Foundations of Data Science

The volume of the hemisphere below the plane x 1 = √ 1
d−1
is a lower bound on the
entire volume of the upper hemisphere and this volume is at least that of a cylinder of
√
1
height √
d−1
and radius 1 − 1
1
. The volume of the cylinder is V (d−1)(1− ) d−1
2
d−1 d−1
Using the fact that (1 − x) a ≥ 1 − ax for a ≥ 1, the volume of the cylinder is at least
V (d−1)
2 √ d−1
for d ≥ 3.
√ 1
d−1
.
Thus,
ratio ≤
V (d−1)
upper bound above plane
lower bound total hemisphere =
c √ d−1 e− c2 2
V (d−1)
2 √ d−1
= 2 c e− c2 2
One might ask why we computed a lower bound on the total hemisphere since it is one
half of the volume of the unit ball which we already know. The reason is that the volume
of the upper hemisphere is 1 V (d) and we need a formula with V (d − 1) in it to cancel the
2
V (d − 1) in the numerator.
Near orthogonality. One immediate implication of the above analysis is that if we
draw two points at random from the unit ball, with high probability their vectors will be
nearly orthogonal to each other. Specifically, from our previous analysis in Section 2.3,
with high probability both will have length 1 − O(1/d). From our analysis above, if we
define the vector in the direction of the first point as “north”, with high probability the
second will have a projection of only ±O(1/ √ d) in this direction. This implies that with
high probability, the angle between the two vectors will be π/2 ± O(1/ √ d). In particular,
we have the theorem:
Theorem 2.8 Consider drawing n points x 1 , x 2 , . . . , x n at random from the unit ball.
With probability 1 − O(1/n)
1. |x i | ≥ 1 − 2 ln n
d
for all i, and
2. |x i · x j | ≤ √ 6 ln n √
d−1
for all i ≠ j.
Proof: For the first part, for any fixed i, by the analysis of Section 2.3 Prob ( |x i | <
)
1 − 2 ln n
d ≤ e
−( 2 ln n
d )d = 1/n 2 . So, by the union bound, the probability there exists i such
that |x i | < 1 − 2 ln n is at most 1/n. For the second part, there are ( n
d
2)
pairs i and j, and
for each such pair, if we define x i as “north”, the probability that the projection of x j onto
the “North” direction is more than √ √6 ln n
d−1
(a necessary condition for the dot-product to
be large) is at most O(e − 6 ln n
2 ) = O(n −3 ) by Theorem 2.7. Thus, this condition is violated
with probability at most O (( n
2) ) n
−3
= O(1/n) as well.
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