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Following up from an answer by @Royi on adding weights to BPDN problem , I would like to use CVX to test this approach. How can we formulate in CVX the regularized LS L1 norm with weights given by a vector $c$, as follows:
$$ \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| A \boldsymbol{x} - \boldsymbol{y} \right\|}_{ {C}^{-1} }^{2} + \lambda {\left\| \boldsymbol{x} \right\|}_{1} $$

Where $ C $ is the covariance matrix $ \operatorname{diag} \left( \boldsymbol{c} \right) $ ?

Here's a minimal example using Matlab:

% problem data
A  = [1    0    0   0.5;...
      0    1  0.2   0.3;...
      0  0.1    1   0.2];
x0 = [1 0 1 0]';    % original signal
y  = A*x0;          % measurements with no noise
w  = randi(1e3,1,numel(y))'; % random weights vector
y  = y +  1./(sqrt(w)).*randn(numel(y),1); % measurements with weighted noise

CVX that does not include the weights info would be:

lambda = 0.01;      % regularization parameter
cvx_precision high
cvx_solver  SeDuMi
cvx_begin quiet
    variable x(size(A,2),1);
    minimize(norm(A*x-y)+lambda*norm(x1,1))
cvx_end

x =
    0.9864
   -0.0281
    1.0108
         0
bla
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1 Answers1

2

A MATLAB code which implements the problem as defined and solve it using CVX is given by:

%% General Parameters
close('all');
clear('all');

%% Simulation Parameters
numRows = 6;
numCols = 10;


varianceFctr    = 3;
paramLambda     = 2.75;


%% Generate Data
mA  = randn(numRows, numCols);
vX0 = rand(numCols, 1) >= 0.65;
vC  = varianceFctr * rand(numRows, 1);


mCInv = diag(1 ./ vC);


vY = (mA * vX0) + (sqrt(vC) .* randn(numRows, 1));


%% Solving by CVX
cvx_begin('quiet')
    % cvx_precision('best');
    variable vX(numCols);
    minimize( 0.5 * quad_form(mA * vX - vY, mCInv) + (paramLambda * norm(vX, 1)) );
cvx_end
```
Royi
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  • I dont have InitScript.m is that important? is mCInv similar to 1./(sqrt(w)) in my example? – bla Nov 14 '21 at 05:35
  • I removed the need for InitScript.m. vC is the vector of variances. Hence the noise added is by it. In order to create the inverse of the matrix $ C $ I just used the inverse of each value (As it is a diagonal matrix). – Royi Nov 14 '21 at 06:10
  • maybe I dont understand from your code, the noiseless signal mA * vX0 and the noisy signal vY , but you solve for vX so should I compare vX to vX0 ? (because the values there are not close) – bla Nov 14 '21 at 06:18
  • Yes, vX0 is the reference and vX is the solution. They are not close even in the case vC = 1 as the issue is with the sparse reconstruction. But the question is about solving the problem, not if the sparse model solution is perfect. – Royi Nov 14 '21 at 06:52