Stein's method proof of the Berry-Esséen theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:41:14Z http://mathoverflow.net/feeds/question/19651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19651/steins-method-proof-of-the-berry-esseen-theorem Stein's method proof of the Berry-Esséen theorem John Jiang 2010-03-28T19:16:18Z 2010-03-29T17:56:47Z <p>The relevant paper is <a href="http://www.springerlink.com/content/w2557152464xm8u3/" rel="nofollow">"An estimate of the remainder in a combinatorial central limit theorem" by Erwin Bolthausen</a>. I would like to understand the estimate on page three right before the sentence "where we used independence of $S_{n-1}$ and $X_n$":</p> <p><code>\begin{align}E|f'(S_n) - f'(S_{n-1})| &amp;\le E \bigg(\frac{|X_n|}{\sqrt{n}} \big(1 + 2|S_{n-1}| + \frac{1}{\lambda} \int_0^1 1_{[z,z+\lambda]} (S_{n-1} + t \frac{X_n}{ \sqrt{n}}) dt\big)\bigg) \\ &amp;\le \frac{C}{\sqrt{n}} \big(1 + \delta(\gamma, n-1) / \lambda\big)\end{align}</code></p> <p>that is, where $\delta(\gamma, n-1)/\lambda$ shows up, which is the error term in the Berry-Esséen bound. </p> <p>Here $S_n = \sum_{i=1}^n X_i / \sqrt{n}$ and $X1, \ldots, X_n$ are iid with $E X_i =0$, $E X_i^2 = 1$, and $E|X_i|^3 = \gamma$. Furthermore, denote $\mathcal{L}_n$ to be the set of all sequences of $n$ random variables satisfying the above assumptions, then</p> <p>$\delta(\lambda, \gamma,n) = \sup { |E(h_{z,\lambda} (S_n)) - \Phi(h_{z,\lambda})|: z \in \mathbb{R}, X_1, \ldots, X_n \in \mathcal{L}_n }$</p> <p>and $h_{z, \lambda}(x) = ((1 + (z-x)/\lambda) \wedge 1) \vee 0$ and $\delta(\gamma, n)$ is a short hand for $\delta(0,\gamma, n)$, and $h_{z,0}$ is interpreted as $1_{(-\infty, z]}$. I am mainly interested in verifying the second inequality, so I don't need to reproduce the definition of $f$ here, but it is related to $h$. </p> <p>This paper is freely available online through springer. thanks in advance.</p> http://mathoverflow.net/questions/19651/steins-method-proof-of-the-berry-esseen-theorem/19727#19727 Answer by Mark Meckes for Stein's method proof of the Berry-Esséen theorem Mark Meckes 2010-03-29T13:38:37Z 2010-03-29T17:56:47Z <p>If you take expectation first with respect to $S_{n-1}$, then by Fubini's theorem the last term gives $$E \left[\frac{|X_n|}{\sqrt{n}}\frac{1}{\lambda} \int_0^1 P\left(z-t\frac{X_n}{\sqrt{n}} \le S_{n-1} \le z-t\frac{X_n}{\sqrt{n}} + \lambda\right) dt\right].$$ Now if $Y$ is a standard Gaussian random variable and $a\in \mathbb{R}$, then $$P(a\le S_{n-1} \le a+\lambda) \le P(a\le Y \le a+\lambda) + 2\delta(\gamma,n-1) \le \frac{\lambda}{\sqrt{2\pi}} + 2\delta(\gamma,n-1),$$ so the expectation above is bounded by $\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{2\pi}}+2\frac{\delta(\gamma,n-1)}{\lambda}\right)$.</p>