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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.