# Does a subset with small cardinality represent the whole set?

Assume that we have heavy-tailed distribution $$F(x)$$ such that \begin{align} F(x)=\mathbb{P}[X\geq x]=x^{-0.5}. \end{align} Then, we produce $$N$$ independent samples $$X_1,X_2,\ldots,X_N$$ from this distribution. Assuming that $$n(N)\leq N$$ is a function of $$N$$, we pick the $$n(N)$$ largest amounts among $$X_1,X_2,\ldots,X_N$$ and produce the summation as \begin{align} S_{n(N)} = \sum_{\ell=1}^{n(N)} X_{i_\ell}, \end{align} Here, $$i_1, \ldots, i_{n(N)}$$ represent the indices corresponding to the $$n(N)$$ largest values among the $$N$$ variables. The question is whether there exists a specific choice of $$n(N)$$ such that as $$N$$ approaches infinity, the ratio $$n(N)/N$$ tends to zero, and we have \begin{align} \lim_{N\rightarrow\infty}\frac{S_{n(N)}}{S_{N}}= 1. \end{align} In other words, does a very small subset of variables represent the whole of them? How can I think about these kinds of problems to solve them?

• $F(x)=\mathbb{P}[X\leq x]=x^{-0.5}$ is not here a typo? Jul 6, 2023 at 10:11
• If it should read as $X\geq x$, then we have a large number of order $n^2$ with high probability, thus the terms which are greater than $\sqrt{n}$ dominate the sum, and we have $o(n)$ such numbers Jul 6, 2023 at 10:18
• Thank you. You are right, I fixed it. I will be grateful if you write the answer in detail. I would like to know the details of such an analysis. Jul 6, 2023 at 18:04
• I would have written $F(x)= \Pr(X>x) = x^{-0.5} \text{ for } x\ge 1$ (being explicit about that lower bound). And also "a sample of $N$ independent observations from this distribution" rather than "$N$ samples from this distribution." $\qquad$ Jul 6, 2023 at 20:32
• Does there exist a distribution that satisfies this?
– usul
Jul 7, 2023 at 14:42

The probability that all samples are less than $$N^{19/10}$$ is $$(1-N^{-19/20})^{N}$$ that tends to 0. The expected number of samples greater than $$N^{1/2}$$ is $$N^{3/4}$$, thus, the probability that we have more than $$N^{4/5}$$ such samples is by Chebyshev inequality at most $$N^{-1/20}$$,also tends to 0. Therefore, with probability tending to 1 the sum of all but $$N^{4/5}$$ largest samples does not exceed $$N^{3/2}$$ while the sum of $$N^{4/5}$$ largest samples (and even one single largest sample) is at least $$N^{19/10}$$. It yields that you may take $$n(N)=N^{4/5}$$. Of course this is not optimal.
• I think the $N^{19/20}$ towards the end is meant to be $N^{19/10}$. Jul 7, 2023 at 8:22
• The expected number of samples less than $N^{1/2}$ is $NP(X<N^{1/2})\sim N$. Or, perhaps, I misunderstood what you meant by the expected number of samples less than $N^{1/2}$. Jul 7, 2023 at 14:30
• Also, I guess, instead of $N^{19/20}$, you meant $N^{-19/20}$. Jul 7, 2023 at 14:31