Is it possible Fisher information matrix be indefinite?

I`m using the Newton-Raphson method for obtaining MLE for parameters for maximizing my objective function.
At each iteration, I want to check that is the Hessian matrix negative definite or not and I see the Hessian matrix is not negative definite at some iterations. So, I want to use the Fisher scoring algorithm (https://en.wikipedia.org/wiki/Scoring_algorithm#Sketch of derivation)
In this situation, the Fisher information matrix is indefinite!! Is it possible? What are the reasons for this? Are the second-order derivatives wrong? Or are values of expectations incorrect?

• Could you clarify what you mean by "Fisher information matrix"? As usually defined, this doesn't depend (in any way) on "iterations" towards a solution: it depends only on the likelihood function at the optimal parameter.
– whuber
Jun 2, 2021 at 16:04
• I can refer you to this link : en.wikipedia.org/wiki/Scoring_algorithm. I mean expected the observed information matrix calculated at obtained parameter value in previous iteration. Jun 2, 2021 at 16:35
• Thanks. That's a gradient descent algorithm. Refer to posts about these algorithms for the (many) problems they can have.
– whuber
Jun 2, 2021 at 19:10
• @whuber Thank you. Jun 3, 2021 at 5:48
• You need to be aware of the difference between the Fisher Information and the Observed Fisher Information, see for discussion stats.stackexchange.com/questions/154724/… and stats.stackexchange.com/questions/188251/… Jun 3, 2021 at 10:12