I am currently investigating the following mental exercise:

  • An urn is filled with N balls. Each ball possesses a number and it is either red or green.
  • There are M color detectors. Each detector possesses a number and they tell whether a ball inserted is either red or green. If used, a color detector always provides an answer however sometimes this answer is wrong meaning there are two conditional probabilities $P(\textrm{detector says red}|\textrm{ball inserted is red})$ and $P(\textrm{detector says green}|\textrm{ball inserted is green})$ associated with each detector.

Now the following procedure is repeated over and over again: without looking at the color, a ball is taken out of the urn at random and inserted into a random detector. The detection result is noted down and the ball is put back into the urn. In this way a table is obtained similar to the following example:

| Ball Identifier | Detector Identifier | Detection Result |
|               1 |                   1 | red              |
|               2 |                   2 | green            |
|               4 |                   2 | red              |
|               1 |                   1 | green            |
|               3 |                   2 | green            |
|               1 |                   2 | red              |
|             ... |                 ... | ...              |

My question is: is it possible by using such a table, to estimate both the detection probabilities and the true color of the balls?

I already considered the following topics:

  • The problem resembles a Hidden Markov model. But in contrast to those, there are no state transitions here.
  • I was also thinking about using an Expectation Maximization algorithm. However I could not recognize a likelihood function with free parameters.

Could anybody point me to a right direction? Any help is appreciated.

Question and Answers

1. How is the detector identifier created?

The detector identifier is just a unique number to distinguish the detectors. Similar to that, the ball identifier is unique number to distinguish the balls from each other.

2. Is the misclassification random noise?

Yes, in case a red ball is inserted it says red with probability $P(\textrm{detector says red}|\textrm{ball inserted is red})$ and green with probability $1 - P(\textrm{detector says red}|\textrm{ball inserted is red})$. In case a green ball is inserted it says green with probability $P(\textrm{detector says green}|\textrm{ball inserted is green})$ and red with probability $1 - P(\textrm{detector says green}|\textrm{ball inserted is green})$. In addition, the detection result does not depend on whether the ball was already inserted previously in the same detector or not.

3. Do you know what proportion of the Urn is red/green?

The exact proportion is not known.

  • 1
    1. How is the detector identifier created? 2. Is the misclassification random noise? 3. Do you know what proportion of the Urn is red/green? – AdamO Feb 5 at 22:31

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.