THE ARIMOTO-BLAHUT ALGORITHM FOR COMPUTATION OF CHANNEL CAPACITY. William A. Pearlman. References: S. Arimoto - IEEE Trans. Inform. Thy., Jan.

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1 THE ARIMOTO-BLAHUT ALGORITHM FOR COMPUTATION OF CHANNEL CAPACITY Wllam A. Pearlman 2002 References: S. Armoto - IEEE Trans. Inform. Thy., Jan R. Blahut - IEEE Trans. Inform. Thy., July 1972 Recall the defnton of capacty for a dscrete, memoryless channel (DMC). Gven channel transton probabltes P (/), = 1, 2,..., m, = 1, 2,..., n and let Q = (Q(1), Q(2),..., Q(m)) be the nput probablty vector, the capacty s C = max Q m n =1 =1 P (/)Q() log P (/) Q() (1) where P (/) = The quantty P (/)Q() k P (/k)q(k). P (/)Q() log P (/) Q() s the average mutual nformaton I(X; Y ) between the nput and output ensembles of the channel. For a fxed channel,.e., fxed P (/), t s a convex functon of the nput probabltes Q. To emphasze the functonal relatonshp on Q, let us call t J(Q): J(Q) P (/)Q() log P (/) Q() = I(X; Y ) and C = max J(Q) (2) Q 1

2 Now let Φ(/) be any set of condtonal probabltes of nput gven output ( = 1, 2,..., m; = 1, 2,..., n). Defne now a generalzaton of J(Q) by J(Q, Φ) = P (/)Q() log Φ(/) Q() (3) The functon J(Q, Φ) s a functon of the nput probabltes Q and condtonal probabltes Φ. Through ths functon and a seres of lemmas, we shall formulate an algorthm for calculatng the channel capacty C. For the sake of smplcty, assume the logarthmc base s always e. The lemmas wll frst be stated wthout proof, so as not to clutter the development. They wll be proved as a fnal order of busness. Lemma 1 For any fxed Q, J(Q, Φ) J(Q) wth equalty ff Φ(/) = k P (/)Q() = P (/) P (/k)q(k) By ths lemma, J(Q) = max Φ J(Q, Φ). Substtuton nto (2) gves Lemma 2 C = max Q max J(Q, Φ) (4) Φ For fxed Φ(/), consder Q varable 2

3 J(Q, Φ) log( r() = exp[ r()) P (/) log Φ(/)] wth equalty ff Q() = r() k r(k). The double maxmum n (4) can be taken n any order. Therefore, C = max Φ max Q J(Q, Φ) = max log( Φ r()) = max log( Φ exp P (/) log Φ(/)) (5) The pont of ntroducng J(Q, Φ) s that C can be calculated through a double maxmzaton procedure (40 each step of whch can be solved n closed form, as gven n Lemmas 1 and 2. The maxmzaton of J(Q) wth respect to Q has no closed-form soluton. We can doubly maxmze J(Q, Φ) n any order. For example, let us fx Q to some ntal value Q o. By Lemma 1, Φ o (/) = P (/)Qo () k P (/k)qo (k) maxmzes J(Qo, Φ) when Φ = Φ o. Now consder J(Q, Φ o ) for Φ o fxed and vary Q to produce a maxmum. In Lemma 2, that maxmum s J(Q, Φ o ) = log( r()) = log exp P (/) log Φ(/) for Q() = Q 1 () = r() k r(k). Then for that Q1 we can fnd a maxmum of J(Q 1, Φ) for Φ = Φ 1 as gven n Lemma 1 for Q 1 and P (/). Fx Φ 1 and maxmze J(Q, Φ 1 ) for Q = Q 2 as gven n Lemma 2. Now maxmze J(Q 2, Φ) and so on. At each step, J(Q, Φ) s ncreasng untl t eventually reaches capacty C. So, we can now formulate the Armoto-Blahut algorthm to accomplsh the double maxmzaton n (4). 3

