2014
DOI: 10.1109/tit.2014.2360698
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A Lower Bound on the Sum Rate of Multiple Description Coding With Symmetric Distortion Constraints

Abstract: We derive a single-letter lower bound on the minimum sum rate of multiple description coding with symmetric distortion constraints. For the binary uniform source with the erasure distortion measure or Hamming distortion measure, this lower bound can be evaluated with the aid of certain minimax theorems. A similar minimax theorem is established in the quadratic Gaussian setting, which is further leveraged to analyze the special case where the minimum sum rate subject to two levels of distortion constraints (wit… Show more

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Cited by 17 publications
(18 citation statements)
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“…The introduction of (V 1 , V 2 ) is partly inspired by Ozarow's converse argument for the Gaussian multiple description problem [14] (see also [15]- [17]). …”
Section: Lower Boundmentioning
confidence: 99%
“…The introduction of (V 1 , V 2 ) is partly inspired by Ozarow's converse argument for the Gaussian multiple description problem [14] (see also [15]- [17]). …”
Section: Lower Boundmentioning
confidence: 99%
“…Then if λ c,i −τ < 1/2r s/2 , the error in the reconstruction is at least 1/2r s/2 . If σ 2 X2−X1 = Θ(r 2 0 ) and r0 rc is constant as the limits in (22) are approached, the probability that Q s/2 (λ c,1 ) = Q s/2 (λ c,2 ) = τ and λ c,1 = λ c,2 does not approach 0, thus the central distortion cannot satisfy relation (29), while the proposed RDSC scheme can.…”
Section: Resultsmentioning
confidence: 99%
“…Notice that (M ν s ) 2 n = (M 2 Kν c ) 2 n . By plugging (28) in (24) and using the fact that K and κ 0 are constants, relation (29) follows. Further, equalities (23) and (29)…”
Section: Proof Of Corrollarymentioning
confidence: 99%
“…The desired conclusion that r ≥ r (k) (d k ) and that d j ≥ d (k) j (d k ) when r = r (k) (d k ) follows from the corresponding result for the quadratic Gaussian multiple description problem [26], [35]. Note that (k − 1)λ 2 X,2 (λ (k) S,1 )µ (k) (µ (k) − 1) + k(λ (k) X,1 ) 2 λ 2 S,2 = 0 (consequently, condition (11) is satisfied) when λ (j) S,1 > λ S,2 = 0.…”
Section: Proof Of Theoremmentioning
confidence: 90%