# why in this chance constraint we have convexity?

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?

• 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 6:36
• 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
• it follows from math.stackexchange.com/questions/2280341/… – LinAlg Jun 12 '18 at 15:07