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I am looking to have People rate certain items on a scale of 0-10 based on those items perceived utility to the particular Person performing the rating. Fractional values are permitted. Giving no rating on an item is also permitted; however, I prefer to rank "highly rated" items with more ratings higher than "highly rated" items with fewer ratings; so, I know I do not want to use the arithmetic mean. [Note: While this question may seem similar to one asked a few years back, the scenario is distinctly different in the fact that prior question limited ratings to 4 discrete values where this scenario does not involve discrete values.] I also want to take the ratings and come up with an average based on the Laplace rule of succession. However, the rule, as far as I can tell presumes discrete possible values, as opposed to the infinite choices in this particular case. Does a method exist for extending the rule to cover such a scenario?

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  • $\begingroup$ Laplace's law of succession is derived as a Bayes estimator to the success probability, p, in the binomial distribution using a uniform prior on (0,1). Why do you want to use the law of succession? Why not just the arithmetic average? $\endgroup$ – user31668 Jan 8 '14 at 16:43
  • $\begingroup$ @Eupraxis1981: Thanks for commenting. A previously asked question had a similar scenario using discrete ranking values and the rule of succession was suggested as a solution. I have also edited the question to explain why I do not want to use the arithmetic average. $\endgroup$ – xuinkrbin. Jan 8 '14 at 16:50
  • $\begingroup$ @Eupraxis1981: A clarification=> weighting the discrete values in accordance with the rule of succession was suggested as a solution. $\endgroup$ – xuinkrbin. Jan 8 '14 at 16:56
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    $\begingroup$ See below for a suggested extension. $\endgroup$ – user31668 Jan 8 '14 at 17:17
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    $\begingroup$ I made a couple edits to my answer if you are reviewing it $\endgroup$ – user31668 Jan 8 '14 at 17:24
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A simple extension would be as follows:

Laplace estimator: $\hat p = \frac{s+1}{n+2}$. Note that this is simply the sum n Bernoulli(p) r.v.s plus 1 divided by n + 2 total trials.

To extend this, we need to make sure it holds in its limits (No ratings, infinite number of ratings). We can get this agreement using the following score (S) formula for item $i$, given a vector of rankings, $r_i$ and the number of ranks given $n_i$:

$S_i(\mathbb{r}_i,n_i)=\frac{\sum\limits_{j=1}^{n_i}r_{ij}\space +\space 10}{n_i + 2}$ This way, if you have no observations, you get 5 as your estimated ranking. The more ratings an items gets, the closer its rank approaches the arithmetic average of its ranks. If two items (A,B) have the same average ranking (say 6), but $n_A>n_B$, then $S_A>S_B$.

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  • $\begingroup$ Very nice. From where does the "10" come? $\endgroup$ – xuinkrbin. Jan 8 '14 at 17:33
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    $\begingroup$ @xuinkrbin. its a simplification from the more fundamental form of it: $10\cdot\frac{\sum\limits_{j=1}^{n_i}r_{ij}\space +\space 10}{10n_i + 20}$. The bottom gives the maximum possible score achievable from N+2 ratings, while the top gives the sum of actual ratings, plus one perfect score. This is in direct analogy to the one dervied for the binomial estimator. $\endgroup$ – user31668 Jan 8 '14 at 17:36

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