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I have doubt with the mathematical tools used in a problem of Signal Analysis:

I have a complex observable series $Y(t)$ which is the result of summing two complex r.v $X(t)$ (unobservable) and a $\epsilon(t)$ (unobservable).

$$Y(t)=X(t)+\epsilon(t)$$

Assume that $X$ and $\epsilon$ are uncorrelated, and that $\epsilon$ is serially uncorrelated and has constant mean and variance.

I construct a Hankel Matrix of the observed series and then I create a sort of variance-covariance matrix by multiplying the later by its conjugate transpose.

$$ R_Y=\mathcal{Y}\cdot \mathcal{Y^*}=\mathcal{X}\cdot \mathcal{X^*}+\mathcal{E}\cdot \mathcal{E^*}$$

After that I make the eigendecompositon of that matrix and I obtain:

$$U\cdot (\Lambda_X +\sigma^2\cdot I)\cdot U^T$$ Where $U$ is the orthogonal matrix of eigenvectors, $\Lambda_X$ is a diagonal matrix containing the eigenvalues (in decreasing order) of $\mathcal{X}\cdot \mathcal{X^*}$ and $\sigma^2$ represents the "variance" of $\epsilon$ (under my assumptions $\mathcal{E}\cdot \mathcal{E^*}$ is a diagonal matrix with $\sigma^2$ in its diagonal.

Assume that the matrix $\mathcal{X}\cdot \mathcal{X^*}$ has not full rank, and because of that it has some null eigenvalues.

By the eigendecomposition of $R_Y$ we could select which eigenvalues correspond to the $\mathcal{X}\cdot \mathcal{X^*}$ matrix, (by considering the first $d$ eigenvalues, HOW TO SELECT THE ORDER OF THE MODEL?) and then we can estimate $\sigma^2$.

Knowing $\sigma^2$ we could estimate $R_X=\mathcal{X}\cdot \mathcal{X^*}$.

Is there any way to factorize $R_X$ in order to obtain $\mathcal{X}$ and then by the one-to-one relation between Hankel matrices and Series, obtain $X(t)$? if this is possible, under which additional conditions? If the variables are real-valued, is it the same?

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    $\begingroup$ Is this the kind of idea you are looking for? $\endgroup$ – AnonSubmitter85 Oct 29 '18 at 18:49
  • $\begingroup$ Indeed it looks very promising, I'll give it a in-detail look. Thanks! $\endgroup$ – RScrlli Oct 29 '18 at 19:00

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