Estimate of closeness of distributions of sums to normal distribution (Q2759394)

Let \(\xi_1,\ldots,\xi_{n},\ldots\) be a sequence of independent random variables with \(M\xi_{i}=0,\;D\xi_{i}=\sigma_{i}^2<+\infty\), distribution functions \(F_{i}(x)\). Let \(\Phi(x)\) be a distribution function of standard normal law, \(\Phi_{n}(x)\) is a distribution function of random variable \((\xi_1+\ldots+\xi_{n})/B_{n}\), where \(B^2_{n}=\sum_{i=1}^{n}\sigma^2_{i}\). The authors prove the following results: If NEWLINE\[NEWLINE\lambda_{in}^{(1)}=\int\limits_{|x|\leq B_{n}}\max(1,|x|^3)|dH_{i}(x)|,\;\lambda_{in}^{(2)}=\int\limits_{|x|> B_{n}} x^3|dH_{i}(x)|,\;\text{where} H_{i}(x)=F_{i}(x)-\Phi\left({x\over \sigma_{i}}\right),NEWLINE\]NEWLINE NEWLINE\[NEWLINE\lambda_{n}^{(1)}=\max\limits_{1\leq i\leq n}\lambda_{in}^{(1)},\;\lambda_{n}^{(2)}=\max\limits_{1\leq i\leq n}\lambda_{in}^{(2)},\;\lambda_{n}=\sigma^{-2}\left({1\over\sigma\sqrt{n}}\lambda_{n}^{(1)}+\lambda_{n}^{(2)}\right),\;\sigma=\min(1,\sigma_1,\ldots,\sigma_{n}),NEWLINE\]NEWLINE then there exists a constant \(C\) such that for all \(n\geq 1\): \(\rho_{n}=\sup_{x}|\Phi_{n}(x)-\Phi(x)|\leq\lambda_{n}C\). NEWLINENEWLINENEWLINEIf NEWLINE\[NEWLINE\bar\sigma_{i}=\min(1,\sigma_{i}),\;\bar B^2_{n}=\bar\sigma^2_1+\ldots+\bar\sigma_{n}^2,\;b_{n}=\min(\bar\sigma_{1},\ldots,\bar\sigma_{n}),NEWLINE\]NEWLINE NEWLINE\[NEWLINELambda_{n}^{(1)}={1\over B^2_{n}}\sum\limits_{i=1}^{n}\lambda_{in}^{(1)},\;\Lambda_{n}^{(2)}={1\over B^2_{n}}\sum\limits_{i=1}^{n}\lambda_{in}^{(2)},NEWLINE\]NEWLINE then for all \(n\geq 1\): \(\rho_{n}\leq C\left(\bar B^{-1}_{n}\Lambda_{n}^{(1)}+\Lambda_{n}^{(2)}\right)b_{n}^{-3}\), where \(C\) is a constant.