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

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 $$(\cdot)_{+}$$?

• Yes, your proof looks just fine to me. Commented Jun 30, 2020 at 8:26

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.