Question

The relevant paper is "An estimate of the remainder in a combinatorial central limit theorem" by Erwin Bolthausen. I would like to understand the estimate on page three right before the sentence "where we used independence of $S_{n-1}$ and $X_n$":

$$\begin{align}E|f'(S_n) - f'(S_{n-1})| &\le E \bigg(\frac{|X_n|}{\sqrt{n}} \big(1 + 2|S_{n-1}| + \frac{1}{\lambda} \int_0^1 1_{[z,z+\lambda]} (S_{n-1} + t \frac{X_n}{ \sqrt{n}}) dt\big)\bigg) \\ &\le \frac{C}{\sqrt{n}} \big(1 + \delta(\gamma, n-1) / \lambda\big)\end{align}$$

that is, where $\delta(\gamma, n-1)/\lambda$ shows up, which is the error term in the Berry-Esséen bound.

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

$ \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 }$

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 $h$.

This paper is freely available online through springer. thanks in advance.

Answer 1

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 $\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{2\pi}}+2\frac{\delta(\gamma,n-1)}{\lambda}\right)$.