5
Use a normal distribution to generate $n$ values and Normalize to get a point on the sphere. Make sure that the last coordinate always has the same sign using Abs. Generate millions of these points and estimate the mean distance between pairs:
n = 3;
topt[p_] := MapAt[Abs, Normalize[p], -1]
points = topt /@ RandomVariate[NormalDistribution[0, 1], {1000000, n}...
4
Maybe
SeedRandom[1]
n = 300; d = 30;
p = RandomPoint[Sphere[d], {n, 3}];
xyz = Apply[EuclideanDistance, Subsets[#, {2}] & /@ p, {2}];
Mean[Map[Max, xyz]]
1.5164
If we use Ball[d] instead of Sphere[d] we get 1.47179.
And a modification of your code (eliminating repeated invocations of Subsets[#,3] on p):
SeedRandom[1]
n = 300; d = 30;
p = ...
3
Seems no so effective.
Here we use RandomPoint to select uniform points in Ball[] and use RandomSample to select three points to construct a random triangle.
SeedRandom[1];
n = 300;
d = 30;
pts = RandomPoint[Ball[d], n];
Table[Max @@
EuclideanDistance @@@ Subsets [RandomSample[pts, 3], {2}], {i,
200000}] // Mean
1.46926
3
Not an exact answer but a Monte-Carlo way of checking the exact answers.
Generate a random point on the unit $n$-sphere:
P[n_Integer?Positive] := Normalize[RandomVariate[NormalDistribution[], n]]
Measure the mean distance between a random point $P_0$ and another random point on the unit $n$-sphere, by averaging over $m$ random points:
M[n_Integer?Positive, ...
2
For $n=3$:
The PDF is $f(d) = d/2$ or in Mathematica TriangularDistribution[{0, 2}, 2] - a ramp shaped distribution. We can test this numerically and we get a high $p$-value of about 0.31 so it's a good fit:
points = RandomPoint[Sphere[], {1000000, 2}];
distances = EuclideanDistance @@@ points;
testdist = TriangularDistribution[{0, 2}, 2];
...
2
Mathematica often pre-Standardize's data before learning. For example, see my answer here about learning SVM hyperplanes where I also encountered this problem. Therefore things are a bit more involved than my comment about the documentation if you want to extract the right parameters. I hope you can follow along:
(* some sample data, learn a 3 component ...
2
Another way to do the sampling (taking advantage of the built in Sphere function and RandomPoint functionality (modified from a similar question on sampling from the surface of the sphere
distanceDistributionOnHalfSphere[dimensionality_, nSamples_:10^5] :=
With[{
(* take a few extra samples account for loss *)
randomPointsOnSurfaceOfNSphere = ...
1
Clear["Global`*"]
GeometricDistribution is built-in
dist[p_] = GeometricDistribution[p];
PDF[dist[p], x]
(* Piecewise[{{(1 - p)^x*p, x >= 0}}, 0] *)
CDF[dist[p], q]
(* Piecewise[{{1 - (1 - p)^(1 + Floor[q]), q >= 0}}, 0] *)
DistributionParameterAssumptions[dist[p]]
(* 0 < p <= 1 *)
Random samples from the distribution can be ...
1
RandomPoint can be used to sample from arbitrary Region definitions, and Sphere describes the unit sphere in $R^n$ (it can be both a geometric region and a graphics primitive). So @flinty's original solution can be generalized to arbitrary numbers of dimensions. Using this type of approach avoids having to know very much about the problem (as in @Roman's ...
1
Extracting the component distribution parameters from a mixed-normal distribution (for n > 2 normal)
We had a discussion and we came to the conclusion (I think) that what was needed was a "mixed model" rather than a "mixture model". A simplified definition of a mixed model is a regression with more than one error term (resulting in some observations being correlated with other observations). Mathematica does not currently offer a ...
1
Extracting the component distribution parameters from a mixed-normal distribution (for n > 2 normal)
While the estimation process will process any dataset and real datasets never really match the what the model expects, the dataset you created isn't a mixture distribution in the usual sense in that you have the counts of each subpopulation fixed. In a standard mixture distribution, the counts of each subpopulation will vary.
To generate data for your ...
1
With[{sigma = 15, mu = 0},
t = Table[E^(-(x - mu)^2/(2 sigma^2)), {x, -50, 50}];
ListPlot[# Rescale[t] & /@ {5, 10, 15}, PlotRange -> Full,
Filling -> Bottom, DataRange -> {-50, 50}]]
Alternatively,
With[{sigma = 15, mu = 0},
t = Table[{x, E^(-(x - mu)^2/(2 sigma^2))}, {x, -50, 50}];
ListPlot[ScalingTransform[{1, #}][t] & /@ {5, 10,...
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