# Fisher Information matrix as covariance matrix from scores

I've found in a paper here. Observed Fisher information estimated in a way that does not convince me at all. They estimated it as the covariance matrix of the scores, but to me the formula used is wrong:

$$\hat I(\theta)=\sum_{i=1}^n (\mathbf S_i\mathbf S_i^t)-\frac1n\left(\sum_{i=1}^n \mathbf S_i\right)\left(\sum_{i=1}^n \mathbf S_i^t\right)$$

I think that $$1/n$$ should be before the first term, and the second term should be simply the mean of the scores.

Can you confirm that this formula is wrong?

• Here is a MathJax tutorial for typesetting math. Consider using MathJax whenever possible. Sep 1, 2021 at 12:52
• Thanks a lot, I didn't know how to do that. Sep 1, 2021 at 13:00

The formula looks a bit unusual but note that \begin{align} \frac1n\sum_{i=1}^n(S_i-\bar S)(S_i-\bar S)^T &=\frac1n\sum_{i=1}^n (S_iS_i^T-2\bar S S_i^T+\bar S \bar S^T) \\&=\frac1n\sum_{i=1}^n S_iS_i^T-2\bar S\frac1n\sum_{i=1}^n S_i^T+\frac1n\bar S\bar S^T \\&=\frac1n\sum_{i=1}^n S_iS_i^T-\frac1{n^2}\sum_{i=1}^nS_i\sum_{i=1}^nS_i^T. \end{align} This can be seen as a kind of estimator of $$\operatorname{Var}S_i$$. If the observations (and the contributions $$S_i$$ to the score vector) are independent, $$n$$ times this is a kind of estimator of the expected (and I suppose the observed) Fisher information $$I(\theta)=\operatorname{Var}S=\operatorname{Var}(\sum_{i=1}^n S_i)=n\operatorname{Var}S_i.$$ But if the likelihood is available, the observed Fisher information is directly observable so it is a bit strange that the authors try to only estimate it in this way.
Note also that the last term in their expression is not needed as $$S=\sum S_i=0$$ at the MLE of $$\theta$$.