Refined Strong Converse for the Constant Composition Codes Hao-Chung - - PowerPoint PPT Presentation
Refined Strong Converse for the Constant Composition Codes Hao-Chung Cheng 1 glu 2 Bar s Nakibo 1 Department of Applied Mathematics and Theoretical Physics University of Cambridge 2 Department of Electrical and Electronics Engineering
1Department of Applied Mathematics and Theoretical Physics
2Department of Electrical and Electronics Engineering
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1−E′ sp(R) 2
sc(R) 2
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2αe−nD1(wq αw) ≥ P0
2αe−nD1(wq αw)
αq) for an α ≥ 1 and w ≺ q.
sc(R) 2
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n→∞ − 1 n ln P0 e
n→∞ − 1 n ln P1 e
n→∞ − 1 n ln
e
n→∞ − 1 n ln P1 e
α w)
α q)
1 q) = D1(w q)
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n→∞ − 1 n ln P0 e
n→∞ − 1 n ln P1 e
n→∞ − 1 n ln
e
n→∞ − 1 n ln P1 e
α w)
α q)
1 q) = D1(w q)
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n→∞ − 1 n ln P0 e
n→∞ − 1 n ln P1 e
1 wac
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n→∞ − 1 n ln P0 e
n→∞ − 1 n ln P1 e
1 q)
1 w)
1 q)
n→∞ − 1 n ln
e
n→∞ − 1 n ln P1 e
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n→∞ − 1 n ln P0 e
n→∞ − 1 n ln P1 e
1 q)
1 w)
1 q)
n→∞ − 1 n ln
e
n→∞ − 1 n ln P1 e
1 w)
α w)
α q)
1 q)
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αq), there exists an A > 0 such that
2α e−D1(wq αw).
2α e−D1(wq αw) is optimal up to a multiplicative constant; see matching
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α
αw) vs. D1(w q αq) for α ∈ (0, 1)
1 q) dwq 1
α
α
α
α
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αq)
α
2α e−D1(wq αw)
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2α e−D1(wq αw)
αw)−(α−1)τ−(α−1)κ
αw)+D1(wq αq)+τ+α+κ
αw)+τ+α+κ
αq)
αw)+τ
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2α e−D1(wq αw)
αw)−(α−1)τ−(α−1)κ
2 e−D1(wq αw)−(α−1)τ−(α−1)κ
2 e−D1(wq αw)+(1−α)τ
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1 → P(Yn 1 )
1
1 →
1
1
1)}
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1/
α
1/
α
α
−nEsc( 1 n ln M L ,W ,p)
1/ 2α
α−1 α
α
α
2α = n− 1−E′ sc(R,W,p) 2
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α n− 1 2α e−nEsc( 1 n ln M L ,W ,p)
qα,p α
qα,p α
qα,p α
qα,p α
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n ln M L ,W ,p)
1−E′ sc( 1 n ln M L ,W,p) 2
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1+z2 2
e
2α e−n[R+ 1 1+α]
2ln 4 e 1 2n
2α e−n[R+ 1 1+α −ln 2]
2ln 4 e
α 1+α = R 1 2ln 4 e
− ln P(n)
e
n
2ln 4 e )
2ln 4 e , ∞)
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1 q))
1 q) , D1(w q)]
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2α e−nD1(wq αw)
αq)
2α e−D1(wq αw)
αq)
2α e−D1(wq αwq 1 ) n
αq)
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