substituting this ch oice into Bou nd 2 we obtain Re gio n A giv en by , R 0 ≤ 1 − (1 − u ) h  s (1 − p ) + (1 − s ) p  − uh  p  , R 1 ≤ (1 − u ) h  s 2  − 1 2 + u, R 0 + R 1 ≤ h  1 4  − 1 2 . Figure 2 p lots Region A an d Bound 1 for p = 1 4 . Observe that Region A is larger than Bound 1 , an d h ence the Bound s 1 and 2 d o n ot m atch for the 3 -receiver channel shown in Figu re 1. T his implies the following cor ollary . 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 0.05 0.1 0.15 0.2 0.25 0.3 R1 R0 Inner bound Region A Fig. 2. Comparing Bound 1 and Reg ion A for p = 1 4 Cor ollary 2: There exists a class of chann els, given in Figure 1, for which th e in ner and outer bound s ( i.e. Boun ds 1 and 2) do not match. I I I . R E V I S I T I N G O U T E R B O U N D W e n ow show that the inn er b ound is tight for th e channel shown in Figu re 1 . Let π : { 0 , 1 } 7→ { 0 , 1 } ; π (0) = 1 , π (1) = 0 . Consider an ǫ -cod ebook { x n m 0 ,m 1 , 1 ≤ m 0 ≤ 2 nR 0 , 1 ≤ m 1 ≤ 2 nR 1 , A m 0 ,m 1 ⊆ Y n 1 , B m 0 ,m 1 ⊆ Y n 2 , C m 0 ⊆ Y n 3 } , where the disjoin t sets A m 0 ,m 1 , B m 0 ,m 1 , C m 0 represent the decodin g maps. From the skew symmetry of the ch annels X → ( Y 1 , Y 2 ) and the symm etry in chan nel X → Y 3 , it is clear that { π ( x n m 0 ,m 1 ) , 1 ≤ m 0 ≤ 2 nR 0 , 1 ≤ m 1 ≤ 2 nR 1 , π ( B m 0 ,m 1 ) ⊆ Y n 1 , π ( A m 0 ,m 1 ) ⊆ Y n 2 , π ( C m 0 ) ⊆ Y n 3 } represents a valid ǫ -co deboo k as well. From these two codes, construct a new codebook (with erro r bound ed by 1 2 + ǫ ) and size 2 nR 0 × 2 nR 1 +1 as fo llows: T he codewords are in dexed by x n m 0 , ( m 1 ,b ) where b = 0 , 1 . When b = 0 the codeword x n m 0 , ( m 1 ,b =0) = x n m 0 ,m 1 and when b = 1 , we have x n m 0 , ( m 1 ,b =1) = π ( x n m 0 ,m 1 ) . The decod ing maps for this c odebo ok are created as follows: If y n 1 ∈ A m 1 0 ,m 1 1 ∩ π ( B m 2 0 ,m 2 1 ) then th e receiver choo ses one of the two message pairs ( m 1 0 , m 1 1 ) , ( m 2 0 , m 2 1 ) with equ al pro bability . Otherwise it picks t he message p air correspond ing to the uniq ue set A m 1 0 ,m 1 1 or π ( B m 2 0 ,m 2 1 ) that it be longs to. A similar d ecoding strategy applies f or r eceivers Y 2 and Y 3 as we ll. The key feature is the sym metry of th e codeb ook. If x n ∈ C then π ( x n ) ∈ C and correspon d to the same message M 0 . Now o bserve that H ( M 0 , M 1 | Y n 1 ) ≤ H ( M 0 , M 1 , b | Y n 1 ) ≤ 1 + H ( M 0 , M 1 | Y n 1 , b ) = 1 + n ( R 0 + R 1 ) ǫ n . Th erefor e we obtain the same outer bound (Bound 2) using Fano’ s inequ ality and identification of the auxiliary random variables as before. In particular, th e iden tifications of the a uxiliary r andom variables remain the f ollowing: U 1 i = ( M 0 , Y i − 1 31 , Y n 1 i +1 ) and U 2 i = ( M 0 , Y i − 1 31 , Y n 2 i +1 ) . Now fo r the ske w-symmetric channels an d a symm etric co deboo k observe that P “ M 0 = m 0 , Y i − 1 31 = y i − 1 31 , Y n 1 i +1 = y n 1 i +1 , X i = x i ” = X x n 1 \ x i P “ M 0 = m 0 , X n 1 = x n 1 , Y i − 1 31 = y i − 1 31 , Y n 1 i +1 = y n 1 i +1 ” ( a ) = X x n 1 \ x i P “ M 0 = m 0 , X n 1 = x n 1 ” i − 1 Y j =1 P( Y 3 j = y 3 j | X j = x j ) × n Y k = i +1 P( Y 1 k = y 1 k | X k = x k ) ( b ) = X x n 1 \ x i P “ M 0 = m 0 , X n 1 = π ( x n 1 ) ” i − 1 Y j =1 P( Y 3 j = π ( y 3 j ) | X j = π ( x j )) × n Y k = i +1 P( Y 2 k = π ( y 1 k ) | X k = π ( x k )) = X x n 1 \ x i P “ M 0 = m 0 , X n 1 = π ( x n 1 ) , Y i − 1 31 = π ( y i − 1 31 ) , Y n 2 i +1 = π ( y n 1 i +1 ) ” ( c ) = P “ M 0 = m 0 , Y i − 1 31 = π ( y i − 1 31 ) , Y n 2 i +1 = π ( y n 1 i +1 ) , X i = π ( x i ) ” . Here ( a ) follows from the discrete memo ryless property of the chan nel; and ( b ) follows from ( i ) symmetry of the co de, ( ii ) sym metry of the chan nel X → Y 3 with r espect to π ( · ) , and ( iii ) the skew symm etry between receivers Y 1 , Y 2 i.e. P( Y 2 = π ( y ) | X = π ( x )) = P ( Y 1 = y | X = x ); and ( c ) is a con sequence of π ( · ) being a bijection. Therefo re the rand om v ar iables ( U 1 , X ) and ( U 2 , X ) are identical up to re-lab eling. Since the m utual inf ormation and entropy do not depend on the labeling, it follows that I ( U 1 ; Y 3 ) = I ( U 2 ; Y 3 ) I ( X ; Y 2 | U 1 ) = I ( X ; Y 2 | U 2 ) . Remark 3 : This techniq ue can be extend ed to o ther skew- symmetric chann els as well, i.e one for which such a π ( · ) exists. Therefo re we o btain the fo llowing revised outer bound . Bound 3: The u nion over the set of rate pairs ( R 0 , R 1 ) satisfying R 0 ≤ I ( U 1 ; Y 3 ) R 0 + R 1 ≤ min { I ( U 1 ; Y 3 ) + I ( X ; Y 1 | U 1 ) , I ( U 1 ; Y 3 ) + I ( X ; Y 2 | U 1 ) } R 0 + R 1 ≤ min { I ( X ; Y 1 ) , I ( X ; Y 2 ) } over all possible choices of rando m variables ( U 1 , X ) su ch that U 1 → X → ( Y 1 , Y 2 , Y 3 ) form s a M arkov ch ain constitutes an outer bound fo r this chan nel. It is straightfor ward to see (using the boun dary p oints) that Bound 3 matches the inner boun d and forms the capacity region. A C K N O W L E D G E M E N T S The a uthors wish to than k Arvind Ramachandr an for valu- able sug gestions and f eedback on the conten ts of the paper . R E F E R E N C E S [1] J. K ¨ orner and K. Marton, “General broadcast channels with degrade d message sets, ” IEEE T rans. Info . Theory , vol. IT -23, pp. 60–64, Jan, 1977. [2] S. Diggavi and D. 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Theory , vol. IT -53, pp. 350–355, January , 2007. [8] B. Hajek and M. Pursley , “Ev aluati on of an achie vable rate regio n for the broadca st channel, ” IEEE T rans. Info. Theory , vol. IT -25, pp. 36–46, January , 1979. A P P E N D I X A. Pr o of o f Cla im 1 In this section we show that when 1 6 ≤ p ≤ 1 2 , the ratio f (1) ( x ) g (1) ( x ) is a d ecreasing fun ction o f x, x ∈ [0 , 1 2 ] . Recalling the definitions, f ( x ) = h ( x 2 ) + h ( 1 − x 2 ) − 1 , and g ( x ) = h ( x (1 − p ) + (1 − x ) p ) . As f ( x ) and g ( x ) are st rictly inc reasing in x ∈ [0 , 1 2 ] , it suffices to sh ow that f (2) ( x ) f (1) ( x ) ≤ g (2) ( x ) g (1) ( x ) , (2) where f (2) ( x ) , g (2) ( x ) denote the s econd de riv atives of the function . Let J ( x ) = log 1 − x x , U ( x ) = x (1 − x ) an d x ∗ p = x (1 − p ) + p (1 − x ) . Using th is notation and substituting fo r the deriv ativ es, ( 2) reduces to sh owing J ( x ∗ p ) U ( x ∗ p ) 1 − 2 p ≥ 2  J  x 2  − J  1 − x 2   1 U ( x 2 ) + 1 U ( 1 − x 2 ) . (3) Now observe that as x → 1 2 both J ( x ∗ p ) and J  x 2  − J  1 − x 2  tend to zero and all other ter ms rem ain p ositiv e. T hus we hav e an eq uality at x = 1 2 . T o show the inequ ality f or x ∈ [0 , 1 2 ] it suffi ces to prove th at th e de rivative of the left hand side (L.H.S .) of (3) is smaller than derivative o f the right hand sid e (R.H.S.) of (3). The d eriv ati ve of the L.H.S. is given b y d dx J ( x ∗ p ) U ( x ∗ p ) 1 − 2 p = − 1 + J ( x ∗ p )(1 − 2( x ∗ p )) . Let us define R ( x ) to be the deriv ativ e of th e R.H.S., i.e. d dx 2  J  x 2  − J  1 − x 2   1 U ( x 2 ) + 1 U ( 1 − x 2 ) = R ( x ) . W e wish to show that − 1 + J ( x ∗ p )(1 − 2( x ∗ p )) ≤ R ( x ) , (4) for all 1 6 ≤ p ≤ 1 2 and x ∈ [0 , 1 2 ] . Gi ven any x ∈ [0 , 1 2 ] , observe that J ( x ∗ p )(1 − 2 ( x ∗ p )) is a decre asing functio n of p for 0 ≤ p ≤ 1 2 . Thus estab lishing ( 4) for p = 1 6 suffices. Let S ( x ) = − 1 + J ( x ∗ 1 6 )(1 − 2( x ∗ 1 6 )) . Figure 3 plots S ( x ) and R ( x ) . 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 −2 0 2 4 6 8 10 x R(x) S(x) Fig. 3. Comparing R ( x ) and S ( x ) Thus we have S ( x ) ≤ R ( x ) for 0 ≤ x ≤ 1 2 . This com pletes the p roof o f Claim 1.