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I havea chance constraint of the form $$ \mathbb{P}[a^\text{T}x\leqslant b]\geqslant\alpha$$ where $b\in\mathbb{R}$ is fixed, $a\in\mathbb{R}^n$ is a vector whose entries are Independent and identically distributed , and normally distributed with mean $\overline{a}$ and variance $\Sigma$

(that is, $a\sim\mathcal{N}(\overline{a},\Sigma)$), and $\alpha>1/2$ . It is well known that this constraint is equivalent to $$ F^{-1}(\alpha)\|\Sigma^{1/2}x\|_2\leqslant-\overline{a}^\text{T}x+b $$

why if $\alpha>1/2$ , then $\ \ F^{-1}(\alpha)\|\Sigma^{1/2}x\|_2$ is a convex?

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    $\begingroup$ Because $F^{-1}(\alpha) > 0$ for $\alpha > \frac{1}{2}$. $\endgroup$ – madnessweasley Jun 11 '18 at 13:41
  • $\begingroup$ @madnessweasley I know $F^{-1}(\alpha)\>0 if \alpha >0.5 $ , I do not know the convexity of this $|\Sigma^{1/2}x\||_2 $ $\endgroup$ – yaodao vang Jun 12 '18 at 6:36
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    $\begingroup$ Read a proof as to why second order cones are convex. $\endgroup$ – LinAlg Jun 12 '18 at 12:32
  • $\begingroup$ @ LinAlg thanks for your help $\endgroup$ – yaodao vang Jun 12 '18 at 14:08
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    $\begingroup$ it follows from math.stackexchange.com/questions/2280341/… $\endgroup$ – LinAlg Jun 12 '18 at 15:07

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