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Mar 9, 2016 at 13:25 vote accept Rob
Mar 8, 2016 at 16:30 answer added Robert Israel timeline score: 2
Mar 8, 2016 at 13:01 comment added Rob @FedericoPoloni Yes! :-)
Mar 8, 2016 at 12:43 comment added Federico Poloni @Rob Same comment: so, essentially, you want to find the distribution of the entries of $X^{-1}Y$? That is what the coefficients of that linear combinations are.
Mar 8, 2016 at 12:27 comment added Rob @FedericoPoloni I changed the question so that it should be much clearer now what I would really like to know.
Mar 8, 2016 at 12:24 comment added Rob @AnthonyQuas Thanks! I must confess my question was not very accurate. I changed it. In addition I specified the distribution as "absolutely". Is it actually the same as assuming $\mathbb{P}(X=x)=0$ $\forall x$ for the distribution?
Mar 8, 2016 at 12:15 history edited Rob CC BY-SA 3.0
added "absolutely" and modified the question
Mar 7, 2016 at 18:59 answer added Mark Fischler timeline score: 0
Mar 7, 2016 at 18:29 comment added Federico Poloni So, essentially, find the distribution of $X^{-1}y$, where $X$ is obtained by stacking the $x_i$'s horizontally?
Mar 7, 2016 at 18:00 comment added Anthony Quas Are you asking whether the random variables $(a_i)$ are iid? No definitely not (but they are exchangeable). Also, you said that every set of $n$ vectors from a continuous distribution is a.s. linearly independent. Is this a hypothesis? In general it's not true as the one-dimensional distribution along a line is continuous. (It is true if you say absolutely continuous distribution). To get a counter-example to your iid claim, you can consider distributions that are continuous, but a small perturbation of a discrete distribution.
Mar 7, 2016 at 16:59 history asked Rob CC BY-SA 3.0