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What are $y^+$ and $y^-$ supposed to represent? $y$ is a vector.

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    Hard to say unless you give us some further context.2011-03-22

1 Answers 1

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As lhf points out, it depends on what context you are using these.

One possible answer which I have seen is $y^{+}$ is a vector whose elements are same as $y$ if the corresponding element in $y$ is positive and is zero if the corresponding element in $y$ is negative. Similarly, $y^{-}$ is a vector whose elements are same as $y$ if the corresponding element in $y$ is negative and is zero if the corresponding element in $y$ is positive.i.e.

If $y \in \mathbb{R}^{n \times 1}$, then

$y^{+} \in \mathbb{R}^{n \times 1}$ with $y^{+}(i) = \max(y(i),0)$

and

$y^{-} \in \mathbb{R}^{n \times 1}$ with $y^{-}(i) = \min(y(i),0)$

Hence, $y(i) = y^{+}(i) + y^{-}(i)$

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    @Matt: Yes. Some people use that convention as well i.e. $y^{-}(i) = \max(-y(i),0)$ so that both $y^{+}$ and $y^{-}$ are non-negative and then we get $y(i) = y^{+}(i) - y^{-}(i)$2011-03-22