The relevant paper is ["An estimate of the remainder in a combinatorial central limit theorem" by Erwin Bolthausen](https://doi.org/10.1007/BF00533704 "Z. Wahrscheinlichkeitstheorie verw Gebiete 66, 379–386 (1984)"). 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.



Stein's method proof of the Berry–Esséen theorem
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 $\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{2\pi}}+2\frac{\delta(\gamma,n-1)}{\lambda}\right)$.


