The problem of generating a shared secret key S by two parties knowing dependent random variables X and Y, respectively, but not sharing a secret key initially, is considered. An enemy who knows the random variable Z, jointly distributed with X and Y according to some probability distribution P/sub XYZ/, can also receive all messages exchanged by the two parties over a public channel. The goal of a protocol is that the enemy obtains at most a negligible amount of information about S. Upper bounds on H(S) as a function of P/sub XYZ/ are presented. Lower bounds on the rate H(S)/N (as N to infinity ) are derived for the case in which X=(X/sub 1/, . . ., X/sub N/), Y=(Y/sub 1/, . . ., Y/sub N/) and Z=(Z/sub 1/, . . ., Z/sub N/) result from N independent executions of a random experiment generating X/sub i/, Y/sub i/ and Z/sub i/ for i=1, . . ., N. It is shown that such a secret key agreement is possible for a scenario in which all three parties receive the output of a binary symmetric source over independent binary symmetric channels, even when the enemy's channel is superior to the other two channels.
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机译：考虑了由分别知道相关随机变量X和Y但由两个当事方生成共享密钥S的问题，但是最初不共享密钥。知道随机变量Z（根据某个概率分布P / sub XYZ /与X和Y共同分配）的敌人也可以接收双方通过公共通道交换的所有消息。协议的目的是使敌人最多获得有关S的微不足道的信息。根据P / sub XYZ /给出H（S）的上限。对于其中X =（X / sub 1 /，..，X / sub N /），Y =（Y / sub，N）的情况，得出速率H（S）/ N的下界（从N到无穷大）。 1 /，..，Y / sub N /）和Z =（Z / sub 1 /，..，Z / sub N /）是随机产生N个独立实验产生X / sub i /，Y的结果/ sub i /和Z / sub i / for i = 1，。 。可以看出，这种秘密密钥协议对于以下情况是可能的：即使当敌方的通道优于其他两个通道时，所有三方都通过独立的二进制对称通道接收二进制对称源的输出。
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