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I have a problem that requires a distribution on the positive orthant (all dimensions >= 0) of the unit hypercube. Therefore,

$$ {\bf x} \in [0,1],\hspace{0.5cm}x_i = 1 \hspace{0.2cm}\text{ for some }\hspace{0.2cm}i \in \{1,\ldots,d\rbrace. $$

For a $d$-dimensional hypercube, there are $d-1$ degrees of freedom, so I can't simply use, for instance, a $d$-dimensional beta. Are there any distributions defined on this space? Alternatively, are there any dimension reduction techniques that I can transform it into a $d$-1 dimensional space?

I've currently used polar coordinate transformation, and I'm looking at the simplex now.

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  • $\begingroup$ There are many ways to approach this. There is a nice one-to-one continuous map from the portion of the sphere $S^{d-1}$ in the positive orthant and this surface: given positive $y_i$ such that $y_1^2 + \cdots + y_d^2=1,$ let $y$ be the largest of these values and set $x = (y_1/y,y_2/y,\ldots,y_d/y).$ Thus you can use literally any distribution on the sphere in your application (after conditioning it on the positive orthant). Such distributions correspond to distributions on $\mathbb{R}^{d-1}$ (use Stereographic projection, for instance). $\endgroup$
    – whuber
    Jan 11, 2021 at 20:42

1 Answer 1

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Just place all d-1-hypercubes on one plane next to each other, then sample from the constructed hyper-rectangle. There's 2d hypercubes of d-1 dimensions.

Then put back d-1 hypercubes back where they were. So the points will be on the surface of d-hypercube

Here's Python dem for 3D:

import numpy as np

d = 3
n = 1000
r = np.random.random(size=(n,d-1))
k = np.random.randint(d,size=(n,1))
j = np.random.randint(2,size=(n,1))
x = np.zeros((n,d))
for i in np.arange(n):
  x[i, np.arange(d)[np.arange(d)!=k[i]]] = r[i,:]
  x[i, k[i]] = j[i]

from mpl_toolkits.mplot3d import Axes3D  # noqa: F401 unused import
import matplotlib.pyplot as plt

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x[:,0], x[:,1], x[:,2])

plt.show()

enter image description here

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