# Fréchet mean of the spherical shape space

The Fréchet mean of a general subspace is defined as $$F(x)=\int_M dist(x,y)^2d\mu(y),$$ where $\mu$ is the probability measure on a general metric space $(M,dist)$.

I think the sample mean of $y_1,...y_i$ in Euclidean space is defined as $$\int(x-y)^2f(y)dy,$$ which can also be denoted as $$\hat{F}(x)=\sum d^2(x,y_i).$$

I'm not entirely sure if this is correct though.

Also, does a Fréchet mean exist for the spherical shape space? And if it does exist then how would one derive such a mean?

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– user53153 Jan 27 '13 at 21:05

1. The Fréchet mean of a measure $\mu$ on a metric space $M$ with metric $D$ is the point $x$ minimizing $$F(x):=\int_M D(x,y)^2 d\mu(y),$$ if a unique minimum exists. If there is more than one minimum point, there is a "Fréchet mean set".

2. As you say, the ordinary mean in ${\Bbb R}^n$ is a special case of the Fréchet mean, given by taking $M:={\Bbb R}^n$ and $D(x,y):=||x-y||$. This is a restatement of the fact that for a random variable $X$, $x={\Bbb E}X$ is the point which minimizes ${\Bbb E}||X-x||^2$.

3. The Fréchet mean on the $n$-sphere can be defined using the definition above. If the $n$-sphere is coordinatized by embedding it as the unit sphere within ${\Bbb R}^{n+1}$, then you can take $D(p,q)$ to be the length of the shortest geodesic between $p$ and $q$, $\cos^{-1}(p\cdot q).$ If $\mu$ is not too spread out over the sphere, the Fréchet mean of $\mu$ will always exist.

There will always be at least one minimizing point for $F$ when $M$ is a sphere, but it might not be unique; for example, if $\mu$ is the uniform measure, the Fréchet mean set will be all of the sphere.
An example where $F(x)$ has no minimum is $M=(-2,-1)\cup (1,2)$, with the usual metric inherited from $\Bbb R$ and a measure $\mu$ which is uniform over $M$. In this case $$F(x)=x^2+\frac73,$$ but this has no minimum on $M$ as the infimum, $\frac{10}{3}$, is not attained. $F$ can also fail to have a minimum in cases such as the Cauchy distribution over $\Bbb R$, where $F(x)=\infty$ for all $x$. If $M$ is bounded and compact, neither of these problems can occur, so the Fréchet mean set will always exist.