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I have a set of timeseries data $X^B = \begin{bmatrix} {X^B_1},{X^B_2},\dots, X^B_n \end{bmatrix}$ consisting of observations recorded at different spatial locations.

There is crosstalk between the underlying signals $X$, which are also corrupted by additive Gaussian noise:

$$ X^B = M(X + \epsilon) ;\; \epsilon \sim \mathcal{N}(0, \sigma) $$

The mixing matrix $M$ has a known functional form:

$$ M = I + \alpha e^{-(D / \lambda)^2} $$

where $D_{ij}$ is the Euclidean distance between locations $i$ and $j$, $\alpha$ is the amplitude of the crosstalk, and $\lambda$ is the length constant.

I'd like to estimate the values of the unknown parameters $\hat{\alpha}$ and $\hat{\lambda}$, construct $\hat{M}$, then use this to remove the correlations in $X^B$ arising due to the crosstalk.

I thought I might be able to use this identity:

$$ cov(MX) = Mcov(X)M^T \\ cov(X^B) = M(cov(X) + \sigma^2 I)M^T $$

However, the structure of $cov(X)$ is unknown and non-trivial. One thing I can be certain of is that $cov(X)$ is not influenced by $D$.

Given $X^B$, $D$, and the known functional form of $M$, is it possible to estimate $\hat{\alpha}$, $\hat{\lambda}$ and $\hat{X}$?

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  • $\begingroup$ Does $\sigma$ denotes the variance or the standard deviation? Also, is it known, or unknown? $\endgroup$ May 5 '14 at 22:56
  • $\begingroup$ Sorry. To clarify, $\sigma$ is the standard deviation, and it is also unknown $\endgroup$
    – ali_m
    May 5 '14 at 23:00
  • $\begingroup$ Do I understand correctly that the matrix $M$ is of the form (if say it was $2\times 2$) $\endgroup$ May 5 '14 at 23:30
  • $\begingroup$ $$M= \left[ \begin{matrix} 1 & 0\\ 0 & 1\\ \end{matrix}\right] + \alpha\left[ \begin{matrix} \exp{\{-D_{11}/\lambda\}} & \exp{\{-D_{12}/\lambda\}}\\ \exp{\{-D_{21}/\lambda\}} & \exp{\{-D_{22}/\lambda\}}\\ \end{matrix}\right]$$ $\endgroup$ May 5 '14 at 23:30
  • $\begingroup$ That would be $exp[-(D_{ij}/\lambda)^{2})]$, but apart from that you're correct $\endgroup$
    – ali_m
    May 5 '14 at 23:36
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The covariance matrix, $cov(X^B)$, has $\frac{n(n+1)}{2}$ unique elements. That means you have $\frac{n(n+1)}{2}$ equations in your equation, if you see what I mean. How many unknowns do you have? Well, $cov(X^B)$ has $\frac{n(n+1)}{2}$ unknowns. Then there are $\alpha$ and $\lambda$. So, you have $\frac{n(n+1)}{2} + 2$ unknowns. So, there is not a unique solution.

You can see the same thing another way. Suppose you have measured $cov(X^B)$. I tell you that I know that $cov(X^B)=cov(X)$ and that $\alpha=0$. How could you disprove this?

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  • $\begingroup$ OK, perhaps a unique solution is not possible, but what about the fact that $cov(X)$ is not influenced by distance? If you told me that $cov(X^{B}) = cov(X)$, couldn't I look for a relationship between the covariances for each pair of signals in $X^{B}$ and their distance $D_{ij}$, as described above. If I find such a relationship, I would say that it's very unlikely that $cov(X^{B}) = cov(X)$, since I expect the true $cov(X)$ to be flat w.r.t.$D$. $\endgroup$
    – ali_m
    May 6 '14 at 9:20
  • $\begingroup$ Yes, that would work, but lots of other things would work also. All you have to do is put enough structure on $cov(X)$ (that is, make enough assumptions about $cov(X)$) to reduce the number of free parameters in it by at least 2. Maybe you could use a Bayesian approach in which you specify a prior which puts heavy weight on $cov(X)$ being flat in distance. $\endgroup$
    – Bill
    May 6 '14 at 15:26

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