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Jan 23, 2021 at 3:05 history bounty ended King Kong
Jan 18, 2021 at 20:26 comment added King Kong that makes sense, thanks!
Jan 18, 2021 at 20:14 comment added Mateusz Kwaśnicki @KingKong: This is correct. One thing that I am afraid is that one can craft $P$ and $t_P$ in such a way that we necessarily have $t_{P^*} = t_P$. In this case the kind of "iterative" method might fail to finish in finite time. Also, this seems to be extremely inefficient. Finally, people know a lot about extensions of the linear programming problem, I suspect a more efficient approach is feasible (in case you care about time complexity of the solution).
Jan 18, 2021 at 14:50 comment added King Kong @MateuszKwaśnicki. The conditions you describe do accurately capture the problem, but I just wanted to clarify about the last paragraph. I only need to find some t* that `works' (for which we can find a probability distribution that satisfies the conditions). So is this not just a matter of iterating the linear procedure through different values of t*, starting just above 2/3?
Jan 16, 2021 at 17:42 comment added Mateusz Kwaśnicki @KingKong: That is just an abuse of the notation for an indicator function: $\mathbb 1_{A_i \subset X \cap Y}$ is equal to one if $A_i \subset X \cap Y$ and zero otherwise. Perhaps I should have used the Iverson bracket for that and simply write $[A_i \subset X \cap Y]$.
Jan 16, 2021 at 17:37 comment added King Kong @MateuszKwaśnicki, that's correct, $P^{*}$ should be probabilistic, which means that the probabilities of the atoms need to add to 1.
Jan 16, 2021 at 17:36 comment added King Kong Thanks! I've added a linear programming tag. Can I ask what the 𝟙𝐴𝑖⊂𝑋∩𝑌 notation denotes?
Jan 16, 2021 at 8:29 comment added Mateusz Kwaśnicki I think so — if $P^*$ is supposed to be a probability measure.
Jan 16, 2021 at 4:55 comment added Bill Bradley Does the probability of the atoms have to sum to 1?
Jan 15, 2021 at 19:58 history answered Mateusz Kwaśnicki CC BY-SA 4.0