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I hope you guys find this an interesting question. Imagine you want to implement a system that rewards people for their contributions (regardless of quality or any external assessment). What is a good model, considering the following conditions:

  • It shouldn't reward people that contributes "excessively" only to gain whatever you're giving as reward. For example, if you're giving an award for 5 contributions, then you can't give that award again because of the sixth contribution.
  • It shouldn't be too complex to be computationally unfeasible (although, if you know of something complex but interesting, I'd like to know about it).

As a clarification, the situation is such that you want to give a single award (instead of a variety of them).

Maybe you know some established theory dealing with this kind of stuff. Let me know if that's the case.

By the way, I searched in Google Scholar but I didn't find anything relevant.

Thanks!

Asaf Karagila
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r_31415
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  • I do not really think this can be considered a mathematical question until you're more specific about what you mean by "best." – Qiaochu Yuan Jun 09 '12 at 18:35
  • @QiaochuYuan Don't be distracted by "best". After all, it's not an optimization problem. I just want to read good ideas about this. – r_31415 Jun 10 '12 at 03:44
  • Anyway, I changed 'best' to 'good' to avoid any confusion :-) – r_31415 Jun 10 '12 at 03:46

1 Answers1

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Your reward function can follow the Erlang distribution for instance. If $c$ denotes the contribution, then the reward is $r(c)$ given as $$r(c) = r_{\max} \int_{x=0}^c \dfrac{\lambda^k x^{k-1} \exp(-\lambda x)}{(k-1)!} dx$$ where $r_{\max}$ is the maximum reward one can earn. This is the Erlang distribution. The parameters $\lambda$ and $k$ affect the rate of decay and the shape of the distribution respectively.

  • Looks nice. Reading about it... :-) – r_31415 Jun 08 '12 at 04:02
  • Correct me if I'm wrong, but I think the idea is calculating the probability of having c contributions in a given period of time when I know the average number of contributions. Assuming this is the right interpretation, the benefit of using Erland distribution instead of an exponential distribution would be that using Erland I can vary the shape of my distribution to fit better my particular situation or is there other argument here?. Also, I think you were focusing on a situation in which there are several rewards, right?. So, the higher the probability, the greater the award. – r_31415 Jun 08 '12 at 04:58
  • Uhm, now I realize you meant that the time between contributions was Erland distributed instead of the number of contributions. Then the question to ask would be "Given a mean time between contributions, what is the probability of having k contributions in some time span?". Is that right?. – r_31415 Jun 08 '12 at 16:58
  • Why is everyone spelling A.K. Erlang's name with a d? –  Jun 09 '12 at 02:21
  • @RahulNarain I don't know. I think I read it here or somewhere else but I should have written Erlang. Seems like there has been some edition over here, though. – r_31415 Jun 10 '12 at 03:48
  • @RobertSmith Yes, I changed it to Erlang. And I don't understand your previous comments. If a person contributes $c$ times, then reward him with $$r_{\max} \int_{x=0}^c \dfrac{\lambda^k x^{k-1} \exp(-\lambda x)}{(k-1)!} dx$$ points. –  Jun 10 '12 at 04:38
  • @Marvis Yes, but I wanted to deconstruct the meaning of the integrand in that expression. It implies a mean time $\lambda$ between contributions, so you are dealing with intervals of time between contributions which makes a natural choice to use that distribution. That's also why I was asking why not to use the exponential distribution (it's simpler to calculate) but if the time between contributions are supposed to be Erlang distributed, then Erlang of order $k$ would fit nicely. However, if you have another interpretation, please let me know. – r_31415 Jun 10 '12 at 05:19
  • @RobertSmith I don't understand what time you are talking about. There is no time in picture. We are talking about contribution $t$. Think of $\lambda$ as being inversely proportional to mean contribution. –  Jun 10 '12 at 06:48
  • @Marvis Well, now I don't understand what you mean. Did you actually mean integrate over 'contributions'?. How do you do that? What is a mean contribution?. I thought this makes perfect sense in the following sense: "The Erlang random variable describes the time interval between any event and the kth following event". Fundamentals of Applied Probability And Random Processes by Oliver Chukwudi Ibe. – r_31415 Jun 10 '12 at 07:05