7

In practice, the observed information matrix (Newton-Raphson) is usually replaced by its expectation, known as Fisher scoring.

Link: https://en.wikipedia.org/wiki/Scoring_algorithm#Fisher_scoring

What I don't understand is, why the expectation of the matrix is easier to compute than the matrix itself? Otherwise, there wouldn't be any point to use Fisher scoring.

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
SmallChess
  • 6,764
  • 4
  • 27
  • 48
  • 1
    Because you only need to take the first (not the second) derivatives, which can be expressed as the expectation. – Randel Oct 07 '15 at 14:31
  • 1
    @Randel Can you elaborate a bit more? Other users might also find your answer helpful. Most textbooks don't say why Fisher scoring is better computationally. – SmallChess Oct 07 '15 at 14:33
  • 1
    I would refer to Page 88, Section 2.11.2 [Empirical FS algorithm](https://books.google.com/books?id=IWQR8d_UZHoC&pg=PA88&lpg=PA88&dq=empirical+information+matrix&source=bl&ots=_0P72eKhA6&sig=fCECyJWRgzdYKYpt5VABBODTF1k&hl=en&sa=X&sqi=2&ved=0CE0Q6AEwBWoVChMI5pb9vsKxyAIVxWk-Ch2Abg32#v=onepage&q&f=false) of the book by Prof. Demidenko. This is the FS algorithm I usually see, and there are some approximation. – Randel Oct 07 '15 at 23:53

0 Answers0