Author: Robert Pope
Date: 07:18:11 12/31/01
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On December 31, 2001 at 09:44:52, Andrew Dados wrote: > >Suppose I am getting tons of scores for some experiment which outcome will obey >known distribution (In my problem it is Poisson distribution; type of >distribution should not matter). > >I can't store all scores, but I need to know average and mean parameters, so I >could recreate distribution function at some time later. How can I store some >set of data as small as possible to be able to add new scores to it and still >get my mean/sigma right? > >Example: One experiment is 1000 tosses of a coin. In this case outcome is number >of heads. I will collect unspecified number of such results. In this case I >could simply store an array of 1000 counters, but I can't afford it. Average >number can be easily stored and incrementally updated with 2 ints: total sum and >number of experiments. Can some similar trick be done to recalculate mean value >after new score comes in? > >Chess example (closer to my problem): I have a chess position for which I am >getting time-to-solve results from many players. So their rating distribution is >'predefined' here. The more samples I will collect, the more accurately I can >assing a rating for some new player solving this position. I can not collect all >separate times-to-solve. So for each player I need to update some totals to be >able to calculate mean from those totals (average is easy). Can this be >accurately done? > >..and no... while it sounds like that - it is not some school assignment. :) > >-Andrew- I'm not sure about your terminology here. In statistics, mean _is_ the average, the way most people think about it. Do you intend to say standard deviation? The poisson distribution only has one parameter, the mean (sum(Xi)/N). The standard deviation, sigma, is equal to the mean by definition. It sounds like you already know how to update this statistic. E.g. If you know the number of prior observations included in your current sample mean, N, you can update the sample mean with a new observation like this: newMean = (oldMean+(X[i+1]/N))*N/(N+1). Or you can keep a running total Sum(X[i]) and a running total N. Xbar(new) = Xbar(old)
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