Secrecy Analysis Clause Samples
The Secrecy Analysis clause defines the procedures and standards for evaluating and maintaining the confidentiality of sensitive information shared between parties. It typically outlines the types of information subject to secrecy, the methods for assessing potential risks of disclosure, and the responsibilities of each party in safeguarding data. For example, it may require regular reviews of security protocols or mandate specific encryption practices. The core function of this clause is to ensure that confidential information remains protected, thereby minimizing the risk of unauthorized access or leaks.
Secrecy Analysis. We need to show that for the proposed encoder and decoder, the equivocation at the eavesdropper satisfies n
(1) (31) function of (ui, si). Hence we have that where on(1) is a term that goes to zero as n → ∞. Σn H(y n|sn, un) = H(ye,i|si, ui). (35) 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 key is a deterministic functino of un and hence we have Furthermore note that e 1 | H(κ y n) = H(κ, un|y n) − e | H(un y n, κ) Σn e 1 ≤ H(y n) H(yei). (36) e 1 1 n n 1 n n = H(u |ye ) − n H(u |ye , κ) Finally, in order to lower bound the term n; y n|un) we e = H(un|y n) − εn let J to be a random variable which equals 1 if (sn, un) are n jointly typical. Note that Pr(J = 1) = 1 − on(1). where the last step follows from the fact that there are 1 I(sn; y n|un) = 1 H(sn|un) − 1 H(sn|un, y n) e e T0 = 2n(I(u;ye)−εn ) sequences in each bin. Again applying n n n the packing lemma we can show that with high probability 1 n n 1 n n n ≥ H(s |u , J = 1) Pr(J = 1) − H(s |u , y ) the eavesdropper uniquely finds the codeword un(L) jointly n n e typical with y n in this set and hence ▇▇▇▇’s Inequality implies 1 n n 1 n n n ≥ H(s |u , J = 1) − H(s |u , y ) − o
(1) that e 1 n n 1 n n n It remains to show that |ye , κ) ≤ εn. ≥ H(s|u) − n H(s 1 Σn |u , ye ) − on(1) (37) 1 H(un |ye ) ≥ I(u; yr) − I(u; ye) − on(1). ≥ H(s|u) − n i=1 H(si|ui, ye,i) − on(1) (38) n Using the chain rule of the joint entropy we have where (37) follows from the fact that sn is an i.i.d. sequence 1 n n 1 n 1 n n 1 n and hence conditioned on the fact that (sn, un) is a pair of |ye ) = n H(u ) + n H(ye |u ) − n H(ye ) (32) typical sequence there are 2nH(s|u)−non(1) possible sequences 1 n 1 n n n 1 n 1 n n n sn. = H(u ) + n H(ye |u , s ) − n H(ye ) + n I(s ; ye |u ).
Secrecy Analysis. We need to show that for the proposed encoder and decoder, the equivocation at the eavesdropper satisfies 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- e1 eM ej → ∞ 1 H(y n) ≤ 1 H(y ) → H(y ) (60) ei e struction of the wiretap codebook Cj it follows that j ej 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 e ≥ 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 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