0
$\begingroup$

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 sub-linear respect to $N$?

$\endgroup$
  • 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
1
+100
$\begingroup$

(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 Pareto-distributed 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:

Bentkus, V., Juozulynas, A. & Paulauskas, V. Lévy–LePage Series Representation of Stable Vectors: Convergence in Variation. Journal of Theoretical Probability 14, 949–978 (2001)

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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 $N$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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