Average degree of convex hull vertices in a Delaunay triangulation

Question

Let $P \subset \mathbb{R}^2$. The boundary of $DT(P)$, the Delaunay triangulation of the point set $P$, is $conv(P)$. It is also known that the average degree of the vertices of $DT(P)$ is $\lt 6$. My question is there any known bound on the expected degree of the hull vertices of $DT(P)$? Is there an intuitive argument at least for this bound to also be $O(1)$?

Edit: Consider the two cases:

Case 1: $P$ is drawn uniformly and independently from the unit square.

Case 2: $P$ is drawn uniformly and independently from the unit disk.

Are there known results for point sets with other distributions? Any valuable reference on this questions?

Answer 1

I don't know if I understood you correctly, I also don't know the probability space (i.e. if $P$ is random in $\mathbb{R}^2$, and if yes, then what is the distribution). However, if $P$ is fixed, then the answer is probably no, i.e. the average degree (for any Delaunay triangulation) may be arbitrarily high. Just take $P$ to be a regular $(n-3)$-gon and add 3 more points such that the rest will be strictly contained in the resulting triangle, that means the convex hull of $P$ will be those three points. In any triangulation each point of the circle will have to connect to some point of the hull, and there are only $O(1)$ of them, so the average degree must be high.

Answer 2

It is not quite clear for me if you mean the average degree of the convex hull vertices or the sum of the degrees of the convex hull vertices. In the latter case the fact that the expected number of vertices on the convex hull is $O(\log n)$ would make a $O(1)$ bound surprising.