What would be the MLE (Maximum Likelihood Estimator) of the given data:

Assuming the i-th point $ {p}_{i} = ({x}_{i}, {y}_{i}) $.
We have the following measurements of this specific point - $ (x, y, r = \sqrt{{x}_{i}^{2} + {y}_{i}^{2}}) $.
Namely, we have a measurement of the x, y coordinate and the norm of the point with the following distributions:

$$ x \sim N(0, {\sigma}_{x}), y \sim N(0, {\sigma}_{y}), r \sim N(0, {\sigma}_{r}) $$

Given this measurement vector, what would be the MLE of $ {p}_{i} $ coordinates (${x}_{i}, {y}_{i})$?

I know the brute force to calculate it, is there any efficient method for this specific formulation?

What about given a set of points and their corresponding measurements?
Assuming the model of the set of points is linear (polynomial of degree N, etc...).

Has anyone encountered an article or anything about this problem?
What about the Cramer-Rao lower bound, or better bound?

Thank You.

  • 1
    $\begingroup$ You don't have a MLE of data. You have a MLE of some quantity, like a parameter, or a distribution function. Please clarify what you're estimating by ML. $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '13 at 0:44
  • $\begingroup$ You're absolutely right. I didn't mention the quantity to estimate (The Parameter Vector) which is the point coordinate ($ {x}_{i}, {y}_{i}) $. $\endgroup$ – Royi Nov 4 '13 at 5:29
  • $\begingroup$ Sorry, I thought $(x_i,y_i)$ was data. If that's parameters, what's the data? I think perhaps I am confused by the notation. $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '13 at 11:38
  • $\begingroup$ @Glen_b, The parameters to estimate are the point coordinates. The measurements are the coordinates + range (Squared root of the coordinates). $\endgroup$ – Royi Nov 4 '13 at 15:49
  • $\begingroup$ Now I really have no idea what's going on. The more you explain, the less sense I can make of it. $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '13 at 23:43

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