I'm looking for a technique that would help me rank individuals according to 3 scores A, B and C (each representing a separate score in 3 specific fields).

A, B and C are all normalized value between 0 and 1, where 0 is the worst possible score, and 1 is the best possible score. I could simply average the three variables, except the probability densities of A, B and C are different and unknown.

My level of understanding of statistics is just enough to feel that averaging is probably a terrible idea.

What technique, if any, could help me order my individuals from best to worst? Do I need to know the probability density functions of A, B and C to do so?

Any help and/or reading advice is appreciated.

  • 3
    $\begingroup$ The key issue here is to decide how you want to weight the various scores (A, B, & C) relative to each other. Eg, you might think that someone having a better rank on A is 2X as important as having a better rank on B; alternatively, you might think they are equally important. Can you provide that information? $\endgroup$ – gung - Reinstate Monica Sep 9 '14 at 18:03
  • 1
    $\begingroup$ The other issue that may be worth addressing is speed vs. precision. If you want it to process faster, depending on number of records, you may decide on a simpler approach realizing that the ranking of individuals probably won't be drastically different regardless of what method you use (not included weighting different scores, which could cause large differentiation between methods). Ultimately, if you're trying to get 100 quality individuals based on these criteria, any reasonable method you choose will give you that. $\endgroup$ – SocioMatt Sep 9 '14 at 18:07

Since you don't know the distributions, I think all the information is in the placing (think medals, such as in the Olympics). Perhaps give 1 point for every person any given person beat in each of the three scores. So if you had 3 people, you would distribute 9 points (1 each to the three 2nd place finishers and 2 each to the three 1st place finishers). If one person won two and came in 3rd on another, they would have 2x2+0 = 4 'points.' If another had 2 seconds and 1 third, they would have 2 'points.'

  • $\begingroup$ Thanks @MikeP, I'll try this. I'm more looking for an absolute score between 0 and 1, but the placing technique is intuitive and seems to fulfill what's i'm looking for. If I knew the distribution though, would I be able to device an absolute measure? $\endgroup$ – Rodolphe Sep 9 '14 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.