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Let $(X_n)$ be a sequence of i.i.d. random variables uniformly distributed in $[0,1]$; and, for $n\geq 1$, set $$ S_n = \sum_{k=1}^n \frac{1}{\sqrt{X_k}}\,. $$ It follows from the generalized central limit theorem (as in [1] and [2, Theorem 3.1]) that $$ \frac{S_n-2n}{\sqrt{n\ln n}} $$ converges in law to a Gaussian distribution. However, in this case Berry—Esseen fails to give any convergence rate, as the summands are not square integrable.

Moreover, Hall's results for this sort of sum do not apply. (In [3], this would correspond to $\alpha=2$, while the results hold for $0<\alpha<2$.)

What is the rate of convergence of $\frac{S_n-2n}{\sqrt{n\ln n}}$ to Gaussian?


[1] Shapiro, Jesse M., “Domains of Attraction for Reciprocals of Powers of Random Variables.” SIAM Journal on Applied Mathematics, vol. 29, no. 4, 1975, pp. 734–739. JSTOR, www.jstor.org/stable/2100234.

[2] Christopher S. Withers and Saralees Nadarajah, Stable Laws for Sums of Reciprocals. December 2011.

[3] Hall, P. (1981), On the Rate of Convergence to a Stable Law. Journal of the London Mathematical Society, s2-23: 179-192. doi:10.1112/jlms/s2-23.1.179

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  • 3
    $\begingroup$ You can easily get $\frac 1{\log n}$ from above. The difficulty is to get a matching bound from below (if it is the correct rate, of course)... $\endgroup$ – fedja Mar 26 '18 at 17:03
  • $\begingroup$ @fedja I'd hope for some $1/n^c$ rate -- but indeed, if it turns out to be logarithmic, that's interesting as well. (Even though a bit sad.) $\endgroup$ – Clement C. Mar 26 '18 at 17:59
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$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

Let $V_k:=1/\sqrt{X_k}$, $b_n:=\sqrt{n\ln n}$, \begin{equation*} Z_n:=\frac{S_n-2n}{\sqrt{n\ln n}}=\frac1{b_n}\sum_1^n (V_k-\E V_1), \end{equation*} \begin{equation*} \de_n:=\sup_{x\in\R}|\De_n(x)|, \end{equation*} where \begin{equation*} \De_n:=F_n-G,\quad F_n(x):= P(Z_n<x),\quad G(x):=P(Z<x), \end{equation*} and $Z\sim N(0,1)$.

We shall show that for all large enough $n$ \begin{equation*} \frac{\sqrt{\ln\ln n}}{\ln n}\ll\de_n\ll\ep_n:=\frac{\ln\ln n}{\ln n}; \tag{1} \end{equation*} here everywhere, the constants associated with $\ll$, $\gg$, $O(\cdot)$ are universal. Thus, we have a rather tight bracketing of $\de_n$. (Conjecture: $\de_n\asymp\frac{\ln\ln n}{\ln n}$.)

Let $c$ denote various complex-valued expressions (possibly different even within the same formula) such that $|c|\ll1$.

