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Weak secrecy does not imply strong secrecy

Statement

Consider the noiseless finite wiretap channel with $X = Y = Z = {0, 1}$ and $W (y, z ∣ x) = 1$ exactly when $y = z = x$ . At blocklength $n$ , let $k_{n} = ⌊ \sqrt{n} ⌋$ , let $M_{n}$ be uniform on ${0, 1}^{k_{n}}$ , and transmit $X^{n} = (M_{n}, 0^{n - k_{n}})$ . Bob and Eve both observe the transmitted word exactly. This reliable code sequence satisfies weak secrecy but not strong secrecy: its leakage is $I (M_{n}; Z^{n}) = k_{n} \ln 2 \to \infty$ , whereas $n^{- 1} I (M_{n}; Z^{n}) \to 0$ . In general, strong secrecy implies weak secrecy.

Facts & Assumptions

Given: The finite wiretap code model, finite mutual information and the weak/strong secrecy criteria.

[F1]

Finite discrete memoryless wiretap channel and code model: a wiretap code has a uniform message, Bob observes $Y^{n}$ , Eve observes $Z^{n}$ , and error includes erasure.

[F2]

Information leakage and equivocation for finite observations: leakage is $I (M; Z^{n})$ in nats and normalized leakage is $n^{- 1} I (M; Z^{n})$ .

[F3]

Weak and strong secrecy for wiretap code sequences: weak secrecy requires normalized leakage to vanish, while strong secrecy requires unnormalized leakage to vanish.

[F4]

Finite mutual information: $I (M; Z^{n}) = D (P_{M Z^{n}} ∥ P_{M} \otimes P_{Z^{n}})$ for finite variables.

[F5]

Finite entropy and mutual-information chain rules: $I (M; Z^{n}) = H (M) - H (M ∣ Z^{n})$ and the entropy chain rule holds for finite variables.

[F6]

Finite KL divergence is nonnegative with equality exactly at equal laws: finite KL divergence is nonnegative.

[F7]

Finite distributions, entropy, divergence, and stochastic matrices: a uniform law on $N$ points has entropy $\ln N$ .

Proof

1.1F1givenconstruct

For each $n$ , the input string has length $k_{n} + (n - k_{n}) = n$ . The noiseless channel gives $Y^{n} = Z^{n} = X^{n}$ with probability one, so Bob recovers the first $k_{n}$ bits exactly and the error probability is zero. Eve's observation also determines $M_{n}$ exactly, hence $H (M_{n} ∣ Z^{n}) = 0$ .

2.1F5F7step 1.1algebra

By uniformity and [F7], $H (M_{n}) = \ln (2^{k_{n}}) = k_{n} \ln 2$ . Thus [F5] gives $I (M_{n}; Z^{n}) = H (M_{n}) - H (M_{n} ∣ Z^{n}) = k_{n} \ln 2$ .

3.1F2F3step 2.1algebra

Since $\sqrt{n} - 1 < k_{n} \leq \sqrt{n}$ , one has $k_{n} \to \infty$ and $k_{n} / n \to 0$ . Step 2.1 therefore gives unnormalized leakage tending to infinity and normalized leakage tending to zero by [F2]. With zero error this is weak secrecy and failure of strong secrecy by [F3].

4.1F2F3F4F6algebra∎

For any wiretap code sequence, [F4] identifies mutual information with finite KL divergence, and [F6] makes it nonnegative. Hence $0 \leq I (M_{n}; Z^{n}) / n \leq I (M_{n}; Z^{n})$ for $n \geq 1$ . If the right side tends to zero, so does the left side, proving that strong secrecy implies weak secrecy.

Depends on

Used by

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Sources

Paul Cuff, Princeton ELE 538: Information Theoretic Security, Lecture 15 (standard reference, not scraped)
Matthieu R. Bloch and J. Nicholas Laneman, Strong Secrecy from Channel Resolvability, arXiv:1105.5419 (standard reference, not scraped)