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 socalled “canonical
path”.
To get an idea of how the ±rst step works, consider the 2dimensional 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
−