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How would one proceed with proving that there exists none $(x,y,z)$ such that all the following hold :

$$x+2y+7z = 0$$ $$\mathbb{P}(x + 3y + 9z \geq 0) = 1$$ $$\mathbb{P}(x + y + 5z \geq 0) = 1$$ $$\mathbb{P}(x + 13y + 10z \geq 0) = 1$$ $$\mathbb{P}(x + 3y + 9z >0 ) >0$$ $$\mathbb{P}(x + y + 5z > 0) >0 $$ $$\mathbb{P}(x + 13y + 10z > 0) >0$$

This is a "tool" that I need as a step to prove a case in a Mathematical Finance exercise.

Any ideas will be really appreciated.

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    $\begingroup$ It looks like $(x,y,z)=(1,3,-1)$ satisfies all the conditions $\endgroup$ – WW1 Nov 13 '18 at 0:48
  • $\begingroup$ @WW1 Thanks for the comment, there was actually a typo. Updated the question now ! $\endgroup$ – Rebellos Nov 13 '18 at 18:09
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$x=y=z=0$ is a solution to the first 4 constraints That means that there exists $a,b,c\ge 0$ such that $$\begin{align}x+2y+7z&=0\\x+3y+9z&=a\\x+y+5z&=b\\x+13y+10z&=c\end{align}$$ You must prove that the equations are not linearly dependent, so the only solution is $x=y=z=0$ for $a=b=c=0$. But then the last three inequalities are not obeyed (strict inequality)

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  • $\begingroup$ Updated the question now, as the second per inequality equation is : $$x+y+5z$$ and the third one $$x+13y+10z$$ I apologize for that typo! Does the same approach still hold ? $\endgroup$ – Rebellos Nov 13 '18 at 18:11
  • $\begingroup$ Yes. You must prove that you have solution only if $a\cdot b\cdot c=0$ $\endgroup$ – Andrei Nov 13 '18 at 18:19
  • $\begingroup$ I elaborated your given answer using a Gaussian Elimination approach and yielded the proper result. Thanks a lot for your help and sorry for accepting it late ! $\endgroup$ – Rebellos Nov 17 '18 at 13:20
  • $\begingroup$ No problem. Glad that I could help. $\endgroup$ – Andrei Nov 17 '18 at 13:37

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