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I'm not in signal processing, I'm from an another discipline. I've derived a simple result which I presume must be well known in SP and I'd like to know whether there's a paper or textbook that has it that I can cite.

Suppose I have an n-dimensional stochastic process

$\mathbf{x}_{t}\;=\;\left[\begin{array}{cccc} x_{t}^{\left(1\right)} & x_{t}^{\left(2\right)} & \cdots & x_{t}^{\left(n\right)}\end{array}\right]$

where $t$ denotes an observation. Each of the components $j$ is the combination of a signal $x_{t}^{*}$ and some random idiosyncratic noise:

$x_{t}^{\left(j\right)}\;=\;x_{t}^{*}b_{j}+\varepsilon_{t}^{\left(j\right)}\sqrt{1-b_{j}^{2}}$

where $\varepsilon^{(j)}_{t}$ is iid distributed and uncorrelated across the $n$ components. For convenience, all these variables are already standardized (mean 0 variance 1).

$b_{j}$ and therefore the signal-to-noise of each $j$ component ( $b_{j}/(1-b^2_{j})$ ) are not observed.

The simple result is: you observe the correlation matrix $\mathbf{R}$ of the $n$ components of $x_{t}$. Then a simple quadratic system gives you the noise-to-signal ratios of each $j$ component of $\mathbf{x}_{t}$ .

Anybody can tell me whether you've seen this result in a paper or textbook?

Thanks!

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  • $\begingroup$ may I assume the $b_j$ are $\text{const.}$ and your ${}^*$ is just a "marker" for "this is the underlying value common to all $x_t^{(i)},i=1,\ldots, n$"? $\endgroup$ – Marcus Müller Feb 13 at 17:54
  • $\begingroup$ Yes. Precisely. $\endgroup$ – bbecon Feb 13 at 18:16
  • $\begingroup$ ok... then I don't see why you need a correlation matrix to get your result? If both $x_t^*$ and $\epsilon_t^{(j)}$ have variance 1 and mean 0, then the SNR of that component $x_t^{j}$ is $$\frac{b_j^2}{1-b_j^2}$$ per construction. $\endgroup$ – Marcus Müller Feb 13 at 18:49
  • $\begingroup$ Thanks and sorry for not being clear enough. The whole point is that $b_{j}$ is unobserved. I have amended the question to clarify it. Also, I'm not looking for the answer. I have the answer. I want to confirm that this is a known result and where I can find it. $\endgroup$ – bbecon Feb 13 at 19:34
  • $\begingroup$ yes, but $b_j^2$ is just the non-diagonal entries in your $\mathbf R$? $\endgroup$ – Marcus Müller Feb 13 at 20:40

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