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.

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

$$$$\frac{\partial}{\partial p_i } \log p_{i(y)}=\left\{ \begin{array}{@{}ll@{}} 0, & \text{if}\ i(y)\neq i \\ \frac{1}{p_{i(y)}}, & \text{otherwise} \end{array}\right.$$$$

Similarly,

$$$$\frac{\partial^2}{\partial p_i \partial p_i } \log p_{i(y)}=\left\{ \begin{array}{@{}ll@{}} 0, & \text{if}\ i(y)\neq i \\ -\frac{1}{p_{i(y)}^2}, & \text{otherwise} \end{array}\right.$$$$

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.