Deriving Fisher Information Matrix over a Histogram Structure I am reading a classic paper by Barron and Risannen, entitled:

The Minimum Description Length Principle in Coding and Modeling

In this paper they state the following idea: 
On this point I am interested in how they derive that the determinant of the Fisher Information Matrix (FIM) evaluates to the multiplication of the inverse of the histogram probability values.
It's probably a simple idea but I seem to have a conceptual block ....
Thus far, I am convinced that the FIM should be a diagonal matrix, which means that the determinant is just the multiplication of the diagonal elements. However, I can't imagine how the inverse probabilities arise.
 A: We know that the $(i,j)^{th}$ element of the Fisher information matrix is defined as
$$
[\mathcal{I}(p)]_{i,j} = -E \left[ \frac{\partial^2}{\partial p_i \partial p_j} \log f(y|p,m) \right]
$$
$f(y|p,m) = mp_{i(y)}$, so that $\log f(y|p,m) = \log m + \log p_{i(y)}$.
Derivating with respect to $p_i$ yields
\begin{equation}
  \frac{\partial}{\partial p_i } \log p_{i(y)}=\left\{
  \begin{array}{@{}ll@{}}
    0, & \text{if}\ i(y)\neq i \\
    \frac{1}{p_{i(y)}}, & \text{otherwise}
  \end{array}\right.
\end{equation}
Similarly,
\begin{equation}
  \frac{\partial^2}{\partial p_i \partial p_i } \log p_{i(y)}=\left\{
  \begin{array}{@{}ll@{}}
    0, & \text{if}\ i(y)\neq i \\
    -\frac{1}{p_{i(y)}^2}, & \text{otherwise}
  \end{array}\right.
\end{equation}
Also, $\frac{\partial^2}{\partial p_i \partial p_j } \log p_{i(y)} = 0$ for any $j \neq i$.
We finally have to compute the statistical expectation of this quantity:
$$
[\mathcal{I}(p)]_{i,j} = \int_{i(y) = i}dy \frac{1}{p_{i(y)}^2} f(y|p,m) = \int_{i(y) = i}dy \frac{1}{p_{i}^2} m p_{i} = \frac{1}{p_i}
$$
The FIM is thus a diagonal matrix which $i^{th}$ entry is $p_i^{-1}$ ; and, as you mentioned, its determinant will be their product.
