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Oct 14, 2017 at 13:19 comment added Mateusz Kwaśnicki I do not quite understand what Henry.L meant in his answer, but certainly one needs to know the joint distribution of $(X,Y)$ in order to describe the distribution of $X - Y$. Note that it is not always possible to even define $(X,Y)$ with given marginals so that $X > Y$ a.s.; namely, one needs to know that $\mathbb{P}(X < t) < \mathbb{P}(Y < t)$ whenever both quantities are in $(0,1)$.
Oct 14, 2017 at 12:08 comment added Muriel Probably now I see your point. Are you saying that if they are not independent then their joint distribution should not be the one given in the solution? I should go back and look at the Akkouchi paper I mentioned, but you are probably right.
Oct 14, 2017 at 11:39 comment added Mateusz Kwaśnicki How they are coupled then, that is, what is the joint distribution of $(X,Y)$?
Oct 14, 2017 at 10:12 vote accept Muriel
Oct 14, 2017 at 10:12
Oct 14, 2017 at 10:10 comment added Muriel I didn't write that they are independent because in my problem they are not. Indeed, the last assumption I make is that $X>Y$ a.s., so they are dependent.
Oct 3, 2017 at 15:47 comment added Mateusz Kwaśnicki This is very unclear: are $X$ and $Y$ independent and we are asking about $\mathbb{P}(X-Y<t|X>t)$, or are $X$ and $Y$ coupled in some way so that $X>Y$ a.s.?
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May 6, 2017 at 12:23 answer added Henry.L timeline score: 1
May 6, 2017 at 12:23 history edited Henry.L
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May 6, 2017 at 11:54 history edited Stefan Kohl
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Feb 7, 2017 at 12:04 history edited Muriel CC BY-SA 3.0
typo in $\beta$ subscripts
Feb 7, 2017 at 12:02 history edited Gerry Myerson CC BY-SA 3.0
typo in title
Feb 7, 2017 at 11:50 review First posts
Feb 7, 2017 at 12:00
Feb 7, 2017 at 11:44 history asked Muriel CC BY-SA 3.0