Let $d_n, n\in\{1,2,\cdots,N\}$ be $N$ realizations drawn independent and identically from uniform distribution on $(0,L)$ where $L=\gamma\sqrt{N}$ with constant $\gamma$. Suppose that we need to approximate the sum $$\alpha=\sum_{n=1}^{N}d_n^{3},$$ with the restricted sum $$\hat{\alpha}=\sum_{n\in\mathcal I}d_n^{3},$$ where $\mathcal I\subset N$. We define the set $\mathcal I$ as the largest $\mathcal I$ element of random variables $d_1^{3},d_2^{3},\cdots,d_N^{3}$. Now, the question is to find the order of size of subset $\mathcal{I}$ which produce a good approximation ($\hat{\alpha}\simeq\alpha$), i.e., $$\mathbb{P}[\alpha\hat{\alpha}\leq\epsilon]\geq 1 \beta.$$ Is $\mathcal I$ in order of sublinear respect to $N$?

1$\begingroup$ If you miss even a single realization you may have an error as large as $1/\epsilon^3$ with probabiity $\epsilon/L$ that vanishes only linearly in $\epsilon$, so this will never provide a good approximation. $\endgroup$ – Carlo Beenakker Jan 25 at 15:14

$\begingroup$ But, this occurs with a very small probability. By choosing $\mathcal{I}$ largest value of $d_n^{3}$, it seems that by properly choosing $\mathcal{I}$ we have a good approximation. $\endgroup$ – Math_Y Jan 25 at 15:21

$\begingroup$ But the probability that all of $N$ points be in $(0,\epsilon)$ is $(\epsilon/L)^N$ which is so small. $\endgroup$ – Math_Y Jan 25 at 15:39

$\begingroup$ For example, assume that $N$ points are $L/N,2L/N,\cdots,L$. Then, the question is that is it possible to truncate the series $\sum_{i=1}^{N}i^3$ to have a good approximation? $\endgroup$ – Math_Y Jan 25 at 15:43

$\begingroup$ It is computationally efficient for me to choose only a small fraction of them. $\endgroup$ – Math_Y Jan 25 at 15:44
(Edited after noticing an error which completely changes the answer.)
This is not a complete solution; however, it strongly suggests that the answer is positive.
Let $U$ be a random variable with uniform distribution on $[0, 1]$. The random variable $U^{3}$ has tail $\mathbb{P}[U^{3} > x] = x^{1/3}$, and hence it is in the domain of attraction of a stable law with index $\alpha = \tfrac{1}{3}$.
Observe that $d_n = L U_n = \gamma \sqrt{N} U_n$ for an i.i.d. sequence $U_n$ with uniform distribution on $[0, 1]$. By the invariance principle, the processes $$ X^N_t = \frac{1}{N^3} \sum_{n = 1}^{\lfloor N t \rfloor} U_n^{3} $$ converge (in the appropriate Skorokhod topology) to the increasing $\tfrac{1}{3}$stable Lévy process (i.e. the $\tfrac{1}{3}$stable subordinator) $X_t$. Clearly, $$ \sum_{n = 1}^{\lfloor N t \rfloor} d_n^{3} = \frac{N^{3/2} X^N_t}{\gamma^3} \, . $$ Denote $J = \mathcal{I}$, and let $\hat\alpha_N$ be the sum of $J$ largest variables among $d_n^{3}$, $n = 1, 2, \ldots, N$. Then $$ \hat\alpha_N \approx \frac{N^{3/2}}{\gamma} \times (\text{sum of $J$ largest jumps of $X_t$, $t \in [0, 1]$}) . $$ (I am not rigorous here. However, I am rather convinced one can turn this into a completely argument.)
The question thus turns into the following problem, with $\alpha = \tfrac{1}{3}$ and $\delta = \tfrac{3}{2}$:
How many largest jumps of the $\alpha$stable subordinator $X_t$, $t \in [0, 1]$, one has to add in order to get an approximation which is within $\pm\gamma N^{\delta} \times \epsilon$ from $X_1$ with probability $1  \beta$.
This has been studied a lot: it is the question how fast does the Ferguson–Klass–LePage series of a stable subordinator converge. In particular, since the $n$th largest jump of $X_t$ is comparable with $n^{1/\alpha}$, the error should be of the order $J^{1  1/\alpha}$ when $J$ largest jumps are taken into account. This suggests that $J \approx N^{\delta / (1/\alpha  1)}$ should do the job. In our case, this gives $J \approx N^{(3/2) / (3  1)} = N^{3/4}$, and thus it suggests that it is sufficient to take roughly $\mathcal{I} = N^{3/4}$ largest values of $d_n^{3}$.
I bet one can find a reference for what is written above (and in fact at least some authors do work with Paretodistributed jumps $U_n^{1/\alpha}$ rather than the ordered jumps of $X_t$). I am not an expert in this area, though, and I failed to find a reference in a (very) quick Internet search. The closest one that I encountered is:
One may also search in the standard reference for simulation of stable random variables:
Janicki, A., and Weron, A. (1994). Simulation and Chaotic Behaviour of $\alpha$stable Stochastic Processes. Marcel Dekker, New York
If we first choose $d_n=\frac{L}{nN}$ and include the $M$ smallest $d_n$'s in the restricted sum, then the relative error is $$E=\frac{\sum_{n=M+1}^N 1/n^3}{\sum_{n=1}^N 1/n^3}=\frac{\psi ^{(2)}(M+1)}{\psi ^{(2)}(1)}+{\cal O}(N^{2}).$$ This ratio of polygamma functions amounts to less than a 1% error for $M=6$, independent of $N$.
For a statistical test I compared $N=50$ and $N=500$ at a fixed $M=10$: shown below are the two histograms of the cumulative distribution of the error $E$, obtained from $10^3$ realizations of the set of random variables $d_1,d_2,\ldots d_N$. As you can see, the two histograms ($N=50$ on the left, $N=500$ on the right) are nearly the same, with $E<1\%$ happening with probability 0.9, so you can keep $M$ fixed as you scale up $N$.

$\begingroup$ Thank you so much. I only said $nL/N$ as an example. Maybe, it is better to write the problem in probabilistic way. I mean that $\mathbb{P}(\alpha\bar{\alpha}\leq\beta)$. $\endgroup$ – Math_Y Jan 25 at 16:22