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Oct 23, 2012 at 22:59 comment added whuber Now I'm convinced :-).
Oct 23, 2012 at 22:12 comment added gui11aume @whuber actually thinking about it, I stick to my point (see the edit). Did I get something wrong?
Oct 23, 2012 at 22:11 history edited gui11aume CC BY-SA 3.0
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Oct 23, 2012 at 21:01 history edited gui11aume CC BY-SA 3.0
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Oct 23, 2012 at 21:01 comment added gui11aume @whuber absolutely. My bad. By the way, I like your geometric answer (to the duplicate of that question) better :D
Oct 23, 2012 at 20:58 comment added whuber Umm... I'm pretty sure that $T$ has three degrees of freedom, not two. Not that it matters--everything is conditioned on $T$. But maybe I misunderstand the notation. All the work depends on justifying that assertion about the joint distribution.
Oct 23, 2012 at 20:56 history post merged (destination)
Oct 23, 2012 at 20:28 history answered gui11aume CC BY-SA 3.0