On extracting common random bits from correlated sources on large   alphabets
  Suppose Alice and Bob receive strings $X=(X_1,...,X_n)$ and $Y=(Y_1,...,Y_n)$ each uniformly random in $[s]^n$ but so that $X$ and $Y$ are correlated . For each symbol $i$, we have that $Y_i = X_i$ with probability $1-\eps$ and otherwise $Y_i$ is chosen independently and uniformly from $[s]$.   Alice and Bob wish to use their respective strings to extract a uniformly chosen common sequence from $[s]^k$ but without communicating. How well can they do? The trivial strategy of outputting the first $k$ symbols yields an agreement probability of $(1 - \eps + \eps/s)^k$. In a recent work by Bogdanov and Mossel it was shown that in the binary case where $s=2$ and $k = k(\eps)$ is large enough then it is possible to extract $k$ bits with a better agreement probability rate. In particular, it is possible to achieve agreement probability $(k\eps)^{-1/2} \cdot 2^{-k\eps/(2(1 - \eps/2))}$ using a random construction based on Hamming balls, and this is optimal up to lower order terms.   In the current paper we consider the same problem over larger alphabet sizes $s$ and we show that the agreement probability rate changes dramatically as the alphabet grows. In particular we show no strategy can achieve agreement probability better than $(1-\eps)^k (1+\delta(s))^k$ where $\delta(s) \to 0$ as $s \to \infty$. We also show that Hamming ball based constructions have {\em much lower} agreement probability rate than the trivial algorithm as $s \to \infty$. Our proofs and results are intimately related to subtle properties of hypercontractive inequalities. 
Siu On Chan, Elchanan Mossel, Joe Neeman
cs.IT math.IT