The pdf of $V_k$ is $x\mapsto\frac2{x^3}\,I\{x>1\}$, where $I$ denotes the indicator. Note that $|e^{iu}-1-iu|\le2|u|$ and $|e^{iu}-1-iu+u^2/2|\le|u|^3/6$ for real $u$. So, for the characteristic function (c.f.) $f_V$ of $V_k$ and $|t|\le1$ we have \begin{multline*} \frac12\,f_V(t)=\int_1^\infty\frac{e^{itx}}{x^3}\,dx =\int_1^\infty\frac{1+itx}{x^3}\,dx -\int_1^{1/|t|}\frac{t^2x^2/2}{x^3}\,dx \\ +\int_1^{1/|t|}\frac{e^{itx}-1-itx+t^2x^2/2}{x^3}\,dx +\int_{1/|t|}^\infty\frac{e^{itx}-1-itx}{x^3}\,dx \\ =\frac12+it-\frac{t^2}2\,\ln\frac1{|t|} +c\int_1^{1/|t|}\frac{|t|^3x^3}{x^3}\,dx +c\int_{1/|t|}^\infty\frac{|t|x}{x^3}\,dx \\ =\frac12+it-\frac{t^2}2\,\ln\frac1{|t|}+ct^2. \end{multline*} So, for $|t|\le1$ we have $\ln f_V(t)=2it-t^2\,\ln\frac1{|t|}+ct^2$ and hence for the characteristic function $f_n:=f_{Z_n}$ of $Z_n$ we have \begin{multline*} \ln f_n(t)=-i2nt/b_n+n\ln f_V(t/b_n) =-\frac{t^2}{\ln n}\,\ln\frac{\sqrt{n\ln n}}{|t|}+c\frac{t^2}{\ln n} \\ =-\frac{t^2}2-\frac{t^2}{\ln n}\,\Big(\frac12\,\ln\ln n-\ln|t|\Big)+c\frac{t^2}{\ln n} \\ =-\frac{t^2}2+\frac{t^2}{\ln n}\,\ln|t|+c\frac{t^2}{\ln n}\,\ln\ln n \tag{2} \end{multline*} for $|t|\le b_n=\sqrt{n\ln n}$. So, with $\ep_n$ as in (1), \begin{equation*} \ln f_n(t)= \begin{cases} -\frac{t^2}2+c\ep_n|t| & \text{ if }|t|\le1 \\ -\frac{t^2}2+c\ep_n t^2\ & \text{ if }1\le|t|\le\ln n, \end{cases} \end{equation*} whence \begin{multline*} \int_{|t|<\ln n}\frac{|f_n(t)-e^{-t^2/2}|}{|t|}\,dt \\ \le \int_{|t|<1}\frac{|e^{c\ep_n|t|}-1|}{|t|}\,dt +\int_{\R}\frac{|e^{-(1-2c\ep_n)t^2/2}-e^{-t^2/2}|}{|t|}\,dt \le c\ep_n; \end{multline*} the latter integral was bounded using the identity $\int_0^\infty\frac{e^{-at^2/2}-e^{-t^2/2}}t\,dt=\ln(1/\sqrt a)$ for $a>0$. By the Esseen smoothing inequality (see e.g. formula (6.4)), \begin{equation*} \de_n\le c\int_{|t|<\ln n}\frac{|f_n(t)-e^{-t^2/2}|}{|t|}\,dt +c/\ln n. \end{equation*} Now the second inequality in (1) immediately follows.

It remains to prove the first inequality in (1). For real $t$ and real $A>0$, \begin{multline*} \int_0^\infty e^{itx}d\De_n(x)=\int_0^A e^{itx}d\De_n(x)+c(1-F_n(A))+c(1-G(A)) \\ =c\De_n(A)+c\De_n(0)-it\int_0^A e^{itx}\De_n(x)dx+2c(1-G(A))+c\de_n \\ =c\de_n+c|t|A\de_n+ce^{-A^2/2} =c\de_n+c|t|\de_n\sqrt{\ln\frac1{\de_n}} \end{multline*} if $A=\sqrt{2\ln\frac1{\de_n}}$. Similarly estimating $\int_{-\infty}^0 e^{itx}d\De_n(x)$, we have \begin{equation*} f_n(t)-e^{-t^2/2}=\int_{-\infty}^\infty e^{itx}d\De_n(x)=c\de_n+c|t|\de_n\sqrt{\ln\frac1{\de_n}}. \end{equation*} Letting $t=1$ here and in (2), we see from the second line in (2) that $$\de_n\sqrt{\ln\frac1{\de_n}}\gg\frac{\ln\ln n}{\ln n},$$ whence the first inequality in (1) immediately follows.

It appears that similar techniques should work for a somewhat wide class of distributions, like the ones referenced by the OP.

