1

I need a uniform distribution on the unit circle 

(x; y) : x^2 + y^2 <= 1

with the density 

f(x) = (2/Pi) Sqrt[1-x^2]

as a marginal distribution of pairs considered in point e.

I have no idea how to implement this and would appreciate any help.

Michael E2
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mark-markus
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3 Answers3

6

For a uniform distribution on the unit circle you want the angle to be uniform on {0, 2Pi}

thetaDist = UniformDistribution[{0, 2 Pi}];

xDist = TransformedDistribution[
   Cos[theta], theta \[Distributed] thetaDist];

Assuming[-1 < x < 1, PDF[xDist, x] // Simplify]

1/(Pi*Sqrt[1 - x^2])

Verifying that this is properly scaled

Integrate[PDF[xDist, x], {x, -1, 1}]

1

ListPlot[{Cos[#], Sin[#]} & /@
  RandomVariate[thetaDist, 200],
 AspectRatio -> Automatic]

enter image description here

Alternatively,

ListPlot[
 Flatten[
  {{#, Sqrt[1 - #^2]}, {#, -Sqrt[1 - #^2]}} & /@
   RandomVariate[xDist, 200],
  1],
 AspectRatio -> Automatic]

enter image description here

Bob Hanlon
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  • @wolfies - the points on (as opposed to within) the unit circle are the points {Cos[theta], Sin[theta]} with theta uniformly distributed on {0, 2Pi}. The distribution of x is TransformedDistribution[ Cos[theta], theta \[Distributed] thetaDist] For a given x, the corresponding ys on the unit circle are just Sqrt[1-x^2] and -Sqrt[1-x^2]. Hence the mapping of {{#, Sqrt[1 - #^2]}, {#, -Sqrt[1 - #^2]}} onto the list of random x values. – Bob Hanlon Jun 29 '17 at 18:34
3

Since V10.2, there is RandomPoint:

pts = RandomPoint[Disk[], 2000];
Graphics[{
  Circle[],
  Red, Point@pts}]

Mathematica graphics

(If the unit circle was desired, use Circle[] in place of Disk[].)

Michael E2
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0

Just to provide other insights. RandomPoint as per MichaelE2 versus user defined probability distribution and confirmation desired marginal distribution:

pd = ProbabilityDistribution[
  Piecewise[{{1/Pi, 0 < x^2 + y^2 <= 1}}], {x, -1, 1}, {y, -1, 1}]
md = MarginalDistribution[pd, 1]
PDF[md, x]
Show[Histogram3D[pts = RandomPoint[Disk[], 100000], Automatic, "PDF"],
  Plot3D[PDF[pd, {x, y}], {x, -1, 1}, {y, -1, 1}, 
  PlotStyle -> {Blue, Opacity[0.5]}, Mesh -> None]]
Show[Histogram[pts[[All, 1]], Automatic, "PDF"], 
 Plot[PDF[md, x], {x, -1, 1}]]

enter image description here

ubpdqn
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