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Let $X$ be a set of $n$ points in $\mathbb{R}^3$ and $f_m$ be the Fréchet mean, i.e.:

$ f_m= \arg \min_{p \in M} \sum_{i=1}^n w_id^2(p,x_i) $

where $(\mathbb{R}^3,d)$ is a complete metric space, $d$ a distance function (Euclidean $d_E$, log-Euclidean $d_{log}$, Riemannian $d_R$) between any two points and $w$ a weight function.

$ d_E(x_i,x_j)= \| x_i - x_j\| $

$ d_{log}(x_i,x_j)= \| \log(x_i) - \log(x_j)\| $

$ d_R(x_i,x_j)= \| \log(x_i^{-1/2}x_jx_i^{-1/2})\| $

How could I solve this problem? Maybe by Gradient Descent?

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    I was vague bacause I would like to analize the differences between Euclidean, log-Euclidean, and Riemannian metric.2012-06-05

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In the case $d=d_E$ you have $f_m=\sum w_i x_i$ (assuming $\sum w_i=1$, which can always be achieved by scaling the weights). Proof: for any $x\in\mathbb R^3$

$\sum w_i\|x-p_i\|^2 = \sum w_i\|x-f_m+f_m-x_i\|^2 \\= \|x-f_m\|^2 + \sum w_i\|f_m-x_i\|^2 + 2\sum w_i\langle x-f_m, f_m-x_i\rangle$ In the last term we move summation into the bracket to get $\sum w_i\langle x-f_m, f_m-x_i\rangle=\langle x-f_m, f_m-\sum w_ix_i\rangle=0$ Thus, $\sum w_i\|x-p_i\|^2 =\|x-f_m\|^2+\sum w_i\|f_m-x_i\|^2\ge \sum w_i\|f_m-x_i\|^2 $ QED

Also, if your metric is of the form $d(x,y)=\|\Phi(x)-\Phi(y)\|$ where $\Phi\colon\mathbb R^3\to\mathbb R^3$ is some invertible map, then $f_m=\Phi^{-1}(\sum w_i \Phi(x_i))$.

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    @no_name If you can write them in the form given in the last paragraph of my answer, then yes. Otherwise, unlikely. Unfortunately, I don't understand your formulas for either metric: how do you take the logarithm of a vector, for example?2012-06-06