K-wise Independence and -biased K-wise Indepedence

1 Deenitions Consider a distribution D on n bits x = x 1 x n. D is k-wise independent ii for all sets of k indices S = fi 1 x ik = a 1 a k ] = 1 2 k : The idea is that if we restrict our attention to any k positions in x, no matter how many times we sample from D, we cannot distinguish D from the uniform distribution over n bits. We can get a Fourier interpretation of k-wise independence by viewing the distribution D as a function from Z n 2 to R. Let jxj denote the Hamming weight of x. Claim 1 D is k-wise independent (according to the deenition above) ii ^ D(y) = 0 for all y 6 = 0 : jyj k. Proof: (sketch) For the forward direction, recall that ^ D(y) = P x (?1) yx D(x). Let S y be the set of ones in y, and let X Sy = L i2Sy x i. Then, ^ D(y) = X x:XS y =0 For jyj = k, the deenition of k-wise independence guarantees that if we look at just the k positions S y , each of the 2 k possible values is equally likely. In particular, this means that the XOR of these bits is equally likely to be a 0 or a 1, and hence ^ D(y) = 0. Since k-wise independence implies (k ? 1)-wise independence, this same argument shows that ^ D(y) = 0 for any y : 0 < jyj < k. Conversely, suppose all the low-order Fourier coeecients are 0. Since ^ D(100 0) = Prx 1 = 0] ? Prx 1 = 1] = 0, x 1 is equally likely to be 0 or 1, and similarly for the other x i 's. Now let p a1a2 = Prx 1 x 2 = a 1 a 2 ]. Since x 1 , x 2 , and x 1 x 2 are all unbiased, we get the equations p 00 + p 01 = p 00 + p 10 = p 00 + p 11 = 1=2. Furthermore, p 00 + p 01 + p 10 + p 11 = 1. From these equations, we deduce that all of the p a1a2 's are 1=4, and similarly for any pair x i and x j. Proceeding inductively in this manner …