Simple question on calculation of Fisher metric for $p(x,\theta)$

Question

The Fisher information metric is defined as:

$$g_{jk}(\theta)=
\int_X
\frac{\partial \log p(x,\theta)}{\partial \theta_j}
\frac{\partial \log p(x,\theta)}{\partial \theta_k}
p(x,\theta) \, dx.$$

Say $p(x,\theta):= \exp \frac{\theta}{\log x},$ for $x\in(0,1).$

Then $\theta \in (0,\infty)$ gives us a coordinate on a $1-$manifold equivalent to the positive real line $(0,\infty).$

I calculated (confirming with wolfram alpha) the integral to be a Bessel function: $$\frac{2K_1(2\sqrt{\theta})}{\sqrt{\theta}} $$

But this calculation was done without the normalization constant. Also I'm not sure how to write my result in the proper form i.e. as a metric.

Am I on the right track here? What is the correct calculation and proper way to write the metric?

I think I basically got the answer but looking for verification of the solution and additional feedback on how to write the metric.

Edit:

With the normalization constant included, I found the metric to be:

$$ g(\theta)=\frac{K_0(2\sqrt{\theta})^3}{\theta K_1(2\sqrt{\theta})^2}-2K_0(2\sqrt{\theta})^2+\frac{1}{\theta}. $$

I think this is probably the correct form of the metric.

Still unsure of the $jk$ subscripts and where they come into play in writing the metric properly.

Answer 1

I suppose your probability density function is $$ p(x;\theta) = \frac{\exp(\theta/\log x)}{2\sqrt{\theta}K_1(2\sqrt{\theta})}, $$ i.e., proportional to the one you proposed, but with the normalisation constant.

The expression for the Fisher metric as you wrote is general for a distribution parametrised by a vector $\theta = (\theta^1, \dots, \theta^n)$ in some open set $\Theta \subset \mathbb{R}^n$. In that expression, the indices vary as $1 \le j,k \le n$.

You are right in your comment: in this case the parametrisation is one-dimensional, so that the metric has only one element, namely $g_{11}(\theta)$.

I tried quickly computing it with Mathematica, but found something slightly different: $$ g_{11}(\theta) = \frac{K_1(2\sqrt{\theta})^2 - K_0(2\sqrt{\theta})^2}{\theta K_1(2\sqrt{\theta})^2}. $$

This does not seem equivalent to your solution. Maybe you should double check it.