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  • $\begingroup$ That's pretty neat. Do you have any idea what causes the empirical log-log-plot to hint at a polynomial rate of convergence? $\endgroup$ – Clement C. Apr 1 '18 at 18:49
  • $\begingroup$ @ClementC., you took too few values, so it's not visible. I made computations for $n=1.5^k$, $k=5,\dots,20$, with $5000000$ MC samples, and precise computation of the distance between cdfs; the plot is convex (as it should be), not linear. $\endgroup$ – zhoraster Apr 1 '18 at 19:08
  • $\begingroup$ @zhoraster Fair. I couldn't be more precise, Mathematica runs out of memory when I try to do so... $\endgroup$ – Clement C. Apr 1 '18 at 19:17
  • $\begingroup$ @ClementC., Matlab runs out of memory too. So I had split the MC simulations into groups of 1000. $\endgroup$ – zhoraster Apr 1 '18 at 19:28
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    $\begingroup$ @ClementC. : I can only guess why your graph seems linear. One possible reason was offered by user zhohaster, which may especially be the case in this situation, when the rate of convergence is very slow. It appears that the other two answers also establish, by different methods, a $\ln\ln n/\ln n$ rate of convergence of the c.f. The simple argument at the end my answer also shows that one cannot get a rate of convergence of the Kolmogorov distance much faster than that of the c.f. $\endgroup$ – Iosif Pinelis Apr 1 '18 at 19:33
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I guess the answer to this question depends on how we define the rate of convergence. However if we focus on the limiting behavior of characteristic function then the answer is pretty easy. Define by $Z_n := \frac{S_n-2 n}{\sqrt{n \log(n)}}$ as the detrended and normalized random variable and denote by $\kappa_{Z_n}(k)$ the characteristic function of that variable. Note that the characteristic function of $Z_n$ always exists however the higher moments -- as we will see- don't. We have: \begin{eqnarray} \kappa_{Z_n}(k) &:=& \int\limits_{[0,1]^n} e^{\imath k \left( \frac{\sum\limits_{j=1}^n \frac{1}{\sqrt{x_j}} - 2 n}{\sqrt{n \log(n)}}\right)} \prod\limits_{j=1}^n d x_j\\ &=& e^{-\imath \frac{k 2 n}{\sqrt{n \log(n)}}} \cdot \left( e^{\imath \frac{k }{\sqrt{n \log(n)}}}(1+\frac{\imath k}{\sqrt{n \log(n)}}) + \frac{k^2}{n \log(n)}(-\imath \pi +Ei(\frac{\imath k}{\sqrt{n \log(n)}})\right)^n \end{eqnarray}

Now, we have: \begin{eqnarray} \kappa_{Z_n}(k) - e^{-k^2/2} = \frac{1+2 \gamma-\imath \pi +2 \log(k) - \log(\log(n))}{2 \log(n)} \cdot k^2 + O(k^3 \log(k)) \end{eqnarray}

Therefore for "small" values of $k$ we can see from the above what the rate of convergence is. If we want to be more precise and speak about convergence in terms of some norm more work is required. Here I only plot the difference between the characteristic function in question and the limiting function as a function of $k$. The coloring of the curves corresponds is like in the rainbow with violet and red representing $n=2$ and $n=10$ respectively.

Update: It is only now that I got familiar with Berry-Esseen-type of bounds by looking at Terence Tao's blog post https://terrytao.wordpress.com/2010/01/05/254a-notes-2-the-central-limit-theorem/#more-3281 . Let us therefore try to make my answer more rigorous. As stated in the question we are looking to find estimates on the supremum norm of the cumulative distribution functions. Let us denote $G := N(0,1)$ and $\phi(x):= 1_{x>0}$ and write: \begin{align} \lVert \Psi^{<}_{Z_n} - \Psi^{<}_G \rVert &= \mbox{sup}_{a\in {\mathbb R}} \left| P(Z_n < a) - P(G < a)\right| \\ &=\mbox{sup}_{a\in {\mathbb R}} \left| E[ \phi(Z_n-a) ] - E[\phi(G-a)]\right| \\ &=\mbox{sup}_{a\in {\mathbb R}} \frac{1}{2\pi} \left|\int\limits_{{\mathbb R}} \hat{\phi}(k) \cdot e^{\imath k a}\cdot \left( \kappa_{Z_n}(k) - e^{-\frac{k^2}{2}}\right) dk\right|\\ &\leq\frac{1}{2\pi} \left|\int\limits_{{\mathbb R}} \left|\hat{\phi}(k)\right| \cdot 1 \cdot \left| \kappa_{Z_n}(k) - e^{-\frac{k^2}{2}}\right| dk\right| \end{align} where $\hat{\phi}(k) := \int\limits_{\mathbb R} \phi(x) \exp(\imath k x) dx$. As explained in Terence's Tao blog post $\left| \hat{\phi}(k) \right|$ is bounded for large values of $k$.

