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Let $X_n$ be a sequence of $iid$ random variables with zero mean, finite variance $\sigma^2$ and partial sum $S_n:=\sum_{k=1}^n X_k$. Let $0<\delta<0.5$ and $\epsilon >0$ and define $c_n:=\frac{2\delta\sigma}{\epsilon\sqrt{\log\log n}}$ and $b_n:=c_n\sigma\sqrt{n}$. In addition, for each $n\geq 1$ and $1\leq j \leq n$ define the truncated random variables as follows. $$X'_{n, j}:= X_j\mathbb I_{\{|X_j| < 0.5b_n\}}; \ X''_{n, j}:= X_j\mathbb I_{\{|X_j| > \sqrt{n}\}}; \ X'''_{n, j}:= X_j\mathbb I_{\{0.5b_n\leq|X_j| \leq \sqrt{n}\}}.$$ Then to bound these truncated random variables, I read the following lines.

$$ \left|\mathbb E X'_{n, j}\right| = \left|\mathbb E X_j\mathbb I_{\{|X_j| < 0.5b_n\}}\right| = \left|-\mathbb EX_j \mathbb I_{\{|X_j|\geq 0.5b_n\}}\right|\leq \mathbb E |X_j|\mathbb I_{\{|X_j|\geq 0.5b_n\}} \leq \frac{2}{b_n} \mathbb EX^2 \mathbb I_{\{|X_j|\geq 0.5b_n\}}. $$

I do not know how we could get the second equality and the last inequality above. Could anyone point it out, please? Thank you!

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Second equality is due to the condition that $X_n$ are zero mean.

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First inequality: $|E(Z)|\leqslant E(|Z|)$.

Second inequality: For every positive $z$, $|Z|\mathbf 1_{|Z|\geqslant z}\leqslant z^{-1}\,Z^2\mathbf 1_{|Z|\geqslant z}$ with full probability, hence $E(|Z|\mathbf 1_{|Z|\geqslant z})\leqslant z^{-1}\,E(Z^2\mathbf 1_{|Z|\geqslant z})$.

Some other equality: If $E(Z)=0$, then $E(Z\mathbf 1_{Z\lt z})+E(Z\mathbf 1_{Z\geqslant z})=0$ hence $|E(Z\mathbf 1_{Z\lt z})|=|E(Z\mathbf 1_{Z\geqslant z})|=|-E(Z\mathbf 1_{Z\geqslant z})|$.

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  • $\begingroup$ Thanks. But I mean the second equality. Any hint, please? $\endgroup$ – LaTeXFan Oct 5 '14 at 23:13

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