2
$\begingroup$

I with to estimate the parameters of a specified semi-infinite distributional family based on a sample drawn from that distribution. It is plausible that my sample median converges to the population median. More generally I would like to choose parameters such that the observed values of the quantiles from the sample lie near the calculated values of the same quantiles given the distribution parameters, for some metric of distance, appropriately weighted.

However, it is clear that the highest observed value does not represent the 100 percent quantile, and the lowest observation does not represent the zero quantile. As a result, the true values even of the sample quantiles are unknown (although the sample order statistics are of course known). Here the out-of-sample values I am referring to represent only sample variability, not censorship or truncation.

Recognizing this, how does one calculate the sample quantiles to match with the calculated quantiles? Does one discard a fraction of the observations at each end of the distribution? Does one push all the values away from the median by some adjustment factor? Or is there some other procedure that is widely used to adjust for this problem?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Does Wilks' method for the estimation of quantiles answer your question?

Say you have an unknown distribution (at least continuous), and samples drawn from that distribution. Wilks gives you a way of estimating quantiles for a chosen confidence level, from your samples only, and without having to make any other assumption on your distribution.

For instance, if you wish to have a 95% quantile, and if you have at least 59 samples, then your highest value will be higher than the quantile, with a confidence level of 95%. As soon as you have 93 values, you may take the second highest value as an upper bound.

See this page for the details : http://openturns.github.io/openturns/master/theory/data_analysis/quantile_estimation_wilks.html

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks @RomainReboulleau! This answer may work for me. I have two questions. I’ll put them in separate comments. First, the distribution I am drawing from has a Pareto-like tail, and my preliminary investigations suggest that the parameters are in the range where the distribution has a mean but no variance. Does that affect the validity of this method, or its suitability for my intended use? $\endgroup$
    – andrewH
    Commented Apr 9, 2018 at 20:40
  • $\begingroup$ Second question: I am having a little trouble with unfamiliar notation. Is h the quantile function? The CDF? What is the transformation from the Xs to the Ys? What is capital C? I usually see a "sup" sub- or superscript to refer to the supremum of a set. Do I correctly read it here as the upper bound of a confidence interval? Finally, the procedure seems to use a single sample of size N, but the language “sample of N independent samples of the random vector X” makes it sound like a resampling technique. Are the samples just individual draws, or is this some version of the bootstrap? $\endgroup$
    – andrewH
    Commented Apr 9, 2018 at 20:43
  • $\begingroup$ First question: as far as I know, the only thing you have to make sure is that the distribution is continuous. But I haven't personally seen any case like that. Maybe that kind of information can be found in Wilks original publication here: projecteuclid.org/euclid.aoms/1177731788 $\endgroup$
    – Romain Reboulleau
    Commented Apr 10, 2018 at 9:36
  • $\begingroup$ Second question: the example given in the page cited in my original answer is a common example, where you have a process (h) that transforms an input vector X into an output vector Y. This example is given because Wilks method is used as a very simple way to propagate uncertainties through complex processes, such as computer codes. It shows that you only need to know the distribution of each component in X, and make no specific assumption on the distribution of Y. In the generic case, you may only have samples of Y, it works the same. C is a notation for binomial coefficients $\endgroup$
    – Romain Reboulleau
    Commented Apr 10, 2018 at 9:49
  • $\begingroup$ (continued) I'm not sure how to understand the "sup" subscript, you may read it without the "sup" actually, it still makes sense to me. Finally, there is no re-sampling in the process, the formulation is tricky but you should understand it just as "X variable sampled N times". $\endgroup$
    – Romain Reboulleau
    Commented Apr 10, 2018 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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