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The Hellinger distance between two probability distributions $f(x)$ and $g(x)$ is given by: $$H(f,g)=1-\int_\Omega(dx\sqrt{f(x)g(x)})$$ My question is:

if $f(x)$ and $g(x)$ are two Weibull distributions defined as $$f(x;\lambda_1,k_1) = \begin{cases}\frac{k_1}{\lambda_1}\left(\frac{x}{\lambda_1}\right)^{k_1-1}e^{-(x/\lambda_1)^{k_1}} & x\geq0 ,\\0 & x<0 ,\end{cases}$$ and $$f(x;\lambda_2,k_2) = \begin{cases}\frac{k_2}{\lambda_2}\left(\frac{x}{\lambda_2}\right)^{k_2-1}e^{-(x/\lambda_2)^{k_2}} & x\geq0 ,\\0 & x<0 ,\end{cases}$$

is it possible to derive a metrics tensor in the space of probability as a function of the parameters $\lambda$ and $k$ using the Hellinger distance?


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up vote 1 down vote accepted

If we consider a new probability distribution defined through $f(x;\theta_i)$ and $g(x;\tilde\theta_i)$ given by $P(x;\theta_i,\tilde\theta_i)=\sqrt{f(x;\theta_i)g(x;\tilde\theta_i)}$, we can generalize the Fisher-Rao metric in the following way.

$$\frac{\partial H}{\partial\theta_i}=-\frac{1}{2}\int_\Omega dxP(x;\theta_i,\tilde\theta_i)\frac{\partial}{\partial\theta_i}\ln f(x;\theta_i)$$

$$\frac{\partial H}{\partial\tilde\theta_i}=-\frac{1}{2}\int_\Omega dxP(x;\theta_i,\tilde\theta_i)\frac{\partial}{\partial\tilde\theta_i}\ln g(x;\tilde\theta_i)$$

and so

$$\frac{\partial^2 H}{\partial\theta_i\partial\theta_j}=-\frac{1}{2}\int_\Omega dxP(x;\theta_i,\tilde\theta_i)\left[\frac{\partial^2}{\partial\theta_i\partial\theta_j}\ln f(x;\theta_i)+\frac{1}{4}\frac{\partial}{\partial\theta_i}\ln f(x;\theta_i)\frac{\partial}{\partial\theta_j}\ln f(x;\theta_i)\right]$$

and similarly

$$\frac{\partial^2 H}{\partial\tilde\theta_i\partial\tilde\theta_j}=-\frac{1}{2}\int_\Omega dxP(x;\theta_i,\tilde\theta_i)\left[\frac{\partial^2}{\partial\tilde\theta_i\partial\tilde\theta_j}\ln g(x;\tilde\theta_i)+\frac{1}{4}\frac{\partial}{\partial\tilde\theta_i}\ln g(x;\tilde\theta_i)\frac{\partial}{\partial\tilde\theta_j}\ln g(x;\tilde\theta_i)\right].$$

One has also cross products as

$$\frac{\partial^2 H}{\partial\theta_i\partial\tilde\theta_j}=-\frac{1}{4}\int_\Omega dxP(x;\theta_i,\tilde\theta_i)\frac{\partial}{\partial\theta_i}\ln f(x;\theta_i)\frac{\partial}{\partial\tilde\theta_j}\ln g(x;\tilde\theta_i).$$

Now, as done by Rao about Fisher information matrix, we can interpret these second derivatives of $H$ as the components of a metric tensor $h_{ij}(\theta,\tilde\theta)$ so that we can write down


and work out Riemann geometry.

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In the case of the Weibull pdf, is it possible to assume theta to be lambda and k? – Riccardo.Alestra Feb 20 '12 at 10:41
In your case you will $\theta=(\mu_1,\lambda_1)$ and $\tilde\theta=(\mu_2,\lambda_2)$ as you are measuring the distance between two distributions that are supposed to have different parameters. – Jon Feb 20 '12 at 10:43

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