# Computational geometry – How to find the 2D distribution of the Voronoi cell area?

@ C.E. The answer is a good start, but without careful a consideration could generate erroneous statistics. Bring together good statistics build a "big" 5000 VoronoiMesh cells within one unit `Disk`:

``````pts = RandomPoint[Disk[], 5000];
mesh = VoronoiMesh[pts, Axes -> True]
``````

We note that the statistics are obviously offset by the presence of "border" cells of a much larger area than that of ordinary internal cells. Exclude large cells by selecting only those from the region of origin of the random point distribution – unit `Disk`:

``````vor = MeshPrimitives[mesh,2];
vor // length
``````

5000

``````vorInner = Select[vorRegionWithin[vorRegionWithin[vorRegionWithin[vorRegionWithin[Disk[], #]&];
vorInner // length
``````

4782

Of course we have fewer elements and they are all nice regular cells:

``````Graphic[{FaceForm[None], EdgeForm[Gray], vorInner}]
`````` You can now see that there is a minimum of distribution in the zero zone. Of course, the cells near the null surface are not very present with the finite number of points per region (or finite density). So, there is some predominant finite mean area here.

``````areas = area / @ vorInner;
hist = histogram[areas, Automatic, "PDF", PlotTheme -> "Detailed"]
`````` You can find a simple analytic distribution very close:

``````dis = FindDistribution[areas]
``````

which corresponds very well to the empirical histogram:

``````Show[hist, Plot[PDF[dis, x], {x, 0, .0015}]]
`````` And now, it's easy to find something like:

``````{Mean[dis], Variance[dis], Kurtosis[dis]}
``````

{0.0006175091492073445`, 1.1396755130437449`* ^ – 7, 4.793269963533813`}

``````Probability[x > .0005, x [Distributed] say]
``````

0.5740480899719699`

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