Suppose $X_1, \cdots, X_n \sim_{\mathrm{iid}} U([-1,1])$, where $U([-1, 1])$ denotes the continuous uniform distribution over the interval $[-1, 1]$ (so $E[X_i] = 0$ and $\text{Var}[X_i]= 1/3$). Let $D_n$ denote the distribution of the following sum, scaled to have unit variance: $$ \sqrt{\frac{3}{n}} \sum_{i = 1}^n X_i \sim D_n.$$ Note that $D_n$ is a scaled and shifted version of the Irwin-Hall distribution. Is it known how quickly $D_n$ converges to $\mathcal{N}(0, 1)$ (i.e. the standard normal distribution) in total variation (TV) distance? Some quick experimentation for small values of $n$ in Mathematica possibly indicates $$ \Delta_{\mathrm{TV}}\left(D_n, \mathcal{N}(0, 1) \right)\stackrel{?}{\le} O\left(\frac{1}{n}\right), $$ but I couldn't find a simple way to try to prove it. The closest thing I could find is the Berry–Esseen theorem, but that only gives a bound on the the Kolmogorov–Smirnov distance (i.e. max difference in CDFs), which does not generally imply a TV distance bound.

I would appreciate any help. Thanks so much!

EDIT: [Only keep reading if you want to see the empirical results for small $n$.]

Below are the empirical values for $ \Delta_{\mathrm{TV}}\left(D_n, \mathcal{N}(0, 1) \right)$ (computed in Mathematica by N[1/2*Integrate[Abs[PDF[NormalDistribution[0, 1], x] - Sqrt[n/3]*PDF[UniformSumDistribution[n, {-1, 1}], x*Sqrt[n/3]]], {x, -Infinity, Infinity}]] for varying values of $n$):

$n$ $\Delta_{\mathrm{TV}}$
$1$ $0.1976779590175315$
$2$ $0.05124700117544534$
$3$ $0.027101879212265846$
$4$ $0.019292677385873307$
$5$ $0.014897694156660687$
$6$ $0.012364908454190667$
$7$ $0.010492076525242137$
$8$ $0.009125408176875193$
$9$ $0.008072492963113527$
$10$ $0.007237608640375671$
$11$ $0.00655926922222608$
$12$ $0.005997229294516531$
$13$ $0.0055239287588410195$
$14$ $0.005119885056770444$
$15$ $0.00477093097015108$

Computing a linear regression on top of the log-log version of the table for $5 \leq n \leq 15$ gives $\ln(\Delta_{\mathrm{TV}}) \approx -2.53629 - 1.03812 \ln(n)$ with $R^2 \ge 0.9999$. For this reason (i.e. because $- 1.03812 < -1$ and $R^2$ is very close to $1$), it seemed like $\Delta_{\mathrm{TV}} \le O\left( \frac{1}{n} \right)$ is empirically possible.

  • 1
    $\begingroup$ If the pdf of $D_n$ is greater than the standard normal pdf in $k$ connected regions, then the total variation distance is at most $2k$ times the K-S distance. So graphing those $k$ regions (and computing $k$) for a few different $n$ might help solve the problem quickly. For example: Is $k\le 3$ for all $n$? $\endgroup$
    – Matt F.
    Sep 3 at 2:03
  • 1
    $\begingroup$ (i) Berry-Esseen type bounds in relative entropy, and hence TV (by Pinsker's inequality), are thoroughly discussed in the following PTRF paper link.springer.com/content/pdf/10.1007/s00440-013-0510-3.pdf. (ii) Are you sure about the empirically observed $O(1/n)$ rate? Can you please share some of these empirical results in your post? In particular, this seems to defy the usual Monte Carlo error $O(1/\sqrt{n})$. $\endgroup$ Sep 3 at 10:17
  • 1
    $\begingroup$ Thanks for the comments and the reference! Indeed, it looks like the PTRF paper gives a $O(1/\sqrt{n})$ bound by Theorem 1.1. Thanks for sharing! I just added experimental justification for the possible $O(1/n)$ bound. Is there any hope for (validity or proof) of that? $\endgroup$
    – anon
    Sep 4 at 22:26
  • $\begingroup$ Thanks for the Mathematica code snippet! Here is a slight modification to help "see" the rate: ListLogLogPlot[{Table[ NIntegrate[ Abs[PDF[NormalDistribution[0, 1], x] - Sqrt[n/3]* PDF[UniformSumDistribution[n, {-1, 1}], Sqrt[n/3] x]], {x, -Infinity, Infinity}], {n, 1, 15}], Table[0.1/n, {n, 1, 15}]}, PlotStyle -> {Red, Blue}] $\endgroup$ Sep 5 at 18:06
  • 1
    $\begingroup$ Here is one answer with precise estimates and references that I gave several years ago. $\endgroup$
    – cardinal
    Sep 25 at 18:01

1 Answer 1


$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=o(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \tag{2}\label{2} \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence, slightly modifying the above reasoning, we see that \begin{equation*} \dee(D_n,N(0,1))\sim\frac cn, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$:

enter image description here

Following the lines of the proof, it easy to see that \eqref{2} will hold whenever, say, the $X_i$'s are iid with $EX_i=EX_i^3=0$ and light enough distribution tails (in your case, the $X_i$'s are iid symmetric random variables with no distribution tails), provided that there is some natural $k$ such that $S_k$ has an absolutely continuous pdf with integrable derivative (in your case, $k=2$ will do).

  • 1
    $\begingroup$ (+1) Uspensky (1937) provides a similar result, in flavor, based on a straightforward analysis of characteristic functions, as detailed in this answer. Empirically, the bound seems reasonably tight. $\endgroup$
    – cardinal
    Sep 25 at 18:06
  • $\begingroup$ @cardinal : Thank you for your comment. I actually own the Uspensky book, but had not noticed that result (which is for the Kolmogorov distance, though). $\endgroup$ Sep 25 at 20:53

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