I have a simple (maybe not) issue about the interpretation of the link between Fisher information matrix and its inverse which is the covariance matrix.
How to formulate that a line of Covariance matrix by Element-Wise Multiplication with the correponding column of Fisher matrix will give the value 1 (product of both matrices gives identity matrix).
In other words, how to prove that, starting from the Fisher information definition :
$$I(\theta)=-\mathbb{E}_{\mathbb{Y}}\left[\frac{d^{2}}{d \theta^{2}} \log p(Y \mid \theta)\right]$$
we have : $$\sum_{j=1}^{n} I_{ij}(\theta)\,\text{Cov}(\theta_{j},\theta_{i})=1$$
It looks like the result of a probability equal to 1 for example with a $\text{PDF}=f(\theta)$ :
$$P(-\infty < \theta < +\infty) = \int_{-\infty}^{+\infty} f(\theta) \text{d}\theta = 1$$
But I don't know how to better explicit this product wised-element between a row and a column which gives a value of 1.
If someone had a rigorous demonstration about the quantities involved (even if I know there are variance, covariance and Hessian terms of factors but I make confusions to demonstrate that we get a value equal to 1 for all diagonal elements, i.e getting an identity matrix).