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Given a set of real numbers $X$ and a subset $Y \subset X$, I would like to demonstrate that $Y$ is not a result of a uniform sampling from $X$.

One solution that I thought recall this, that is:

  1. Perform a uniform random sampling from $X$ to produce a subset $X_{unif} \subset X$ such that $|X_{unif}| = |Y|$.
  2. Perform a Kolgomorov-Smirnov test on $X_{unif}$ and $Y$ to assess that they come from two different distributions.

By the way I think that since I have the original set $X$ is somewhat useless/wrong to perform an additional random sampling.

Do anyone know something that could solve this problem?

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    $\begingroup$ in my opinion probability to sample a subset $Y_0$ is equal to probability of sampling any other set $Y_i$. In general you can not prove that this particular subset $Y$ was selected non-randomly - unless you have a prior knowledge of what type of bias was used during subsampling. If you have such knowledge - construct test statistic based on this knowledge, take 1000 subsamples of size |Y| and check how many test statistics from these 1000 subsamples exceed (or equal) to yours. This will be your p-value. $\endgroup$ – German Demidov Apr 10 at 13:18
  • $\begingroup$ I am not sure to have understood, you are proposing to use the solution I show in the question? $\endgroup$ – BalrogOfMoria Apr 10 at 13:25
  • $\begingroup$ If you change "produce a subset" to "produce subsets" - I would say yes, and Kolmogorov-Smirnov test looks like terribly bad choice if you want to use p-values from it - but even if you use just W I doubt if it is a good choice of statistic especially if |Y| is small and |X| is quite complex. $\endgroup$ – German Demidov Apr 10 at 13:31
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    $\begingroup$ Why not just perform a K-S test to compare $Y$ directly to the uniform distribution on $X$? There's no need to construct a sample from $X$ in order to analyze your data! $\endgroup$ – whuber Apr 10 at 13:32
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    $\begingroup$ @German I would just like to remark that the subtleties you were alluding to in earlier comments derive from the fact that under the null, every subset (of a given size) of $X$ has equal chances of being sampled. Thus, there's absolutely no way to distinguish this $Y$ from any other subset $Y^\prime.$ One makes progress by specifying an alternative to the null hypothesis: that is, exactly how could uniform sampling be violated? Might it consist, for instance, of unusually large or small gaps between the values of $Y$? Until the OP tells us this, it will be hard to justify any answer. $\endgroup$ – whuber Apr 10 at 14:34
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I would combine your idea of the KS-test and the answer below your question, if the size of the $X$ and/or $Y$ is small:

Sample (uniformly) $B$ times (e.g. $B=1000$) a subset $X_b$ of $X$ of size $|Y|$ and calculate the Kolmogorov-Smirnov statistic $D_b$ from $X_b$ and $X$. This gives you an empirical distribution of the values $D_b$.

Calculate $D^{\ast}$ as the Kolmogorov-Smirnov statistic from $Y$ and $X$ and assess how many of your $D_b$, $b=1,\ldots,B$ are larger than or equal to your $D^{\ast}$, this gives you the approximate $p$-value: $p=\#\lbrace D_b\geq D^{\ast} \rbrace/B$.

If $p$ is smaller than a pre-defined $\alpha$, you can reject the null hypothesis that $Y$ was sampled uniformly with error $\alpha$.

If $|X|$ and $|Y|$ are large, you can calculate the KS-statistic $D^{\ast}$ directly and use tabellarized values to assess significance.

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  • $\begingroup$ $|X| \sim 3k$ and $|Y| \sim 6M$. Do you think that the two sizes allow to directly perform KS-test between $X$ and $Y$? $\endgroup$ – BalrogOfMoria Apr 10 at 13:47
  • $\begingroup$ I guess it is the other way around, $|X|\approx6\cdot10^6$ and $|Y|\approx3\cdot10^3$? Yes, 3000 data points should be enough. $\endgroup$ – Edgar Apr 10 at 14:54
  • $\begingroup$ Yes sorry, it is as you say. Thanks for the clarifications. $\endgroup$ – BalrogOfMoria Apr 10 at 14:57

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