Talk:Fisher information metric

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I suspect the appropriate fate of this article is to redirect it to Fisher information, incorporate everything here into that article, and greatly expand it so it's no longer a stub, putting in the usual topics on Fisher information, which at this point are missing. Michael Hardy 01:14, 7 Apr 2004 (UTC)

  • Let's emphasize that one more time, eh? "STUB" You are yet only a student of the obvious. The *normal* way to do this is to add a msg:stub header at the top of the page.
  • This is a metric, in the differential geometry sense of the word. Read the books cited for more information. Kevin Baas 16:13, 7 Apr 2004 (UTC)


Anyone know what a "statistical manifold" is? Or a "statistical differential manifold"? Melcombe (talk) 14:15, 18 September 2009 (UTC)

Statistical manifold = smooth parametric model = a finite-dimensional family of probability distributions that can be viewed as a smooth manifold = it can be smoothly parametrized by finite-dimensional θ, at least locally. stpasha » 17:40, 9 February 2010 (UTC)
And the article doesn't say anything like that, at present. Melcombe (talk) 10:54, 10 February 2010 (UTC)

Another Question[edit]

Why doesn't this article mention the Cramér-Rao bound? Seems to this reader that belongs in the first paragraph. It is clearly the most important application, and the link will lead readers to a more accessible treatment of the concept in a simpler situation. — Preceding unsigned comment added by Robroot (talkcontribs) 23:20, 1 March 2014 (UTC)

Robroot 23:22, 1 March 2014 (UTC)


The article doesn’t define a metric. For a reference, metric is a function d:M×MR+, where M is the smooth manifold in question. The quantity g described in the article fails in many respects:

  • g takes values in the space of k×k matrices (where k is the dimension of the “parameter vector”), not in the space R+
  • g is a function of a single argument, instead of being a function of two arguments. So it's not even possible to check if it defines a symmetric function, and if it is equal to zero only when the distributions are equal
  • Using parameter θ is not adequate when talking about smooth manifolds. By definition of a manifold, it may be Riemannian-parametrized only locally, there does not exist any θ for the entire manifold.

... stpasha » talk » 16:59, 21 September 2009 (UTC)

I see that Information geometry#Fisher information metric as a Riemannian metric duplicates what is here. Melcombe (talk) 09:38, 22 September 2009 (UTC)
I tried to delete this because it is NOT dubious (and how could it be, it's a definition). Someone put it back up, not sure why. This is not a matter of opinion. The person who wrote the information below is not familiar with a Riemannian metric or information geometry. 17:31, 22 December 2009 Hotnewrelease (talk | contribs)
Standard wikipedia conventions are not to delete stuff from talk/discussion pages but to retain it for other peoples edification, and "for the record". Change the article so that it is correct, but leave the discussion intact and explain on the doscussion page why confusion has arisen or what is wrong. Also, add at the end of threads. Melcombe (talk) 17:59, 22 December 2009 (UTC)

For the record, there are MANY different notions of metric.

It doesn't define a metric in the sense you are mentioning, but for a general semi-Riemannian manifold, it does define one, see Riemannian metric. These are very well studied because of their relevance to general relativity.

g doesn't actually take a single argument, it is a matrix. The indices represent the components of a matrix (tensor). The importance is this: We can construct an infinitesimal distance between two points that is locally valid and the way this distance changes in time defines the intrinsic curvature of the space. The scalar product on of this metric is invariant with respect to reparametrizations - so no matter how we relabel our coordinate space, we still have a valid description of distance between two points.

No one ever said the parameters had to be globally valid. —Preceding unsigned comment added by (talk) 04:01, 10 December 2009 (UTC)

I corrected the wikilink in the above. (I see it was put correctly in the article.) Unfortunately, that redirects to something that is generally incomprehensible. But then that matches the present version of this article, which totally fails to explain what it is about, or why it might be important. Melcombe (talk) 10:22, 10 December 2009 (UTC)

I reclaim my previous comment, it was erroneous.  … stpasha »  17:40, 9 February 2010 (UTC)


In this edit, User:Melcombe asked for clarification.

I'll clarify here, rather then mess up the article: yes, the expressions are equivalent; plug one into the other and turn the crank. If you get stuck, recall that

\int p(x,\theta) dx = 1

and that the second derivative of 1 is zero. (and that the derivatives commute with the integral). linas (talk) 02:29, 12 July 2012 (UTC)