3
$\begingroup$

I am trying to obtain numerically the fisher information. Given a likelihood function $$ f(X,\theta),$$ with $X \in [0,1]$. The fisher information is given by $$ \mathbb{I}(\theta)=\mathbb{E}\left[\left. \frac{\partial^2 \text{log }f( X|\theta)}{\partial \theta^2}\right|_{\theta=\theta^*} \right].$$

To calculate this numerically in Matlab is use this formula: $$\mathbb{I}(\theta) = \int_0^1 \frac{\partial^2 \text{log }f( X|\theta)}{\partial \theta^2} \cdot f(X,\theta) \quad dX$$

Am I doing this correct?

$\endgroup$
4
  • $\begingroup$ @Learner Yes ,it is. I corrected my question. $\endgroup$
    – Stephen
    Commented Mar 20, 2013 at 12:44
  • 1
    $\begingroup$ There is a negative sign missing and $\theta^\star$ should be equal to $\theta$. $\endgroup$
    – Learner
    Commented Mar 20, 2013 at 12:52
  • $\begingroup$ @Learner, thanks a lot. $\endgroup$
    – Stephen
    Commented Mar 20, 2013 at 12:57
  • $\begingroup$ What's the distribution of X? i.e. what is the functional form of f(x)? Since most Fisher Information problems are defined in terms of the parameters, you may find that you need symbolic methods, not numerical ones. $\endgroup$
    – wolfies
    Commented Apr 26, 2013 at 14:38

1 Answer 1

1
$\begingroup$

The formula i've always used is, \begin{equation} \mathbb{I}_{ij}(\theta) = \int\limits_\mathbb{R} f(x,\theta) \frac{\partial \ln f(x,\theta)}{\partial \theta_{i}} \frac{\partial ln \ f(x,\theta)}{\partial \theta_{j}} dx \end{equation}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .