Timeline for Techniques/diagnostics for gaining confidence in normality assumptions and resulting confidence intervals
Current License: CC BY-SA 4.0
23 events
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| Jul 13, 2023 at 20:17 | comment | added | Ute | Good to build up confidence by simulations! Hope you really challenged the bootstrap method. It has become quite trusted - and there are proofs. They are asymptotic, for sample size towards $\infty$. That is the same for many classical methods that build upon asymptotic distribution, such as $\chi^2$ tests. | |
| Jul 13, 2023 at 18:56 | comment | added | travelingbones | I'm unsure of the rigorous statement or proof, but the theorem to be proved that drives this seems to be " if $n$ is sufficiently large, the bootstrapped sample statistics' empirical distribution is approximately the sample statistic's distribution." I have runs some simulations to verify this. pretty awesome that this works our. Hoping for a rigorous statement and a breadcrumbs of the proof. | |
| Jul 13, 2023 at 18:20 | comment | added | Ute | Oh yes, it is a bit handwavy, the bootstrap. Like all inferential statistics: trying to say something about the population. The models that classical statistics uses are in that sense also approximarions only. Your new question is well narrowed down :-) . Let's see if some good explanations come in - I might also take time and write in the next days. | |
| Jul 13, 2023 at 18:07 | comment | added | travelingbones | Thank you for the tips. I've read the Wikipedia pages and other refs. on bootstrapping. I'm still not understanding why this diagnostic procedure works. It is clear that if $n$ is large enough, then the bootstrapped sample means will appear N(\mu, sigma^2/n). What this answer puts forth is the converse, which begs: if the bootstrapped sample means appear normal, then why do we know that we have large enough $n$? I've formulated this as a separate question: stats.stackexchange.com/questions/621332/… | |
| S Jun 21, 2023 at 5:21 | history | suggested | travelingbones | CC BY-SA 4.0 |
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| S Jun 20, 2023 at 0:53 | history | suggested | travelingbones | CC BY-SA 4.0 |
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| Jun 20, 2023 at 0:47 | review | Suggested edits | |||
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| Jun 20, 2023 at 0:44 | comment | added | Ute | Try to google "bootstrap method" or "bootstrapping" :-) | |
| Jun 20, 2023 at 0:37 | vote | accept | travelingbones | ||
| Jun 20, 2023 at 0:36 | comment | added | travelingbones | Thank you--but why? Really what I'm looking for is the theory under pinning this technique. Can you point me to some treatment of it? E.g., it is easy to prove the subsamples with replacement are IID from the original dataset, but these nuanced details (like the sizes and the iterations done), understanding how they affect things would be helpful in use and for gaining intuition. | |
| Jun 20, 2023 at 0:32 | comment | added | Ute | size has to be equal to the original sample size. It is a feature that some values are multiply included. | |
| Jun 20, 2023 at 0:25 | comment | added | travelingbones |
Does it matter if size = n (the size of the subsample) is equal to/less than/larger than the size of the sample? Seems like it should be smaller than the size of the sample else you're going to get some values included multiple times guaranteed.
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| Jun 20, 2023 at 0:17 | comment | added | Ute | You should usually be fine with 10000 subsamples - if this does not take too long. Run the thing a couple of times and check if you see big differences between the runs. If not, the number of subsamples is enough. | |
| Jun 20, 2023 at 0:10 | comment | added | travelingbones | How do you know how large to choose: - how many subsamples with replacement to use when computing a sample mean (size = n)? - how many times to compute a sample mean using subsamples with replacement (nboot)? | |
| Jun 19, 2023 at 22:03 | comment | added | travelingbones | Ok, just tried something similar. It basically shrinks the original distribution in the x-direction but reproduces it--still tri modal. Makes sense that these $\bar{X}_{n-1}(i) $ are not IID, as all pairs share $n-1$ variables. | |
| Jun 5, 2023 at 14:53 | comment | added | Ute | Tx :-) - Did you try it out? your suggestion sounds very reasonable, much more systematic. But it is in a way "too systematic" - it just reproduces exactly what you see in your sample. CLT concerns all possible samples. Bootstrap tries to reflect all possible samples as far as possible. This implies in particular, that the items are sampled independently, which is not the case when you leave one out, but it is the case, when you sample with replacement. (To be precise: conditionally independent, given the observed sample). | |
| Jun 5, 2023 at 14:48 | comment | added | travelingbones | Why are you suggesting bootstrapping with replacement, as opposed to sampling without replacement? One could consider my original suggestion (leave one out) as sampling without replacement. I'm guessing this help with the "near constant" potential problems you suggested. | |
| Jun 5, 2023 at 14:46 | comment | added | travelingbones | awesome explanation. I'll try it out. | |
| Jun 3, 2023 at 11:36 | history | edited | Ute | CC BY-SA 4.0 |
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| Jun 3, 2023 at 7:14 | history | edited | Ute | CC BY-SA 4.0 |
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| Jun 3, 2023 at 1:15 | history | edited | Ute | CC BY-SA 4.0 |
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| Jun 3, 2023 at 1:06 | history | answered | Ute | CC BY-SA 4.0 |