Which distance gives rise to the geometric mean?

Question

The arithmetic mean has the nice property of minimising the sum of squares, or in other words, minimising the sum of quadratic-euclidean distances. Formally, given a set of points $x_0, \dots, x_n \in \mathbb{R}^d$, the arithmetic mean, $\mu = \frac{1}{n} \sum_{i=1}^n x_i$, is the unique solution to the equation $$\mu = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(x_i, x) \, ,$$ where $d$ denotes the Euclidean distance. I was wondering if there exists a distance measure on $\mathbb{R}$ that has a similar property for the geometric mean, i.e.

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Question

Is there a distance measure $d$ on $\mathbb{R}$ such that for any set of points $x_1, \dots, x_n \in \mathbb{R}$ we have $$ \sqrt[n]{x_1 \cdot \dots \cdot x_n} = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(x_i, x) \; \textbf{?}$$

Answer 1

Let $\alpha = (x_1x_2...x_n)^\frac{1}{n} $. Then, $\ln(\alpha) = \frac{1}{n} \sum{\ln(x_i)}$ so because of the property you stated, $$\ln(\alpha) = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(\ln(x_i), x)$$ Which implies that $$\alpha = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(\ln(x_i), \ln(x))$$

Then you can substitute to find the new metric $$\delta^2(x, y) = d^2(\ln(x), \ln(y)) = (\ln(x) - \ln(y))^2$$ $$\delta(x, y) = \displaystyle\left\lvert \ln{\frac{x}{y}} \right\rvert$$

Answer 2

Consider the metric $d:(0,\infty)^2 \rightarrow [0,\infty)$ defined by $$d(x,y)=\Bigg|\ln\Big(\frac{y}{x}\Big)\Bigg|$$ This is the metric you're looking for. Given a dataset $\{x_1,\ldots ,x_n\}\subseteq (0,\infty)$ define a function $f$ on $(0,\infty)$ by $$f(x)=\sum_{i=1}^nd^2(x,x_i)=\sum_{i=1}^n(\ln(x)-\ln(x_i))^2$$ Taking a derivative yields $$f'(x)=\frac{2}{x}\sum_{i=1}^n(\ln(x)-\ln(x_i))=0 \iff \ln(x)=\frac{1}{n}\sum_{i=1}^n\ln(x_i)$$ This shows $x=(x_1 \times \dots \times x_n)^{1/n}$ is a critical point of $f$. Now since $f$ is continuous on $(0,\infty)$, is bounded below by $0$, and $f(x)\rightarrow \infty$ as $x \rightarrow 0^+$ or as $x \rightarrow \infty$, we see that $f$ attains a global minimum at its critical point.