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I have this integral: $\int(\epsilon \wedge \|x\|^{2})\nu(dx)$

The $\wedge$ symbol means that I have to integrate $\|x\|^{2}$ when $\|x\|>\epsilon$ or when $\|x\|>1$?

Arnaud Mortier
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MkD
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  • Related: https://math.stackexchange.com/questions/1839344/why-is-wedge-a-minimum-and-vee-a-maximum – Arnaud Mortier Mar 08 '19 at 09:21
  • Simply put, it refers to the "minimum", i.e. $a\land b = \min(a,b)$. – Minus One-Twelfth Mar 08 '19 at 09:22
  • It is a notation borrowed from Lattice theory, occasionally used in some approaches to measure theory: specifically $$\begin{split}\wedge\equiv\text{ "meet"}&\implies a\wedge b=\min(a,b)\ \vee\equiv\text{ "join"},,&\implies a\vee b=\max(a,b)\end{split}\text{ if }a,b\in\Bbb R.$$ – Daniele Tampieri Mar 08 '19 at 09:46

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The wedge stands for minimum. $a\wedge b=\min\{a,b\}$. The integral is $\int_{\{\|x||^{2} \geq \epsilon\}} \epsilon \nu (dx)+\int_{\{\|x||^{2} < \epsilon\}} \|x\|^{2} \nu (dx)$.

Kavi Rama Murthy
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