Now, all we need to do is to find some upper bound on the log-characteristic function of $Z_n$ for large values of $n$. We have: \begin{align} \log(\kappa_{Z_n}(k)) &= -\imath \frac{k 2 n}{\sqrt{n \log(n)}}+n \log\left( e^{\imath \frac{k }{\sqrt{n \log(n)}}}(1+\frac{\imath k}{\sqrt{n \log(n)}}) + \frac{k^2}{n \log(n)}(-\imath \pi +Ei(\frac{\imath k}{\sqrt{n \log(n)}}) \right)\\ &=-\frac{k^2}{2} + \frac{k^2}{2} \cdot \frac{1+2 \gamma-\imath \pi +2 \log(k) - \log(\log(n))}{\log(n)} \\ &\qquad+ \frac{(\imath k)^3}{3} \cdot \frac{-2+6 \gamma -3 \imath \pi+6 \log(k) - 3 \log(n)-3 \log(\log(n))}{\sqrt{n} [\log(n)]^{3/2}} + O(\frac{k^4 [\log(k)]^2}{n}) \end{align} where in the last line I used Mathematica's Series[] command to expand the log in a series to the fourth order. Now clearly we have: \begin{align} \left| \kappa_{Z_n}(k) - e^{-\frac{k^2}{2}}\right| &= e^{-\frac{k^2}{2}} \cdot \left| e^{\frac{k^2}{2} \cdot \frac{1+2 \gamma-\imath \pi +2 \log(k) - \log(\log(n))}{\log(n)}+\frac{(\imath k)^3}{3} \cdot \frac{-2+6 \gamma -3 \imath \pi+6 \log(k) - 3 \log(n)-3 \log(\log(n))}{\sqrt{n} [\log(n)]^{3/2}} + O(\frac{k^4 [\log(k)]^2}{n})} - 1\right|\\ &\leq e^{-\frac{k^2}{2}} \cdot \left(\frac{k^2}{2} \cdot |\frac{1+2 \gamma-\imath \pi +2 \log(k) - \log(\log(n))}{\log(n)}|+ O\!\left(\frac{k^3 [\log(k)]^1}{\sqrt{n \log(n)}} \right)\right) \end{align} To summarize we have the following: \begin{equation} || \Psi^{<}_{Z_n} - \Psi^{<}_G ||\le \frac{1}{2\pi} \cdot \int\limits_{{\mathbb R}} |\hat{\phi}(k)| \cdot e^{-\frac{k^2}{2}} \cdot \frac{k^2}{2} dk \cdot \frac{\log(\log(n))}{\log(n)} + O\left(\frac{1}{\log(n)}\right) \end{equation} as $n$ goes to infinity.

Update': As we can see from the above the rate of convergence tends to be very slow. Of course it may be that our upper bound is not tight and there exists a more accurate upper bound that diminishes much faster with $n$. For the time being I am not quite sure how to check whether this is the case or not. However some insight can be achieved by doing a Monte Carlo simulation. Below I plot the sample cdf (Blue) along with the limit cdf (Purple) for different values of $n$. Here I took $n=10^1,\cdots,10^4$ and in each case I used $m=10000$ instances of random variables. I used the following Mathematica code to run the simulation:

    SetOptions[ListPlot, ImageSize -> 500, 
  LabelStyle -> {15, FontFamily -> "Arial"}, 
  BaseStyle -> {15, FontFamily -> "Bold"}];
SetOptions[ListLogLogPlot, ImageSize -> 500, 
  LabelStyle -> {15, FontFamily -> "Arial"}, 
  BaseStyle -> {15, FontFamily -> "Bold"}];
ns = Floor[10^Array[# &, 100, {1, 4}]];
m = 10000; myList = {};
Do[
  n = ns[[i]];
  X = RandomReal[{0, 1}, {m, n}];
  ll = (Total[1/Sqrt[#]] & /@ X - 2 n)/Sqrt[n Log[n]];

  delta = 1/10;
  bins = Table[-5 + delta/2 + j delta, {j, 1, (10 - delta)/delta}];
  emp = EmpiricalDistribution[ll];
  DD = CDF[emp, bins];
  limD = CDF[NormalDistribution[0, 1], bins];
  myList = 
   Join[myList, {n, Max[#], 
       First[bins[[Ordering[#, 1, #1 > #2 &]]]]} & /@ {Abs[ 
       DD - limD]}];

  pl = ListPlot[Transpose[{bins, #}] & /@ {DD, limD}, Joined :> True, 
    PlotMarkers -> Automatic, PlotLabel -> "n=" <> ToString[n]];
  If[MemberQ[{10, 100, 1000, 10000}, n], 
   Export["CDFs_n_" <> ToString[n] <> ".jpg", pl, "JPEG"]];
  PrintTemporary["n=", n, "done"];
  , {i, 1, Length[ns]}];
MatrixForm[myList];
pl1 = GraphicsGrid[{{ListLogLogPlot[{#[[1]], #[[2]]} & /@ myList], 
     ListPlot[{#[[1]], #[[3]]} & /@ myList]}}];
Export["LimitBehavior.jpg", pl1, "JPEG"];
Import["LimitBehavior.jpg"]

Here are the plots:

CDFs at n=10

CDFS at n=100

CDFs at n=1000

CDFs at n=10000

Below we also show both the Kolmogorow distance and the location of the maximum as a function of the sample size (left and right respectively).

