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I face a problem, where I have a total number of $c$ samples $S^{c\times r}$ of $r$ features. These are split at a position $p\in{1...c}$ into two subsets $S_{left}^{p\times r}$ and $S_{right}^{c-p\times r}$. For both of these subsets, I have to calculate the determinant of the co-variance matrix $\Sigma$. This I achieve through:

  1. Computing the mean feature vector over all samples $\mathbf{\mu}_{left}$ with $|\mathbf{\mu}_{left}| = r$
  2. Centering the matrix $S_{left,c} = S_{left} - \mathbf{\mu}_{left}$
  3. Computing the co-variance matrix with $\Sigma(S_{left,c}) = \frac{1}{p-1}S_{left,c}^\top S_{left,c}$
  4. Computing the determinant $\det(\Sigma(S_{left,c}))$

So far, so good. Now the split position $p$ is changing over time (monotonically increasing) and samples have to be swapped from the right to the left subset. My question: Is there any way to update the $\Sigma$ without minimum computational costs?

The problem is, that in my case $c>>r$ and the above approach depends strongly on the size of $c$. I've figured out how to update $\mathbf{\mu}$ with $\mathcal{O}(r)$. Equally, I can update $\Sigma$ with $\mathcal{O}(r^2)$, but only under the assumption of a static $\mathbf{\mu}$. How can I achieve at least an estimated $\det(\Sigma)$ after adding/removing a sample?

Additional questions:

  1. Can I use Cholesky decomposition to compute the determinant? The Wiki entry on determinants states that the $\Sigma$ would have to be positive definit for this, but as far as I understand, its only positive semi-definite. I ask, because there has been an interesting answer to a similar problem, that frankly, I do not get, especially where the Sylvester's determinant theorem fits in.
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The problem you pose is a variant of the problem solved by the Kalman Filter. There, you have a (vector valued) current estimate of some actual parameters and of their covariance matrix. Now you add an additional measurement with some measurement covarince matrix. The Kalman Filter tells you what your new best-estimate values and covarinace matrix become.

If you wish to track the change in the covariance matrix as you progress through your splittings, for each step, this is the best you can do (although since your problem is a bit simpler than the fully general case, you may be able to improve by a small factor on just slavishly applying the KF formulas).

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  • $\begingroup$ Thank you. I've read about KF, but figured it would not apply directly to my problem, as I do not really estimate anything through a model, as you've pointed out. I will try to figure out, how the KF deals with updating the $\Sigma$. Any reading/tutorial suggestions? $\endgroup$ – loli Feb 13 '15 at 23:20
  • $\begingroup$ math.stackexchange.com/questions/840662/… is a question on stack exchange math where one of the answers is a fine tutorial on KF. $\endgroup$ – Mark Fischler Feb 15 '15 at 18:08
  • $\begingroup$ Thanks. I've read this (cut short) and others, and believe I understood the working of the KF now, albeit not how to apply it to my problem. I assume, the base idea is to predict a new observation $\hat{z}$, then calculate the difference to the real measurement $z$ and update the internal state, from which to extract my current $\Sigma$. But the $P_{k|k-1}$ used by the KF is not my co-variance $\Sigma$. It come nearer to the measurement co-variance matrix $Q$, but this is never updated. Frankly, I do not really now how to create a state transition $F$ to predict a new observation $\hat{z}$ $\endgroup$ – loli Feb 15 '15 at 19:12
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Solution

I've found a solution through paper-work and tested it.

Formulation

Given a sample of $n$ independent obseravtions $\mathbf{x}_1,...,\mathbf{x}_n$ of length $m$, the sample co-variance matrix of $X\in R^{n\times m}$ is given by

$$ Q = \frac{1}{n-1} \sum_{i=1}^n(\mathbf{x}_i - \mathbf{\hat{x}})(\mathbf{x}_i - \mathbf{\hat{x}})^T $$

where $\mathbf{x}_i$ denotes the $i$-th observation and

$$ \mathbf{\hat{x}} = \frac{1}{n}\sum_{i=1}^n\mathbf{x}_i $$ is the sample mean. Re-organizing the first formula, we get

$$ Q = \frac{1}{n-1}\left[\sum_{i=1}^n\mathbf{x}_i\mathbf{x}_i^T - \mathbf{\hat{x}}\left(\sum_{i=1}^n\mathbf{x}_i\right)^T - \left(\sum_{i=1}^n\mathbf{x}_i\right)\mathbf{\hat{x}}^T + n\mathbf{\hat{x}}\mathbf{\hat{x}}^T\right] $$ which is essentially the application of $(a-b)^2 = 2a^2 - 2ab - b^2$. Substituting the sums, we get $$ Q = \frac{1}{n-1}\left[A - \mathbf{\hat{x}}\mathbf{b}^T - \mathbf{b}\mathbf{\hat{x}}^T + n\mathbf{\hat{x}}\mathbf{\hat{x}}^T\right] $$

Updating

From this form, we can derive an efficient update of $Q_{n+1}$ when a new sample $\mathbf{x}_{n+1}$ becomes available. First we update $A$ and $\mathbf{b}$ $$ A_{n+1} = A_n + \mathbf{x}_{n+1}\mathbf{x}_{x-1}^T\\ \mathbf{b}_{n+1} = \mathbf{b} + \mathbf{x}_{x+1}\\ \mathbf{\hat{x}}_{n+1} = \frac{1}{n+1}\mathbf{b}_{n+1} $$ to finally compute the updated co-variance matrix $$ Q_{n+1} = \frac{1}{n}\left[A_{n+1} - \mathbf{\hat{x}}_{n+1}\mathbf{b}_{n+1}^T - \mathbf{b}_{n+1}\mathbf{\hat{x}}_{n+1}^T + (n+1)\mathbf{\hat{x}}_{n+1}\mathbf{\hat{x}}_{n+1}^T\right] $$

Speed

The update is considerably cheaper than the computation of the complete sample co-variance matrix, not only when $n>>m$:

  • complete: $\mathcal O(nm^2)$
  • update: $\mathcal O(m^2)$

Relation to Kalman Filter

I am not sure if this solution has any relation to the Kalman Filter. But since I do not use prediction, innovation or the Kalman gain, I do not think so. Maybe someone can correct me or confirm this observation?

Implementation

https://github.com/loli/dynstatcov

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