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?



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