4 Algorthm l s the teraton ndex 1. Set l = 0 and choose ntal set of nput probabltes Q o () > 0, all. 2. Compute Φ l (/) = r l () = exp Ql ()P (/) k Ql (k)p (/k) all, P (/) ln Φ l (/) J(Q l+1, Φ l ) = ln( r l ()) 3. Set l = l + 1 and go to 2. Q l+1 () = rl () k rl () Ths algorthm accomplshes what was already explaned, but not how to stop the recurson. The queston now s when do we know that we are close to capacty. We now have to ntroduce one more lemma. For any l, J(Q l+1, Φ l ) = ln( rl ()) C. Let us defne Then c l () rl () Q l () J(Q l+1, Φ l ) = ln( Q l ()c l ()) So J(Q l+1, Φ l ) s the logarthm of the average of the c l () s. Lemma 3 C max ln c l () 4

5 Therefore, channel capacty C s bounded from below and above as follows: ln( Q l ()c l ()) C max ln c l () In order to fnd C wthn accuracy ɛ, stop the teraton when max ln c l () ln( Q l ()c l ()) < ɛ or J(Q l+1, Φ l ) > max ln c l () ɛ Insert n algorthm: 2. Calculate J(Q l+1, Φ l ) + max ln rl () Q l () = T. If T > ɛ, go to 3. If T < ɛ, go to C = J(Q l+1, Φ l ) It s nterestng to note that ln c l () = I(; Y/Q l ); the average nformaton that the output ensemble Y gves about the nput event, when the nput probabltes are Q l. Recall that f I(; Y/Q) = γ for all s.t. Q() > 0 I(; Y/Q) γ for all s.t. Q() = 0 (Kuhn-Tucker condtons). J(Q) = I(; Y/Q) = C and C = γ. So, when we are nearng capacty C, the nformaton gven by the output ensemble Y about each nput event of nonzero probablty s nearng equalty, snce 5

6 Q l () ln c l () ln Q l ()c l () C max ln c l () by the convex property of ln. So we have I(; Y/Q l ) C max I(; Y/Q l ) We now prove the lemmas. Proof of Lemma 1: J(Q, Φ) J(Q) = = P (/)Q() log Φ(/) P (/) [ ] Φ(/) P (/)Q() P (/) 1 [P ()Φ(/) P (/)Q()] = 1 1 = 0 wth equalty ff Φ(/) = P (/) for all and. Proof of Lemma 2: 6

7 J(Q, Φ) = = = J(Q, Φ) = = P (/)Q() log Φ(/) Q() 1 P (/)Q() log Q() + P (/)Q() log Φ(/) 1 Q() log Q() + Q() P (/) log Φ(/) ( ) exp P (/) log Φ(/) Q() log Q() ( ) r() Q() log, r() = exp P (/) log Φ(/) Q() = Q() log( k r(k)) + Q() log r()/ k r(k) Q() log( k r(k)) The last step results from ln x x 1 n the second sum wth equalty ff Q() = r() k r(k). Proof of Lemma 3. Suppose Q acheves capacty. Then C = C = P (/)Q () log P (/) P (), P () = k P (/)Q () log P (/) P l () P l () P () P (/k)q (k) where P l () = k P (/k)q l (k) are the output probabltes for Q l (k). Contnung, 7

8 C = = P (/)Q () log P l () P () + P () log P l () P () + Q () P (/)Q () log P (/) P l () P (/) log P (/) P l () Usng ln x x 1, the frst term s 0. In the second term, we can overbound the P (/) log P (/) P l () by ts maxmum over. Therefore, C max P (/) log P (/) p l () = max I(; Y/Q l ) Recall r l () = exp P (/) ln Φ l (/) = exp P (/) ln Ql ()P (/) P l () ln c l () = ln r l () ln Q l () = P (/) ln Ql ()P (/) P l () P (/) ln Q l () = P (/) ln P (/) P l () = I(; Y/Ql ). Therefore, the concluson of the lemma can be expressed as C max ln c l (), as gven. Note the condton for equalty. We must have P l () = P () or Q l () = Q () for nonzero Q () and I(; Y/Q l ) be the same maxmum value for all such that Q l () > 0. These condtons are both necessary and suffcent for equalty. So we have corroborated the Kuhn-Tucker condtons. 8

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