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Can I use the Fisher information matrix to derive estimators for variances of unknown parameters?

I know that for the Fisher information (non-matrix form) the variance of, say, $\theta$ is given by the inverse of $J(\theta)$ (the Fisher information function).

But what about for multiple (e.g. 2) variables and the Fisher information matrix form?

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    $\begingroup$ In case the fisher information $J(\theta)$ is a matrix the size $n \times n$ with $n > 1$ the variance of the parameters are still given by the inverse of the fisher information. i.e. $J(\theta)^{-1}$. However, inverting a matrix is slightly more tricky than inverting a scalar. You need to find the matrix $B$ whose matrix-product with $J(\theta)$ results in the identity matrix $I$. $\endgroup$
    – Harald Thomson
    Commented Feb 24, 2016 at 10:17

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The variances of the parameters (of a distribution that's compatible with the requirements of Fisher information) are found from the diagonal of the inverse $J(\theta)^{-1}$ of the Fisher information matrix.

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    $\begingroup$ Variances of what precisely? $\endgroup$
    – Scortchi ♦
    Commented Feb 24, 2016 at 12:28
  • $\begingroup$ "Variables of a distribution" isn't a standard term: if you mean "parameters of a distribution", as suggested by the $\theta$ notation in your question, then these are unknown constants & don't have a variance. Fisher information relates to the variance of estimators, but how? & to which estimators? $\endgroup$
    – Scortchi ♦
    Commented Feb 24, 2016 at 13:27
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    $\begingroup$ Not really: as I said, parameters don't have variances because they're not random variables. $\endgroup$
    – Scortchi ♦
    Commented Feb 24, 2016 at 14:14

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