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Having a "measured" vector $\mathbf{y}$ with its statistics (counts or variance per element), one can use weighted least squares approach to solve the linear system $$\mathbf{A}\mathbf{x} = \mathbf{y}$$ by minimizing $$(\mathbf{y} - \mathbf{A}\mathbf{x})^T{\rm diag}(\mathbf{c})(\mathbf{y} - \mathbf{A}\mathbf{x}),$$ where $\mathbf{c}$ contains either counts or inverse variances.

When this linear problem is ill-posed but the model is sparse I can use the Basis Pursuit Denoising approach looking for a sparse solution:
$$ \mathbf{x}^{\ast} = \arg \min_{\mathbf{x}} \left\{ \frac{1}{2} {\left\| \mathbf{A} \mathbf{x} - \mathbf{y} \right\|}_{2}^{2} + \lambda {\left\| \mathbf{x} \right\|}_{1} \right\} $$

Logically i am tempted to just modify the $\mathbf{A}$ term to $\mathbf{A}^T {\rm diag}(\mathbf{c}) \mathbf{A}$ to include the weights for each element, but I am not sure this is a sound approach for the $L_1$ case. Is this approach valid or is there a different way to include the measurement statistic for such regularization?

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  • $\begingroup$ Why can't you just replace ${\left\| \mathbf{A} \mathbf{x} - \mathbf{y} \right\|}_{2}^{2}$ with the term using $\mathbf{c}$? It just changes the problem to a weighted least squares. $\endgroup$
    – Peter K.
    Commented Nov 10, 2021 at 23:27
  • $\begingroup$ that is what I ask more or less... I am not sure it makes sense. $\endgroup$
    – bla
    Commented Nov 11, 2021 at 22:06
  • $\begingroup$ I think it would relate to dsp.stackexchange.com/questions/30147. $\endgroup$
    – Mark
    Commented Nov 12, 2021 at 6:37

1 Answer 1

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Your formulation:

$$ \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| A \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \lambda {\left\| \boldsymbol{x} \right\|}_{1} $$

Has 2 elements:

  1. The Fidelity Term
    This is basically measurements term with the model of AWGN with IID noise.
  2. The Regularization Term
    This is a sparse promoting model by using the Laplace Distribution as a prior.

Since your measurement model is not IID but with different variance per measurement what you need is use Mahalanobis Distance in the measurement model (Which matches the Multivariate Gaussian Distribution).

Using some variation of the norm (See Norm with Symmetric Positive Definite Matrix) which is defined as:

$$ {\left\| \boldsymbol{x} \right\|}_{W}^{2} = \boldsymbol{x}^{T} W \boldsymbol{x}, \; \boldsymbol{x} \in \mathbb{R}^{n}, W \in \mathbb{S}_{++}^{n} $$

You may formulate your problem as:

$$ \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| A \boldsymbol{x} - \boldsymbol{y} \right\|}_{ {C}^{-1} }^{2} + \lambda {\left\| \boldsymbol{x} \right\|}_{1} $$

Where $ C $ is your covariance matrix (Basically $ \operatorname{diag} \left( \boldsymbol{c} \right) $ on your question).
The above is easy to solve using many methods as it is basically transformed LS by using Cholesky Decomposition of the Covariance Matrix.

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  • $\begingroup$ I am trying to understand if I can articulate your answer in CVX (does adding weight render the problem to be non-convex?)... First, if I understand correctly the norm to be minimized $C^{-1}$ is the inverse(?) of the covariance matrix given by $C= (y−Ax)^T diag(c) (y−Ax)$ ?and to obtain the inverse I can use Cholesky Decomposition or etc? $\endgroup$
    – bla
    Commented Nov 11, 2021 at 20:36
  • $\begingroup$ @bla Adding the weighting does not make the problem non-convex. $\endgroup$
    – Peter K.
    Commented Nov 11, 2021 at 22:39
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    $\begingroup$ @bla, Since $ {C}^{-1} = \operatorname{diag} \left( \boldsymbol{c}^{-1} \right) $ you can easily define $ {C}^{-1} $ and use quad_form() from CVX. You may mark this as answered and open a question specifically how to solve my form in CVX and I will post code. $\endgroup$
    – Royi
    Commented Nov 12, 2021 at 5:48
  • $\begingroup$ @Royi, thank you will do. $\endgroup$
    – bla
    Commented Nov 12, 2021 at 8:42
  • $\begingroup$ dsp.stackexchange.com/questions/79087/… $\endgroup$
    – bla
    Commented Nov 12, 2021 at 8:54

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