1

Consider a rank k matrix, call it M, of size nxm. All the elements are non-negative. Now do a noisy observation of it and assume independent Poissonian errors (the error on element $M_{ij}$ is Gaussian with a variance equal to the value of the element, i.e. if $M_{ij}=3$ then the variance on that number is 3.). The problem is determining what the original rank of the matrix was.

I believe you can formulate the problem as a multivariate regression. The design matrix would be identity. The likelihood function would then be a weighted least squares (the weights coming from the fact that the errors are Gaussian but with different variances). Could you then minimize this quantity but only over matrices of rank less then $r$. This would be solvable the same way as a weighted low rank approximation (WLRA, http://www.aaai.org/Papers/ICML/2003/ICML03-094.pdf). Then apply AIC (Akaike information criterion) to select the most likely rank? And if so, how would you be able to calculate the finite sample corrected AIC (CAIC)? I assume the CAIC would be needed if the actual rank of you matrix is close to the dimension of the matrix.

If other ways exist to solve this problem I would also be curious.

Patrick
  • 103
  • 4
  • As you've written it, the errors have non-zero mean (zero mean only if variance = 0, i.e., zero error). Is that what you mean? Are errors independent across elements? Are all element means and variances known? Kind of confusing that you use $M_{ij}$ to refer both to element and mean and variance of its error. – Mark L. Stone May 11 '18 at 22:57
  • Sorry yes, the errors have zero mean. I meant that the value of the element is the same as that elements variance. It's been updated now. All elements are known, since we assume that variance of the error is equal to elements value, we thereby also know all the variances. – Patrick May 11 '18 at 22:59
  • So we know that the true underlying matrix, $M$, must have all elements non-negative? How are you determining the weights, because you don't know the variances, otherwise you would know the element truths? – Mark L. Stone May 11 '18 at 23:09
  • Yes, the elements are non-negative. Essentially the elements are probabilities of an event. And we do know the variances, they are determined by the probabilities. – Patrick May 11 '18 at 23:17
  • Sorry, I don't understand how you know the variances. If you knew them, you'd know the elements, and the statistical aspect would be removed. Do you mean you have estimates of the variances? If so, how? What is the variance of the element which is observed to be -1.242? – Mark L. Stone May 11 '18 at 23:27

0 Answers0