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Suppose that for each $n\geq1$ there is a given sequence of independent random variables $\xi_{n1},\xi_{n2},\ldots,\xi_{nn}$ with $E\xi_{nk}=0$, $V\xi_{nk}=\sigma_{nk}^2$, $\sum_{k=1}^n\sigma_{nk}^2=1$. Let $S_n=\xi_{n1}+\ldots+\xi_{nn},$ $F_{nk}(x)=P\{\xi_{nk}\leq x\},\ \Phi(x)=(2\pi)^{-1/2}\int_{-\infty}^x e^{-y^2/2}dy,\ \Phi_{nk}(x)=\Phi\left(\frac{x}{\sigma_{nk}}\right)$ On page 339 there is an inequality: $\sum_{k=1}^n\left| t\int_{-\infty}^{\infty}(e^{itx}-1-itx)(F_{nk}(x)-\Phi_{nk}(x))dx\right|\leq\frac{|t|^3}{2}\varepsilon\sum_{k=1}^n\int_{|x|\leq\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx+2t^2\sum_{k=1}^n\int_{|x|>\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx$ How can I get it?

Then the book says we can use $E|\xi|^n=\int_{-\infty}^{\infty}|x|^ndF(x)=n\int_0^\infty x^{n-1}[1-F(x)+F(-x)]dx$ to prove $\int_{|x|\leq\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx\leq2\sigma_{nk}^2$ But I can't figure why.

Thank you!

2 Answers 2

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For the second question, rewrite $ \mathbb E|\xi|^n=\int_{-\infty}^{\infty}|x|^n\mathrm dF(x)=\int_{0}^{\infty}x^n\mathrm dF(x)+ \int_{0}^{\infty}x^n\mathrm dG(x), $ with $G:x\mapsto-F(-x)$ and use an integration by parts formula to compute $ \int_0^{+\infty}u(x)\mathrm dv(x),\qquad u:x\mapsto x^n,\quad v=F+G-1.$ To use this identity, note that for any random variables $\zeta_1$ and $\zeta_2$ with PDF $G_1$ and $G_2$, $ \int_{-\infty}^{+\infty}|x|\,|G_1(x)-G_2(x)|\mathrm dx\leqslant(*), $ with $ (*)=\int_{-\infty}^0(-x)\,(G_1(x)+G_2(x))\mathrm dx+\int_0^{+\infty}x\,(1-G_1(x)+1-G_2(x))\mathrm dx. $ One sees that $(*)=A(G_1)+A(G_2)$, with $ A(G)=\int_0^{+\infty}x\,(1-G(x)+G(-x))\mathrm dx. $ Applying the first identity one gets $A(G)=\frac12\mathbb E(\zeta^2)$ for every random variable $\zeta$ with PDF $G$, hence $(*)=\frac12(\mathbb E(\zeta_1^2)+\mathbb E(\zeta_2^2))$. In particular, the inequality mentioned in the post holds, without the restriction to $|x|\leqslant\varepsilon$ (this was expected) and without the factor $2$ on the RHS (this was less expected).

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    Thank you very much! And how can I use the equation you proved$E|\xi|^n=\int_{-\infty}^{\infty}|x|^ndF(x)=n\int_0^\infty x^{n-1}[1-F(x)+F(-x)]dx$to prove$\int_{|x|\leq\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx\leq2\sigma_{nk}^2?$2012-12-30
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We have that $|e^{itx}-1-itx|\leqslant \frac{|tx|^2}2$, which can be obtained by Taylor's formula, and this gives an upper bound for $\int_{-\varepsilon}^{\varepsilon}$.

For the other part, note that $|e^{itx}-1-itx|\leqslant|e^{itx}-1|+|tx|\leqslant\left|\int_0^{|tx|}ie^{is}ds\right|+|tx|\leqslant 2|tx|,$ and we can control $\int_{\{|x|>\varepsilon\}}$.

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    And how about my second question? I still got no idea. Thank you very much!2012-12-27