why in this chance constraint we have convexity?

Asked Jun 11 '18 at 12:22

Active Jun 11 '18 at 12:22

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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?

asked Jun 11 '18 at 12:22

yaodao vang

2

Because $F^{-1}(\alpha) > 0$ for $\alpha > \frac{1}{2}$. – madnessweasley Jun 11 '18 at 13:41
@madnessweasley I know $F^{-1}(\alpha)>0 if \alpha >0.5 $ , I do not know the convexity of this $|\Sigma^{1/2}x||_2 $ – yaodao vang Jun 12 '18 at 06:36
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Read a proof as to why second order cones are convex. – LinAlg Jun 12 '18 at 12:32
@ LinAlg thanks for your help – yaodao vang Jun 12 '18 at 14:08
@LinAlg i searched " proof second order cones are convex." but i cant find anything , can you give me a link , site or book ,? – yaodao vang Jun 12 '18 at 14:20
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it follows from https://math.stackexchange.com/questions/2280341/why-is-every-p-norm-convex – LinAlg Jun 12 '18 at 15:07
@LinAlg thank you , so if $\alpha >0.5$ and we have every norm in $ ||\Sigma ^{0.5 } x||_p$
We have convexity too .
– yaodao vang Jun 14 '18 at 05:04

why in this chance constraint we have convexity?

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