Timeline for Can deleting a random entry from an iid sequence destroy the iid property?
Current License: CC BY-SA 4.0
11 events
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| Oct 3, 2022 at 23:56 | vote | accept | Andras Farago | ||
| Oct 3, 2022 at 23:25 | comment | added | Andras Farago | @IosifPinelis Yes, you are right, the positive dependence indeed stays for the 0-1 case, for every $n$. But as $n$ grows large, this dependence appears to be vanishing, Thus, we may say that for large $n$ if we delete a random 1 from an iid 0-1 sequence, then the leftover remains "almost" iid. I wonder, if there is any case where this vanishing does not happen, even as $n$ grows large. Or could it be true that any elimination rule leaves a leftover sequence which approaches an iid one as $n$ grows? Of course, one would have to precisely define what "approaches" means. Also, how fast is it? | |
| Oct 3, 2022 at 21:18 | comment | added | Iosif Pinelis | @AndrasFarago : That for the 0-1 case the maximum is almost always $1$ does not contradict the positive-dependence heuristics. Indeed, the straight $0$'s are relatively more likely after the removal of $X_\nu$ than before the removal. Have you checked the case $n=3$, as I suggested? I did, and heuristics does work in that case. I believe it will work for all $n\ge3$, but the (positive) dependence for large $n$ will be small. Also, as I said, the expressions for the 0-1 case are messy. Do you want to insist on me working with them nonetheless? | |
| Oct 3, 2022 at 21:06 | comment | added | Iosif Pinelis | @StanleyYaoXiao : Thank you for your comment. | |
| Oct 3, 2022 at 19:12 | comment | added | Andras Farago | @IosifPinelis The intuitive argument at the end of your answer does not seem to work for the 0-1 case, because the maximum is almost always the largest possible value (1). The only exception is when all terms are 0, but if the entries take value 1, say, with probability 1/2, and $n$ is large, then this happens with exponentially small probability. We may say that let's start with a conditionally iid sequence, where the condition is that not all terms are 0. Is it true that leaving out a random 1, the leftover will not be iid even in this case? | |
| Oct 3, 2022 at 18:44 | comment | added | Stanley Yao Xiao | I love this question and answer! Very interesting! | |
| Oct 3, 2022 at 14:47 | comment | added | Iosif Pinelis | @AndrasFarago : Yes, of course. However, the expressions there are messy, because of ties, as the index $\nu$ of the maximum is not unique in that case. In that example, I suggest you consider $Y_1$ and $Y_2$ for $n=3$, which is quite elementary. | |
| Oct 3, 2022 at 14:31 | comment | added | Andras Farago | Nice example! Could the argument be carried over to the example mentioned in the last paragraph of the original question? | |
| Oct 3, 2022 at 14:11 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 3, 2022 at 13:53 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 3, 2022 at 13:35 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |