Secrecy Analysis Sample Clauses

Secrecy Analysis. We need to show that for the proposed encoder and decoder, the equivocation at the eavesdropper satisfies i=1 The second step above follows from the fact that the channel is memoryless and the symbol xi at time i is generated as a n e r e n 1 H(κ|y n) = I(u; y ) − I(u; y ) + o

Secrecy Analysis. We need to show that for the proposed encoder and decoder, the equivocation at the eavesdropper satisfies The second step above follows from the fact that the channel is memoryless and the symbol xi at time i is generated as a function of (ui, si). Hence we have that n H(κ|y n) = I(u; yr) − I(u; ye) + on(1), (31) H(y n sn, un) = | n Σn i=1 H(ye,i|si, ui). (35) → ∞ where on(1) is a term that goes to zero as n . Note that while the key κ in general can be a function of (sn, mx ) as indicated in (1), in our coding scheme the secret Furthermore note that Σ n 1 H(y n) ≤ H(y

Secrecy Analysis. First, consider splitting y n = and finally from the chain rule (y n1 , . . . , y nM ) where the subsequence y nj is obtained by grouping the symbols of ye when s = sj. From the con- n n e1 eM ej → ∞ 1 H(y n) ≤ 1 H(y ) → H(y ) (60) ei e struction of the wiretap codebook Cj it follows that n j ej n j n 1 H(κ |y nj ) ≥ 1 H(κ ) − ε , j = 1, . . . , M (49) as n . Substituting (55), (58), (59) and (60) into (54) completes the claim. The secrecy analysis can be completed by combining (51) Next since the messages are selected independently and the encoding functions are also independent it follows that and (52) as shown below. 1 H(κM , κ |y n) = 1 H(κM |κ , y n) + 1 H(κ |y n) (61) e H(κj|κ1, . . . , κj−1, κj+1, . . . , κM , y n, sn) n 1 s e n 1 s e n s e n ≥ H(κM |sn, y n) + H(κs|y n) (62) = H(κj|y n ) ≥ 1 n 1 H(κj) − εn (50) e n e 1 n ej n n e n ≥ I(u; yr|s) − I(u; ye|s) + H(κs|y ) − on(1) (63) Thus by the chain rule we have that 1 n n 1 n n e 1 n H(κ1, . . . , κM |y n, sn) ≥ R0 − εn (51) ≥ I(u; yr|s)−I(u; ye|s)+ n H(s |ye ) − n H(s |ye , κs)−on(1) (64) | − | where R0 = H(κ1, . . . , κM ) = I(u; yr s) I(u; ye s). To complete the secrecy analysis we require the following addi- ≥ I(u; yr|s) − I(u; ye|s) + H(s|ye) − n H(s |ye , κs) − on(1) n n

Secrecy Analysis. R− = 2 log 1 + P + Q + 1 + ∆ + 2ρ√PQ , (11) We need to show that for the proposed encoder and decoder, where ρ < 1 is the largest value that satisfies the equivocation at the eavesdropper satisfies − ≥ − P(1 ρ2) 1 1 P + Q + 1

Secrecy Analysis. First, consider splitting y n = and finally from the chain rule (y n1 , . . . , y nM ) where the subsequence y nj is obtained by grouping the symbols of ye when s = sj. From the con- n n e1 eM ej → ∞ 1 H(y n) ≤ 1 H(y ) → H(y ) (60) ei e struction of the wiretap codebook Cj it follows that 1 nj 1 n j ej n j n H(κ |y ) ≥ H(κ ) − ε , j = 1, . . . , M (49) as n . Substituting (55), (58), (59) and (60) into (54) completes the claim. The secrecy analysis can be completed by combining (51) Next since the messages are selected independently and the encoding functions are also independent it follows that and (52) as shown below.