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I saw this asked on glassdoor as an interview question:

How do you use three 1-D random variables to simulate 3-D random variables with a given covariance structure?

Is this a well-posed question? I do not understand the question the question in its current state, but I am not well-versed in this area so I'm wondering I'm misunderstanding something.

roulette01
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  • Maybe they mean "you have three independent, square integrable random variables $X,Y,Z$; how can you transform them to get three random variables $A,B,C$ such that $\operatorname{Cov}(A,B)=a,\operatorname{Cov}(A,C)=b, \operatorname{Cov}(B,C)=c$ for given $a,b,c$ ?" – Maximilian Janisch Sep 07 '20 at 14:49
  • @MaximilianJanisch Ah that could be. Is there a general case for this for any distribution of $X,Y,Z$? I believe if $X,Y,Z$ are standard normal, then I can use Cholesky to find A,B,C with that specified covariance structure, but I don't recall this being applicable to non-normal $X,Y,Z$. – roulette01 Sep 07 '20 at 14:56

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