I am working with a small data set that is clearly non-Gaussian. This data is bound within a fairly narrow range. I have been asked to estimate the quantiles of the population that this data is from. I used the Wikipedia for my reference (method R-4). This all went well. However, I was then asked to estimate the probability of a value occurring that is larger than any of the data in my sample set. It seems that quantile analysis gives you no information distribution of the population outside of your sample range. Is that correct? Is there another approach?
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2$\begingroup$ Say you have a sample of 10 data points. What is the chance that an 11th data point is greater than the 10 you have collected? $\endgroup$– soakleyCommented Sep 26, 2014 at 21:28
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1$\begingroup$ The point made by @soakley might be even a little clearer if it is understood that it does not refer to the chance that the 11th point is the greatest one conditional on the data (the 10 points). That chance cannot be estimated, because these 10 points could be any subset of the population. But if you contemplate this event before data collection and ask for its chance, that can be estimated because all 11 values are random, independent, and identically distributed. Moreover, if you also assume a continuous distribution, the chance can be computed exactly. $\endgroup$– whuber ♦Commented Sep 26, 2014 at 21:52
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$\begingroup$ Do you need the probability that a newly observed point is beyond the already observed range, or you also need to know by how much? $\endgroup$– JamesCommented Sep 26, 2014 at 22:19
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2$\begingroup$ This is actually a tolerance issue. So I have 10 data samples that are well within the threshold. I am now asked what is the probability of exceeding the threshold value. Intuitively, it seems like I cannot estimate that number without a parametric model. I am just trying to see if I am missing something here. $\endgroup$– ziggle314Commented Sep 26, 2014 at 22:30
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1 Answer
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In a comment to the question, @soakley writes
Say you have a sample of 10 data points. What is the chance that an 11th data point is greater than the 10 you have collected?
That points towards a solution, but there is a subtlety. The chance in this quotation does not refer to the chance that the 11th point is the greatest one conditional on the data (the 10 points), even though that is how we might like to think of the situation. That chance cannot be estimated, because these 10 points could be any subset of the population. But if before you collect any data you were to contemplate the possibility that the 11th data point is (uniquely) the greatest, you could infer some useful things about the chance of that unconditional event. The entire point is that the data are exchangeable: each has exactly the same chance of being the largest as any of the others.
If you assume a continuous distribution (so that a tie for largest has zero chance) the chance can be computed exactly. Consequently each of the 11 values has the same chance of being the largest as any of the others and those 11 chances sum to unity. The answer in this case therefore is $1/11$.
When ties in the data have nonzero chances, those 11 chances plus the chance of a tie sum to unity. Therefore the chance of any given value being uniquely the greatest will be less than $1/11$ but the chance that is either greatest or tied for greatest must exceed $1/11$. Without knowing the chances of ties, that's about all that can be deduced, because the chance of being tied for greatest can approach $1$. Imagine, for instance, a population of just zeros and ones, with many more ones than zeros: the chance that the 11th value equals 1, and therefore is among the greatest in the sample, can be arbitrarily close to $1$.
If we resolve ties randomly and equitably, the distribution of results becomes continuous and the preceding result again applies: the chance of being declared the greatest will be exactly $1/11$.