Kolmogorow distance and location of maximum

As above for $m=500000$ realizations.

Conclusions:

  1. My Monte Carlo simulation matches the results given by Clement C, despite the smaller number of realizations which I took to avoid running out of memory.
  2. Interestingly enough the location of the maximum seems to be corresponding to $x=0$ and to be independent of the sample size(it would be interesting to find some theoretical explanation of that fact).

  3. With this number of samples($m=10000$) it is hard to tell whether the line in the double logarithmic plot is a straight line or instead concave or convex and more simulations are required to give a definite answer. However if we increase the number of samples to $m=500000$-- which we can achieve by averaging the relevant piece of code over an addition fifty realizations -- we clearly end up with a line that is convex (see the bottom figure above). This rules out a power-law behavior.

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  • $\begingroup$ Thanks for the answer! As hinted by the question (and the reference to the previous works), I'm mostly looking for Berry-Esseen-type bounds, i.e., convergence rates of the distribution function. $\endgroup$ – Clement C. Mar 26 '18 at 18:00
  • $\begingroup$ Thank you for the update. So, as far as I can tell you obtain an $O((\log\log n)/\log n)$ upper bound on the rate of convergence (in Kolmogorov distance). Any intuition on whether this is tight? (In a broad sense. Roughly speaking, is the convergence inverse-polylogarithmic, or inverse-polynomial like $1/n^c$ for some $c>0$?) $\endgroup$ – Clement C. Mar 27 '18 at 19:32
  • $\begingroup$ This will be much easier to read if you abbreviate $k/\sqrt{n\log n}$ as $u$. $\endgroup$ – Matt F. Mar 27 '18 at 20:53
  • $\begingroup$ It seems clear that the rate is sharp; moreover, a slightly sharper approximation is $N(2n, n \log n \color{red}{+ n\log \log n})$. The skewedness is very interesting. It probably comes out of the term $\frac{k^2\log k}{\log n}$ in the expansion of the log-characteristic function. $\endgroup$ – zhoraster Mar 28 '18 at 19:48
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    $\begingroup$ @zhoraster But the fact that the convergence rate for the (log) characteristic function is tight does not imply that the rate of convergence of the CDF in supremum norm is tight... or I fail to see why. There are at least a few places where there may be some significant slack, starting with the use of the triangle inequality to bound the absolute value of an integral. $\endgroup$ – Clement C. Mar 29 '18 at 20:10
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We shall give an asymptotic formula with error term for its character function $$Z_n:=\frac{1}{\sqrt{n\log n}}\sum_{k=1}^n\left(\frac{1}{\sqrt{X_k}}-2\right)$$ as $n$ tends to infinity. Let $n$ be sufficiently large. Note that $X_k$ is $i. i. d.$ we can compute its character function that \begin{align} \varphi_{Z_n}(z)&=\left(\int_{0}^1\exp\left({\rm i}z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\,d x\right)^n\\ &=\left(\int_{0}^1\,d x\left(\cos\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)+{\rm i}\sin\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\right)\right)^n\\ &:=\left(C_n+{\rm i}S_n\right)^n. \end{align} Let $\delta>0$ be fixed. We estimate that \begin{align} |S_n|&=\left|\left(\int_0^{\frac{\delta}{n\log n}}+\int_{\frac{\delta}{n\log n}}^1\right)\sin\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\,dx\right|\\ &\le \frac{\delta}{n\log n}+|z|\left|\int_{\frac{\delta}{n\log n}}^1\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\,dx\right|+O\left(|z|^3\int_{\frac{\delta}{n\log n}}^1\left|\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right|^3\,dx\right)\\ &=O\left(\frac{\delta}{n\log n}+\frac{\sqrt{\delta}|z|}{n\log n}+\frac{\delta^{-1/2}|z|^3}{n\log n}\right) \end{align} uniformly for $z\in\mathbb{R}$ with $|z|\le \sqrt{\delta}$. Similarly, \begin{align} C_n&=\left(\int_0^{\frac{\delta}{n\log n}}+\int_{\frac{\delta}{n\log n}}^1\right)\cos\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\,dx\\ &=1+O\left( \frac{\delta}{n\log n}\right)-\frac{z^2}{2}\int_{\frac{\delta}{n\log n}}^1\left(\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)^2\,dx+O\left(|z|^3\int_{\frac{\delta}{n\log n}}^1\left|\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right|^3\,dx\right)\\ &=1+O\left(\frac{\delta}{n\log n}+\frac{\delta^{-1/2}|z|^3}{n\log n}\right)-\frac{z^2}{2}\frac{1}{n\log n}\left(\log((n\log n)/\delta)+O(1)\right)\\ &=1-\frac{z^2}{2n}-\frac{z^2\log \log n}{2n\log n}+O\left(\frac{\delta+\delta^{-1/2}|z|^3+|z|^2(1+|\log \delta|)}{n\log n}\right). \end{align} Therefore, \begin{align} \varphi_{Z_n}(z)&=\left(1-\frac{z^2}{2n}-\frac{z^2\log \log n}{2n\log n}+O\left(\frac{\delta+\sqrt{\delta}|z|+\delta^{-1/2}|z|^3+|z|^2(1+|\log \delta|)}{n\log n}\right)\right)^n\\ &=\exp\left(n\log\left(1-\frac{z^2}{2n}-\frac{z^2\log \log n}{2n\log n}+O_{\delta}\left(\frac{1}{n\log n}\right)\right)\right)\\ &=\exp\left(-\frac{z^2}{2}-\frac{z^2}{2}\frac{\log\log n}{\log n}+O_{\delta}\left(\frac{1}{\log n}\right)\right). \end{align} Namely, \begin{align} \varphi_{Z_n}(z)=\exp\left(-\frac{z^2}{2}-\frac{z^2}{2}\frac{\log\log n}{\log n}\right)\left(1+O_{\delta}\left(\frac{1}{\log n}\right)\right) \end{align} holds for all $z\in\mathbb{R},|z|\le \sqrt{\delta}$ as $n\rightarrow \infty$.

