l06 - 6.896 Sublinear Time Algorithms February 22, 2007...

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6.896 Sublinear Time Algorithms February 22, 2007 Lecture 6 Lecturer: Ronitt Rubinfeld Scribe: Ning Xie In this lecture, we continue studying the problem of testing whether a monotone distribution is uniform. We will restrict to distributions over the Boolean cube. First recall some notation. Let p : {− 1 , 1 } n R 0 be a monotone distribution over the Boolean cube. De±ne δ ( x )=2 n p ( x ) 1, that is, the weight deviation from uniform distribution. Note that since p ( x ) is monotone, so is δ ( x ). As we proved last time, x δ ( x )=0and x | δ ( x ) | =2 n ± ,where ± is the ` 1 -distance between p and uniform distribution, ± = | p U | 1 . We also de±ne, for each x ∈{− 1 , 1 } n , the bias of x to be bias ( x )= n i =1 x i . It is easy to see that for uniform distribution E U [ bias ( x )] = 0. Our main result for this lecture is the follow lemma: Lemma 1 Let p be a monotone distribution over Boolean cube and ± = x ∈{− 1 , 1 } n | p ( x ) 1 2 n | ,then E p [ bias ( x )] ±. Proof Let S = E p [ bias ( x )], our goal is to obtain a lower bound for it in term of ± .S ince p ( x )= 1 2 n + δ ( x ) 2 n and the expectation of bias(x) over the uniform distribution is zero, we can rewrite S as S = X x n X i =1 x i p ( x )= 1 2 n X x n X i =1 x i δ ( x ) . Our proof strategy is to ±rst transform this sum over nodes to a sum over the edges in the Boolean cube. Then we get a lower bound of the sum over the edges by considering a so-called “canonical path”. To get an idea of how the ±rst step works, consider the 2-dimensional Boolean cube with A =(1 , 1) ,B =(1 , 1) ,C =( 1 , 1) and D =( 1 , 1). Then we have 2 n S = δ A 1+ δ A 1+ δ B 1+ δ B ( 1) + δ C ( 1) + δ C 1+ δ D ( 1) + δ D ( 1) =( δ A δ B )+( δ A
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This note was uploaded on 04/02/2010 for the course CS 6.896 taught by Professor Ronittrubinfeld during the Fall '04 term at MIT.

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l06 - 6.896 Sublinear Time Algorithms February 22, 2007...

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