# Estimation accuracy of precision matrix

I have a couple of questions related to estimation of high-dimensional precision matrix (inverse of the covariance matrix) in the case where p is close to 100 and n < p. As a measure of estimation accuracy I am calculating a 'relative squarred error' defined in the following manner: $$RSE(\Sigma^{-1},\hat{\Sigma^{-1}}) = \frac{||{\Sigma^{-1}-\hat{\Sigma^{-1}}}||}{||\Sigma^{-1}||}$$

where, $||A|| = \sum_{i,j}a_{ij}^2$ is the Frobenius norm of $A$. Now, my questions are,

a) Is there any known result for lower bound of Frobenius norm of a precision matrix? or for that matter, any general matrix?

b) What are the common ways of showing the estimation accuracy of a method. I think we can of course calculate various norms such as the spectral norm etc, but, is there a graphical way of doing this?

I appreciate any kind of help and apologies in advance if this is not the right place or way to ask such questions.

-
I think we'll need some more information about what you're really interested in to be helpful. Here are some initial comments: (1) You've written the squared Frobenius norm on the RHS of your definition of $\|A\|$. (2) Consider the diagonal matrices. That should tell you everything you need to know about possible values that $\mathrm{RSE}$ can take on (knowing nothing further about the structure of your problem). (3) Answering your question (b) will require knowing more about your interests and application. – cardinal Jul 17 '12 at 0:01
Thanks @cardinal. Yes, it's the squarred Frobenius norm. About the lower bound, I am trying to see how good is the RSE I am getting for my method, having a lower bound should help with that. I think considering diagonal matrices is a way of doing that. For part b), I am applying this method of estimation for a data-set where I know the population precision matrix should have a block diagonal structure, then the estimated precision matrix should have a similar structure. How do I demonstrate that using graphical outputs? Also, does having a low RSE imply preservation of the matrix-structure? – VitalStatistix Jul 17 '12 at 0:17
The direction I was trying to point you in regarding diagonal matrices was that (1) the lower bound is, of course, zero and that (2) the upper bound is infinity. The first is seen simply by taking $\Sigma^{-1} = {\hat\Sigma}^{-1}$ and for the second, just fix $\Sigma$ at any arbitrary positive semidefinite matrix and let $\hat\Sigma^{-1}$ be a diagonal matrix with entries that grow without bound. – cardinal Jul 17 '12 at 4:34