Codebook Generation Sample Clauses

Codebook Generation. Assume that the input distribu- tion is such that I(u; yr) > I(u; s) as required in Theorem 1. Let εn be a sequence of non-negative numbers that goes to zero such that 2εn < I(u; yr) − I(u; s). • Generate a total of T = 2n(I(u;yr)−2εn ) sequences. Each sequence is sampled i.i.d. from a distribution p (·). Label available power is used for transmitting the secret-message. As the signal-to-noise ratio increases more information regarding − − 1 T them un, . . . , un . sr gets leaked to the eavesdropper and to compensate for this effect, a non-zero fraction of power is transmitted when sr = 0.

Codebook Generation. We now appropriately bound each term in (20). First note that since the sequence un is uniformly distributed among the set of all possible codeword sequences, it follows that 1 n 1 Generate a total of 2n(I(u;ye )−2εn) sequences. Each se- · quence is sampled i.i.d. from a distribution pu( ). • − − Select a rate R = I(u; yr) I(u; ye) εn and randomly partition the set sequences in the previous step into 2nR bins so that there are 2n(I(u;yr)−εn) sequences in each bin. n H(u ) = n log2 |C | − εn = I(u; yr)− εn (21) Next, given (un, sn), as verified below, the channel to the x • xp xd Received Point Transmitted Point P + Q + 2q PQ Figure 2: Secret-key agreement codebook for the dirty paper channel. The transmitter signal xn is selected so that un = xn + sn is a sequence in the random codebook. The legitimate receiver can decode un (with high probability) and map it to the secret- key. The eavesdropper’s noise-uncertainity sphere includes are possible key values. Note that unlike the traditional dirty-paper code, the transmiter signal xn has a component along sn. The achievable rate, does depend on the interference power and hence it is beneficial to amplify it using part of the transmit power. Also note that unlike a dirty-paper code we do not scale down sn before quantizing but use α = 1. Finally, in order to lower bound the term I(sn; yn|un) we pyn|un,sn (yn|un, sn) let J to be a random variable which equals 1 if (sn, un) are jointly typical. Note that Pr(J = 1) = 1 − on(1). = ∑ pyn|un ,sn,xn (yn|un, sn, xn)p(xn|un, sn)(xn|un, sn) e e ∈ xn X n ∈ = ∑ ∏ xxx |u,s,x (ye,i|ui, si, xi)px|u,s (xi|ui, si) 1 I n n n 1 n n 1 x X i=1 n (s ; ye |u ) = n H(s |u )− n H(s |u , ye ) = ∏ ∑ xxx|u,s,x (ye,i|ui, si, xi)px|u,s (xi|ui, si) 1 n n 1 n n n ≥ H(s |u , J = 1) Pr(J = 1)− H(s |u , y ) i=1 xi∈X 1 n n 1 n n n = ∏ xxx|u,s (ye,i|ui, si) ≥ n H(s |u , J = 1)− n H(s |u , ye )− on(1) 1 n n n ≥ H(s|u)− n H(s |u , ye )− on(1) (25) i=1 The second step above follows from the fact that the channel 1 n is memoryless and the symbol xi at time i is generated as a function of (ui, si). Hence we have that ≥ H(s|u)− n ∑ H(si|ui, ye,i)− on(1) (26) | |n 1 n n n n n n i−1 H(ye s , u ) = ∑ H(ye,i s i=1 , u , ye,1 ) (22) where (25) follows from the fact that sn is an i.i.d. sequence and hence conditioned on the fact that (sn, un) is a pair of | = ∑ H(ye,i si, ui) (23) i=1 ≤n

Codebook Generation. Assume that the input distribu- tion is such that I(u; yr) > I(u; s) as required in Theorem 1. Let εn be a sequence of non-negative numbers that goes to zero such that 2εn < I(u; yr) − I(u; s). • Generate a total of T = 2n(I(u;yr)−2εn ) sequences. Each 1 1 1 sequence is sampled i.i.d. from a distribution pu(·). Label

Codebook Generation. For i = 1, 2, randomly generate Fix probability distribution p(t1, t2, t) = p(t)p(t1|t)p(t2|t)