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By: Marcel Goh

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In reply to Marcel Goh.

Elaborating on my earlier comment, letting (X_1, Y_1) and (X_2, Y_2) be conditionally independent trials of (X,Y), I think I can show that H(X_1, Y_2 | X+Y) \le H(X)/2 + H(Y)/2 + \log K assuming Lemma A.2 as well as (24) and (25) from Theorem 3.1 in the older paper. This is not exactly (28), and I’m not sure how to transform the - into a +, since X_1-Y_2 gives us a lot more information than X_1 + Y_2 from the coupling $X_1+Y_1 = X_2+Y_2$. (At least I think it does.)


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