Figure 1: This figure illustrates Exercise 1.2 for the case d = 2 and = 0.1.
The boundary is the unit square. Points outside the dashed square are within
of the boundary. P
Points between the two parallel dashed lines are within of the center plane j xj = d/2. The shaded region is within of both the
center plane and the boundary. In high dimensions, most of the volume would
be shaded.
1.2 High dimensional geometry tells us that most of the volume in the hypercube
[0, 1]d is near the boundary, when d is large. Specifically, letting
dB (x) = min min(xj , 1 − xj )
16j6d
be the distance from x to the boundary, and A,d = x ∈ [0, 1]d | dB (x) 6 for 0 < < 1/2 and d > 1, then vol(A,d ) = 1 − (1 − 2)d .
On the other hand, for large d, the central limit theorem
Pd tells us that most
of the points x ∈ [0, 1]d are near a hyperplane {x |
j=1 xj = d/2}, which
passes through the center of the cube. Letting
d
1 X
dC (x) = √ (xj − 1/2)
d j=1
be the distance of x to the center plane, B,d = x ∈ [0, 1]d | |dC (x)| 6 has
√
volume nearly 1 − 2Φ(− 12), for large d.
The shaded region in Figure 1 is A,2 ∩B,2 , and so for small d the intersection
of these sets is not large. Use Monte Carlo to estimate vol(A,d ∩B,d ) for = 0.1
and d ∈ {2, 3, 4, 5, 10, 20}.
1