Foundations of Data Science

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1 − ɛ
Annulus of
width 1 d
Figure 2.1: Most of the volume of the d-dimensional ball of radius r is contained in an
annulus of width O(r/d) near the boundary.
we shrink each of the d sides of a d−dimensional cube by a factor f, its volume shrinks
by a factor of f d . Using the fact that 1 − x ≤ e −x , for any object A in R d we have:
volume ( (1 − ɛ)A )
= (1 − ɛ) d ≤ e −ɛd .
volume(A)
Fixing ɛ and letting d → ∞, the above quantity rapidly approaches zero. This means
that nearly all of the volume of A must be in the portion of A that does not belong to
the region (1 − ɛ)A.
Let S denote the unit ball in d dimensions, that is, the set of points within distance
one of the origin. An immediate implication of the above observation is that at least a
1 − e −ɛd fraction of the volume of the unit ball is concentrated in S \ (1 − ɛ)S, namely
in a small annulus of width ɛ at the boundary. In particular, most of the volume of the
d-dimensional unit ball is contained in an annulus of width O(1/d) near the boundary. If
the ball is of radius r, then the annulus width is O ( r
d)
.
2.4 Properties of the Unit Ball
We now focus more specifically on properties of the unit ball in d-dimensional space.
We just saw that most of its volume is concentrated in a small annulus of width O(1/d)
near the boundary. Next we will show that in the limit as d goes to infinity, the volume of
the ball goes to zero. This result can be proven in several ways. Here we use integration.
2.4.1 Volume of the Unit Ball
To calculate the volume V (d) of the unit ball in R d , one can integrate in either Cartesian
or polar coordinates. In Cartesian coordinates the volume is given by
√
√
x 1 =1 x
∫ 2 = 1−x 2 x
∫ 1 d = 1−x
∫
2 1 −···−x2 d−1
V (d) =
· · ·
dx d · · · dx 2 dx 1 .
x 1 =−1
√
x 2 =− 1−x 2 1
x d =− √ 1−x 2 1 −···−x2 d−1
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