Timeline for Why does this test do not follow a chi-square distribution?
Current License: CC BY-SA 4.0
10 events
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| Sep 3, 2022 at 9:39 | history | edited | Gordon Smyth | CC BY-SA 4.0 |
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| Aug 29, 2022 at 9:49 | comment | added | TUC1685 | Yeah. This seems a non-trivial problem. I will keep trying and working on my own. Thanks! | |
| Aug 29, 2022 at 8:57 | comment | added | Gordon Smyth | @TUC1685 I have removed my remark about $n$. I believe that your intended statistic is in fact the score test statistic that is well known in asymptotic statistics, but with observed information substituted for expected information. It is hard to be sure given the ambiguity of your notation in which you don't distinguish between sample averages and population expectations. The chisquare approximation will be less good for such a statistic than it would be for the regular score statistic. | |
| Aug 29, 2022 at 8:45 | comment | added | Gordon Smyth | @TUC1685 The symbol $\mathbb{E}$ is universally understood to mean population average (i.e., expectation), so it is not reasonable to use it to mean sample average. You will be misunderstood by everyone. | |
| Aug 29, 2022 at 8:30 | history | edited | Gordon Smyth | CC BY-SA 4.0 |
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| Aug 29, 2022 at 8:09 | comment | added | TUC1685 | Thanks for your comments, Gordon. I edited some things in the question. When I wrote $\mathbb{E}$, I actually meant the sample average. This is not a score test statistic, it follows a more general principle, which explains the $n$ in the numerator. Finally, it is true the second point you made and I am working on it as well. | |
| Aug 29, 2022 at 5:47 | history | edited | Gordon Smyth | CC BY-SA 4.0 |
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| Aug 29, 2022 at 4:46 | history | edited | Gordon Smyth | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:14 | history | edited | Gordon Smyth | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:08 | history | answered | Gordon Smyth | CC BY-SA 4.0 |