UzzoloDivendetta

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 Mar3 accepted The Geometry of a Linear Transformation Mar2 comment The Geometry of a Linear Transformation Yes, and actually my question descends from my preliminary survey of svd. But let us ignore svd for a moment. There is some other way to obtain the desired equation without SVD? Mar2 accepted Parallel Algorithms for SVD Mar2 asked The Geometry of a Linear Transformation Mar1 comment Parallel Algorithms for SVD Thanks.. I done some googling but missed this! Feb28 asked Parallel Algorithms for SVD Oct25 comment Set in the Complex Plane I did not tag this as "homework" because it wasn't. I found this on a book and simply found it interesting. Oct25 accepted Set in the Complex Plane Oct24 asked Set in the Complex Plane Oct7 comment A Tale of Urns and Balls I got it this way. Now all is perfectly clear. Thanks again, and pardon me for reopening the argument. Oct7 comment A Tale of Urns and Balls I would prefer to solve this exercise whithout using the notion of expectation. Do you refer to total probability theorem? (Decomposition of an event via intersection with a disjoint union of events)? Thanks. Oct7 comment A Tale of Urns and Balls Beg your pardon, but I rethought to exercise. Let's go to the point where we calculate the conditional probability of drawing a black second, GIVEN the urn is the i-th. Correctly, you multiplicate the probability of i-th urn (1/5), by the probability of black in the same urn. This is fine and similar to calculation of an expectation. But a formal argument would require P(C|Ui) (as I already pointed out), and this is equal to P(C intersect Ui)/P(Ui). How we calculate the probability of the above INTERSECTION? P(Ui|C)P(C) does not help, since we don't know P(Ui|C). Oct6 comment Independence of two (discrete) Random Variables Yes, I noticed this later. Thanks a lot. Oct5 comment Urns, Balls, & Cards. This because Cards do partition the entire sample space, right? So we do slight modification and work this way: A remains the same, but B(i) indicates a card from i-th colour. So, we apply bayes formula, as you done. THANKS! [ you can state your comment as answer, so I can accept it ]. Oct5 revised Urns, Balls, & Cards. added 284 characters in body Oct5 asked Urns, Balls, & Cards. Oct4 accepted Independence of two (discrete) Random Variables Oct4 comment Independence of two (discrete) Random Variables Got it. Many thanks. Naturally, if both are equally distributed, it would suffice to test for a single point, right? Oct4 asked Independence of two (discrete) Random Variables Oct3 comment A Tale of Urns and Balls Ok, I think I have understood. Urns perform the disjoint decomposition of the sample space on the first pick as well as on the second. So we apply the so-called "total probability theorem". The rest is done by Bayes formula. Thanks!