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Let $\mathcal{X}\equiv\Bbb{R}^n\times\Bbb{S}_{++}^n$, where $\Bbb{S}_{++}^n$ is the space of all symmetric positive-definite $n\times n$ real matrices. Let $x,y\in\mathcal{X}$, where $$ x=(\mathbf{x},\Sigma_x),\quad y=(\mathbf{x},\Sigma_y). $$ Let $d\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$ given by $$ d(x,y)=d\big((\mathbf{x},\Sigma_x),(\mathbf{y},\Sigma_y)\big) = \frac{1}{8}(\mathbf{x}-\mathbf{y})^\top\Sigma^{-1}(\mathbf{x}-\mathbf{y}) + \frac{1}{2}\ln\Bigg(\frac{\det\Sigma}{\sqrt{\det\Sigma_x\det\Sigma_y}} \Bigg), $$ where $$ \Sigma=\frac{\Sigma_x+\Sigma_y}{2} $$

I want to prove (or disprove) that $d$ defines a metric on $\mathcal{X}$; that is, is $d$ a distance function?

So, in order to prove that $d$ is a distance, we need to prove that the following conditions are satisfied, for any $x,y,z\in\mathcal{X}$:

  1. [Non-negativity / separation axiom] $d(x,y)\geq0$
  2. [Identity of indiscernibles / coincidence axiom] $d(x,y)=0, \iff x=y$
  3. [Symmetry] $d(x,y)=d(y,x)$
  4. [Subadditivity / triangle inequality] $d(x,z)\leq d(x,y)+d(y,z)$

I think that (1)-(3) are easy to be proven true, but what about (4)?

The above "distance" is the so-called Bhattacharyya distance [1], which is defined as a similarity measure between two probability distributions. In our case, the distributions are normal with mean vectors $\mathbf{x}$, $\mathbf{y}$, and covariance matrices $\Sigma_x$, $\Sigma_y$. It is said to be a "distance", but does that really hold true?

In case of negative answer, could it be modified such that it becomes a true distance function? Is there any other (true) distance function that measures similarity of two multivariate normal distributions?

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