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In a proof, I saw the use of the following inequality

$(Y_{n}-a)_{+}\leq (Y_{n})_{+}+\lvert a\rvert(*)$

without any explanation, where $Y_{n}$ is some random variable and $a$ a constant. Note the definition

$(X)_{+}:=\max\{0,X\}$.

I am aware that $(\cdot)_{+}$ as a function is subadditive, but the problem in $(*)$ is that I have a minus rather than a plus, so subadditivity cannot be used directly right?

But rather I can use the monotony of $(\cdot)_{+}$ since clearly $a\leq \lvert a \rvert$ and thus

$Y_{n}-a\leq Y_{n}+\lvert a\rvert$

such that $(Y_{n}-a)_{+}\leq (Y_{n}+\lvert a\rvert)_{+}$. Now I have an upper bound where I can use subadditivity and thus

$(Y_{n}+\lvert a\rvert)_{+}\leq (Y_{n})_{+}+ (\lvert a\rvert)_{+}=(Y_{n})_{+}+ \lvert a\rvert$.

Is my proof/thinking correct? Or is there a more general way to go about this when dealing with $(\cdot)_{+}$?

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  • $\begingroup$ Yes, your proof looks just fine to me. $\endgroup$
    – Jimmy R.
    Commented Jun 30, 2020 at 8:26

1 Answer 1

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Notice that \begin{align} Y_n -a = (Y_n)_+ - (Y_n)_- - a \leqslant (Y_n)_+ - a \end{align} because $(Y_n)_-$ is non negative. Then, triangle inequality says that \begin{align} (Y_n)_+ - a \leqslant \left|(Y_n)_+ - a \right| \leqslant |(Y_n)_+| + |a| \end{align} Remark that $|(Y_n)_+| = (Y_n)_+$ as it is non negative.

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