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Given a compactly supported probability measure $m$ on $\mathbb{R}^n$, we can define its average distance to a point $x$ as $\int_\mathbb{R^n}d(x,y)dm(y)$. In this question I found that for a given measure $m$, the point $x_m$ at minimal average distance from $m$ is almost always unique (the only exception is in some cases where the measure is supported on a line).

However, the original motivation of the question was to find that point at minimal distance from the measure. It seems like for non atomic measures, the point $x_m$ should satisfy $\int_{\mathbb{R}^n} u(x,y)dm=0$, where $u(x,y)=\frac{y-x}{||y-x||}$, but I don't know how useful that is.

The question is, is there a formula that gives you $x_m$ in terms of the measure $m$? (for example expressing $x_m$ as some integral in terms of $m$)

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  • $\begingroup$ Aren't you redefining the barycentre of the measure $\mu$? In that case an implicit formula for the barycentre ($x_m$ in your notation) is $\int exp_{x_m}^{-1}(x) d\mu(x)=0$ in the tangent space $T_{x_m}{\bf{R}}^n$. And that's basically the equation you found. A reference is J. Jöst's "Nonpositive curvature: geometric and analytic aspects" (Chapter 3). The point is you can find $x_m$ coordinate wise, i.e. $(x_m)_i=\int x_i d\mu(x)$. That's a direct formula. The average coordinate is the coordinate that minimizes average distance. $\endgroup$
    – JHM
    Commented Oct 10, 2022 at 22:57
  • $\begingroup$ @JHM The problem is that I am considering distance, not squared distance, so the minimizing point need not be the barycenter of the measure (e.g. for a measure supported in the three points $A=(-1,0),B=(1,0)$ and $C=(0,0.1)$ with measure $1/3$ each, the minimizing point is $C$). But thanks for the reference! I noticed that uniqueness of the minimizing point worked for any convex function (in the same sense as distance) but I didn't know of any reference for this topic $\endgroup$
    – Saúl RM
    Commented Oct 11, 2022 at 0:03
  • $\begingroup$ @JHM, I agree with Saul’s comments; I’d add that for $\mu$ supported on a line, the coordinate minimizing the average distance is the median rather than the mean; and for $\mu$ not supported on a line, analyzing minima one coordinate as a time is not enough — the projection of a minimizer of average distances may not minimize the average distance among the projections. $\endgroup$
    – user44143
    Commented Oct 11, 2022 at 0:35

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There is no nice formula for the point minimizing the average distance, as you can see from my answer for the more specific question about measures spread uniformly over triangles.

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  • $\begingroup$ I also think that there will be no formula. My hope was that, even if calculating the average distance function of the measure is extremely difficult, there would be some way to obtain the infimum without computing the function. My initial hope was that at any point, the gradient of the function (that can be computed with an integral) would point to $x_m$, but that doesn't seem to be the case. But I would also be happy with an indirect method similar to that one $\endgroup$
    – Saúl RM
    Commented Oct 10, 2022 at 21:07

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