baricentric distributions

Let $\mathcal{P}$ be the convex hull of a point set $p_1, \dotsc, p_n$ (for simplicity, assume that no $p \in P$ lies in the convex hull of the other points.) Now, pick a point uniformly at random from the simplex $x_1 + \dotsc + x_n = 1, x_i \geq 0,$ and take $\phi=\sum x_i p_i.$ The point $\phi$ will be contained in $\mathcal{P}$ and in the simple case where $\mathcal{P}$ is a triangle, $\phi$ is uniformly distributed. In general, however, it is not: The two graphics are for the case $\mathcal{P}$ is a square. The first is a simple plot of a million points sampled from the distribution - the distribution looks like it is trying to be supported on the quadrangle spanned by the midpoints of the sides of the square.

The second is the density histogram:

which seems to indicate that the distribution has a bit of a peak at the center of gravity.

Any deep thoughts?

• For the square, the density seems to be proportional to $\min\{x+y,2-x-y\}-\max\{x-y,y-x\}=2\min\{x,y,1-x,1-y\}$ (take the uniform distribution in the tetrahedron spanned by $(0,0,0)$, $(1,1,0)$, $(1,0,1)$, $(0,1,1)$ and project it on the $xy$ plane), which does not agree with your plots. What did I miss? – Mateusz Kwaśnicki Mar 24 '18 at 19:50
• Linearity of expectation implies that $E(\sum X_i P_i)=\sum E(X_i)P_i =\sum \frac{1}{n}P_i=G,$ where $G$ is the center of gravity. Also, you should have a look at this paper: dtic.mil/dtic/tr/fulltext/u2/273207.pdf – Donatien Bénéat Mar 25 '18 at 7:20
• Maybe the fact that barycentric coordinates aren't equal to the $x_i$ for polygons, that are not triangles, plays a rôle, c.f. uniqueness issues for barycentric coordinates – Manfred Weis Mar 25 '18 at 18:18

The problem seems to lie in the way you sample the unit simplex. Let $(U_i)_{1\leq i\leq n}$ be a sequence of i.i.d. uniform random variables, $S=\sum U_i$ and define $X_i=U_i/S.$ Then non-intuitively, $(X_1,\dots,X_n)$ is not uniformly distributed over the unit simplex. The following plot shows 300000 points on the unit square with barycentric coordinates generated using this method:
The method can easily be fixed: if the $U_i$s defined above are i.i.d. exponential random variables, then $(X_1,\dots,X_n)$ is uniformly distributed over the simplex. Again, the following plot shows 300000 points generated according to this second method:
It should be noted that uniformly distributed $X_i$s doesn't imply uniformly distributed $\phi,$ as said by Mateusz Kwaśnicki in a comment.
• As a side remark: Another simple method is to generate $X_1,X_2,\ldots,X_n$ is to arrange $U_1,U_2,\ldots,U_{n-1}$ in a non-decreasing order, $V_1\leqslant V_2\leqslant \ldots \leqslant V_{n-1}$, and set $X_i = V_{i+1}-V_i$ for $i = 1, 2,\ldots, n - 1$, $X_n = 1 + X_1 - X_n$. – Mateusz Kwaśnicki Mar 26 '18 at 7:50