Fact of the day: Brownian centroids

$latex \def\Real{\mathbb{R}}\def\Z{\mathcal{Z}}\def\sg{\mathfrak{S}}\def\B{\mathbf{B}}\def\xx{\mathbf{x}}\def\ex{\mathbb{E}}
$

Consider $latex n$ points in Euclidean space, $latex \xx=\{x_1,\ldots, x_n\}, x_k\in \Real^d, n\leq d+1$.

Generically, there is a unique sphere in the affine space spanned by those points, containing all of them. This centroid (which we will denote as $latex o(x_1,\ldots,x_n)$) lies at the intersection of the bisectors $latex H_{kl}, 1\leq k\lt l\leq n$, hyperplanes of points equidistant from $latex x_k, x_l$.

Assume now that the points $latex x_k,k=1,\ldots,n$ are independent standard $latex d$-dimensional Brownian motions. What is the law of the centroid?

It quite easy to see that it is a continuous martingale. What is the quadratic variation of $latex o$?

If the directions orthogonal to the affine span $latex L(\xx)$ of the tuple, the answer is immediate: it is proportional to
$$
\sum_{k=1}^n p^2_k,
$$
where $latex p_k$ are the barycentric coordinates of $latex o$ with respect to $latex x_1,\ldots,x_n$. Indeed, it is immediate that if $latex dx_k^\perp$ are the components of the Brownian increments orthogonal to $latex L(\xx)$, then
$$
do^\perp(\xx)=\sum_k p_k dx_k^\perp.
$$
It follows -as $latex dx_k^\perp$ are independent and uncorrelated) that the component of $latex o$ orthogonal to $latex L(\xx)$ is a martingale with variation equal to (Euclidean norm on $latex L(\xx)^\perp$ times) $latex \sum_k p_k^2$.

Let’s turn to the motion of $latex o(\xx)$ inside $latex L(\xx)$. Let $latex S(\xx)\subset L(\xx)$ be the sphere centered at the centroid and containing all the points of the sample. Let $latex dx_k^\parallel=\xi_k+\eta_k$ be the decomposition of the $latex L(\xx)$ component of $latex dx_k$ into the vector in the tangent space $latex \xi_k\in T_{x_k}S(\xx)$, and the vector orthogonal to it. If we denote by $latex n_k=(x_k-o(\xx))/|x_k-o(\xx)|$ the unit norm vector from the centroid to $latex k$-th point of the sample, then $latex \eta_k=s_k\cdot n_k$, where $latex s_k$ is the increment of a Brownian motion independent of all all other components (as follows from the orthogonalities hardwired in the construction).

Now, the condition that as the points of the sample move, the distances to the centroid remain constant is equivalent to
$$
\langle \eta_k-\nu,n_k\rangle=\langle \eta_l-\nu,n_l\rangle,
$$
for all $latex 1\leq l,k\leq n$. (Here we denote by
$$
\nu:=do(\xx)^\parallel
$$
the $latex L(\xx)$-component of $latex do(\xx)$.)
Equivalently, this means that $latex \langle \nu, n_k-n_l\rangle=s_k-s_l$.
As $latex s_k, s_l$ are independent, with unit quadratic variation, we conclude that
$$
\langle (n_k-n_l),C (n_k-n_l)\rangle=2
$$
for all $latex k,l$, – here
$$
C:=\ex \nu\otimes\nu’
$$
is the quadratic variation of the centroid in the affine plane spanned by the sample.

Geometrically, an equivalent description is that $latex C=AA’$, where $latex A$ is the linear part of the affine transformation taking the points of the sample to the regular $latex n$-simplex with sides $latex \sqrt{2}$.

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