Timeline for How to sample exactly k indices given the inclusion probabilities of all indices?
Current License: CC BY-SA 4.0
6 events
when toggle format | what | by | license | comment | |
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Jul 23 at 10:31 | vote | accept | Daniel Soudry | ||
Jul 23 at 10:31 | comment | added | Daniel Soudry | 1. Ah, I see, I thought you meant 2 hits for the same $\alpha$ sample. 2. Yeah, I'm not sure yet myself if I need this. Thanks again, I'll approve this nice answer. | |
Jul 23 at 9:24 | comment | added | Karl Fabian | If one of your sectors overlaps with e.g. $\zeta_1$ it is hit by pointer 1 for $\alpha$ close to 0, and by pointer 0 for $\alpha$ close to $2\pi/k$. You are right with the further subdivision, this requires care to make sure that the sectors for each index e.g. lie within a sector of size $2\pi/k$. To answer your question this is not necessary, but I guess you also like to have no unnecessary correlations between the indices. If you do not care, you can leave out the permutation. Maybe permutation is already sufficient to destroy correlation, but I could not prove this so far. | |
Jul 23 at 9:08 | comment | added | Daniel Soudry | Thanks, I'm verifying this now. Two questions: 1. How can a sector be reached by two pointers? The pointer difference is $2\pi / k $ is larger than the sector size $2\pi p_i / k$ since $p_i<1$. So this implies that a sector can be reached only by one pointer, right? 2. If we further sub-divide the sectors and permute (like you suggested in the last sentence), can't this cause several identical indices to be sampled? | |
Jul 23 at 8:22 | history | edited | Karl Fabian | CC BY-SA 4.0 |
Added remark on remaining correlations.
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Jul 23 at 8:16 | history | answered | Karl Fabian | CC BY-SA 4.0 |