Question

Problem 6.2 In this problem, we will look at a simple application of probability to cryptog raphy. Alice would like to communicate a message to Bob but does not want Eve to learn the message. The difficulty is that Eve can hear everything that Alice says. However, Alice and Bob both have access to a shared random key, which will allow Alice to hide the message from Eve. The block diagram below should help you visualize the scenario. We've kept the schemes simple so that it is possible to evaluate everything by hand. However, modern audio compression schemes are designed and analyzed using the same principles. Those of you who have taken EC401 can think about X as one particular sample XIn from a signal X(t) but this is not important for solving the problem Random Key Generator M→| Alice BobM ve To be more precise, let's say Alice's message is represented by a discrete random variable M and the random key is represented by a discrete random variable K. Let M and K be independent with PMFs Px(k)- 0,1,2,3 m 2.3 0 otherwise. 0 otherwise (a) Write the joint PMF PK(m, k) as a table. (b) To send the message M to Bob, Alice first computes X [M + KI mod 4 and then sends X (instead of sending M directly). Recall that, for any integers a, q, the function [a] mod q is the remainder of a after dividing by q, and can take values 0,1, 2,...,q-1. For each possible value of r, write down the pairs (m, k) from the range of (M, K) satisfying [m+k] mod 4 (c) Using your answers from (a) and (b), write down the joint PMF PMx(m, x) as a table. (Hint: From (b), you have that, for each pair (m, x), there is a unique k such that mkmod 4 so you can just look up each probability in the table from (a).) (d) Determine the marginal PMF Px(x) (e) Determine the conditional PMF PMix(mz) and write it as a table (f) Are M and X independent? (g) Briefly argue that Bob can recover M from X and K, but Eve can do no better than randomly guessing M since she only sees X

Answer 1

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