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Jul 29, 2014 at 22:01 history edited Ricardo Andrade CC BY-SA 3.0
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Jan 8, 2013 at 2:36 vote accept gappy3000
Aug 20, 2011 at 22:34 answer added Did timeline score: 7
Aug 20, 2011 at 15:21 answer added Brendan McKay timeline score: 6
Sep 24, 2010 at 16:40 comment added Suvrit Is your integral by any chance related to the Bingham distribution?
Sep 10, 2010 at 19:21 comment added Will Jagy Thanks, Didier. No sorrow necessary.
Sep 10, 2010 at 19:02 comment added Did Will, the formula you wrote above is the natural generalization of the formula for $n=2$. Unfortunately the trick I used to prove the $n=2$ case fails for larger values of $n$ and since, at this moment, I see no "geometrical" reason (whatever that means) pointing to the $\ell^1$ norm, the answer to your question is: I do not know. Sorry.
Sep 10, 2010 at 17:15 comment added Will Jagy Didier, if we demand all $ \lambda_j > 0$ and put $ \lambda_i^2$ in the numerator to make the sum to 1 obvious, do you think $$ E\left( \frac{ \lambda_i^2 X_i^2}{\sum_j \lambda_j^2 X_j^2}\right) = \frac{ \lambda_i}{\sum_j \lambda_j}? $$
Sep 10, 2010 at 16:12 comment added Did Funny: for two random variables $E(X^2_1/(\lambda^2_1X^2_1+\lambda^2_2X_2^2))=1/(|\lambda_1|(|\lambda_1|+|\lambda_2|))$. I wonder why the $\ell^1$ norm enters the picture here...
Sep 10, 2010 at 9:05 answer added Darsh Ranjan timeline score: 1
Sep 10, 2010 at 0:37 comment added Will Jagy Do you have some part answers, say estimates or upper or lower bounds, with only two variables but unequal $\lambda_j$? For instance $\lambda_1=1$ but $\lambda_2$ very large, what happens approximately?
Sep 9, 2010 at 22:00 history edited gappy3000 CC BY-SA 2.5
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Sep 9, 2010 at 16:51 history asked gappy3000 CC BY-SA 2.5