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Update: what is below is seemingly hinting at a $\Theta(n^{-c})$ rate of convergence for some absolute constant $c>0$. However, this appears to be an artifact of the simulation (sample size too small, and $n$ not large enough). See Iosif Pinelis' answer, which establishes a $\tilde{\Theta}(1/\log n)$ rate of convergence in Kolmogorov distance; and the comments below


A small experimental observation: building on the Mathematica code provided by Przemo, I computed the empirical cumulative distribution of $S_n$ (based on $m=500000$ independent samples) for $n$ ranging from $5$ to $399$.

list = {}
For[ k = 5, k < 400, k++,
    n = k; m = 500000;
    X = RandomReal[{0, 1}, {m, n}];
    ll = (Total[1/Sqrt[#]] & /@ X - 2 n)/Sqrt[n Log[n]];
    emp = EmpiricalDistribution[ll];
    err :=  Max[ Table[ Abs[CDF[emp, x] - CDF[NormalDistribution[0, 1], x]], {x, -4, 6, 0.05}]];
    AppendTo[ list, err ]
]

Computing the log-log plot of the result:

listpairs = {}
For[ k = 1, k < 395, k++, AppendTo[ listpairs, {4 + k, list[[k]]} ] ]
ListLogLogPlot[listpairs]

it does look like the convergence rate is of the form $1/n^{\epsilon}$ for $\epsilon \simeq 0.14$.

enter image description here

Of course, there are at least two sources of error in the code above (the sampling error, which translates to a supremum norm error in the empirical CDF of order $1/\sqrt{m}$; and the computation error in the Max due to the gridding by 0.05. But it seems unlikely this could somehow change the trend from logarithmic to inverse polynomial.)


Update: Here are the results of a more thorough experiment, taking $m=10^7$ and $n$ from $20$ to $2000$ by steps of $20$ (so 100 different values). Further, the distance between Gaussian and empirical CDFs are now computed as the max over the interval $[-5,5]$, discretized by a step of $0.0001$ (not $[-4,4]$ by $0.05$ as before). Both the regular and log-log plots are below:

enter image description here

enter image description here

I may be misinterpreting it, but this still seems to hint at an inverse polynomial rate, in spite of the theoretical guarantee (?).

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