TITLE: Determinant of Fisher information
QUESTION [5 upvotes]: In information geometry, the determinant of the Fisher information matrix is a natural volume form on a statistical manifold, so it has a nice geometrical interpretation. 
But what is it in statistics? Does it measure anything meaningful? (For example, I would say that if it is zero, then the parameters are not independent. Does this go any further?)
Also, is there any closed form to compute it?
Thanks.
Update: I posted a similar question on stats.se.

REPLY [3 votes]: You can generalize @Eupraxis1981's answer to all natural exponential families, that is distributions whose density has the form:
\begin{align}
f(x \mid \eta) &= \exp\bigl({T(x) \cdot \eta - g(\eta)} + h(x)\bigr), \qquad x \in \Omega
\end{align}
Their Fisher information is the Hessian of the (always strictly convex and $C^\infty$ differentiable) log-normalizer:
\begin{align}
  \mathcal I(\eta)
  &= -E(\nabla^2_{\eta} \log f(x \mid \eta) \mid \eta) \\
  &= -E\Bigl(\nabla^2_{\eta} \bigl({T(x) \cdot \eta - g(\eta)} + h(x)\bigr) \mathrel{}\Bigm\vert\mathrel{} \eta\Bigr) \\
  &= E\bigl(\nabla^2_{\eta} g(\eta) \mathrel{}\bigm\vert\mathrel{} \eta\bigr) \\
  &= \nabla^2_{\eta} g(\eta).
\end{align}
Therefore the curvature of the log-normalizer tells you how informative realizations of a distribution whose parameters are $\eta$ is.  It makes sense that location parameters don't affect the Fisher information.