Timeline for Unbiased estimator of the ratio of variances
Current License: CC BY-SA 4.0
21 events
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| Nov 25, 2019 at 23:03 | vote | accept | dff | ||
| Nov 25, 2019 at 18:35 | comment | added | StubbornAtom | If your queries are addressed adequately, 'accept' answers to indicate your problem is solved; this will remove the post from the unanswered queue. | |
| Nov 25, 2019 at 11:53 | comment | added | dff | @StubbornAtom Sorry for bothering you. By using F-distribution, I could derive $$\frac{\sum_{i=1}^{n}(Y_i-\bar{Y})^2/(n-1)}{\sum_{i=1}^{m}(X_i-\bar{X})^2/(m-3)}$$is what I wanted to know. Thank you for helping me! | |
| Nov 22, 2019 at 19:16 | comment | added | StubbornAtom | @watA It is not clear to me if you have any difficulty following the argument in my answer. Nevertheless I added one more step. | |
| Nov 22, 2019 at 8:58 | comment | added | dff | @StubbornAtom Thank you for advice. I updated the post. | |
| Nov 22, 2019 at 8:57 | history | edited | dff | CC BY-SA 4.0 |
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| Nov 20, 2019 at 1:51 | comment | added | dff | I found a book that shows the unbiased estimator of the ratio of variances in the same condition as I wrote above. The book says, $$\frac{\sum_{i=1}^{n}(Y_i-\bar{Y})^2/(n-1)}{\sum_{i=1}^{m}(X_i-\bar{X})^2/(m+1)}$$ is the unbiased estimator that I want to know. But I can't show its unbiassedness. I tried to use Jensen's inequality, but my friend pointed out it's not effective. | |
| S Nov 18, 2019 at 19:01 | history | suggested | Steffen Moritz | CC BY-SA 4.0 |
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| Nov 18, 2019 at 17:50 | review | Suggested edits | |||
| S Nov 18, 2019 at 19:01 | |||||
| Nov 10, 2019 at 1:31 | comment | added | dff | @Glen_b I'll try it. Thank you. | |
| Nov 9, 2019 at 10:07 | history | edited | dff | CC BY-SA 4.0 |
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| Nov 9, 2019 at 9:00 | history | tweeted | twitter.com/StackStats/status/1193091007488909313 | ||
| Nov 9, 2019 at 7:32 | answer | added | StubbornAtom | timeline score: 3 | |
| Nov 9, 2019 at 6:50 | comment | added | Glen_b | Assuming you know the basic properties of variances from normal random variables, ;ook up the variance of the F distribution and then you will be able to write down the expectation of the usual estimator of the ratio of variances, and from that seeing how to unbias it is simple. You can derive the expected value of an F-distributed random variable from the formula for the kth moment given there. As for proving that formula, you could attempt it yourself and then if you don't get anywhere, perhaps ask a question about that. | |
| Nov 9, 2019 at 6:19 | answer | added | Kushal Bhattacharya | timeline score: 0 | |
| Nov 9, 2019 at 4:39 | history | edited | dff |
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| Nov 9, 2019 at 4:28 | comment | added | Michael R. Chernick | Add the self-study tag. | |
| Nov 9, 2019 at 3:55 | history | edited | dff | CC BY-SA 4.0 |
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| Nov 9, 2019 at 3:52 | comment | added | Student | I suppose you’re specifically interested in the finite-sample case. So this might not answer your question but you can get an asymptotically unbiased estimator by dividing $\hat\sigma_2^2$ by $\hat\sigma_1^2$. This works because of Slutsky’s theorem. en.wikipedia.org/wiki/Slutsky%27s_theorem | |
| Nov 9, 2019 at 3:50 | review | First posts | |||
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| Nov 9, 2019 at 3:46 | history | asked | dff | CC BY-SA 4.0 |