# Definition of Chernoff distance

In this article the "Chernoff distance" between two data sets is defined as:

$$C D= \frac{1}{2} \alpha_{\mathrm{c}} \alpha_{\mathrm{f}}\left(\mu_{\mathrm{c}}-\mu_{\mathrm{f}}\right)^{ T}\left[\alpha_{\mathrm{c}} \Sigma_{\mathrm{c}}+\alpha_{\mathrm{f}} \Sigma_{\mathrm{f}}\right]^{-1}\left(\mu_{\mathrm{c}}-\mu_{\mathrm{f}}\right)+\frac{1}{2} \log \frac{\left|\alpha_{\mathrm{c}} \Sigma_{\mathrm{c}}+\alpha_{\mathrm{f}} \Sigma_{\mathrm{f}}\right|}{\left|\Sigma_{\mathrm{c}}\right|^{\alpha_{c}}\left|\Sigma_{\mathrm{f}}\right|^{\alpha_{\mathrm{f}}}}$$

where $$\alpha$$, $$\mu$$ and $$\Sigma$$ are the "percentages, means and covariances" of both groups, and the "superscripts $$T$$ and $$−1$$ refer to the transpose vector and to the inverse of the matrix, respectively".

The original article referenced as the source of this expression is Chernoff (1952), but I could not find it inside.

I also could not find a proper definition for this distance anywhere else. Is this expression correct? Is it similar to an Anderson-Darling test, or Kullback–Leibler divergence, or Bhattacharyya distance?

• Why is there an integral sign in the exponent? – innisfree May 23 at 2:53
• It's a typo in the final version of the article, the arXived version shows the proper equation. I've fixed it. – Gabriel May 26 at 20:05

It looks like the Bhattacharya distance for multivariate normal distributions, see wiki, when the factors $$\alpha =1/2$$.
As that distance is $$B= \int \sqrt{f g} dx$$ I speculate that the Chernoff distance is $$\int f^\alpha g^{1-\alpha} dx$$ and that this formulae for multivariate normals would give us your expression.