Timeline for Trying to make sense of claims regarding Rao-Blackwell and Lehmann-Scheffé for sufficient/complete statistics
Current License: CC BY-SA 4.0
30 events
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| Apr 22, 2021 at 6:06 | comment | added | The Pointer | @SextusEmpiricus yes, I will leave this question as is, because, despite the fact that this wasn’t the question I wanted to ask, it attracted a valuable answer and is quite educational. If I decide to ask the intended question sometime in the future, then it is best done as a new question. | |
| Apr 22, 2021 at 6:02 | comment | added | Sextus Empiricus | When I read this question first I thought "what is going on here?". AdamO had the same idea, how do you get to consider two sufficient statistics and only one of them complete? Thomas wrote an excellent answer to explain it. | |
| Apr 22, 2021 at 5:59 | comment | added | Sextus Empiricus | Actually, the phrase needs another addition and that is that $T_1$ must be an unbiased statistic. | |
| Apr 22, 2021 at 5:59 | vote | accept | The Pointer | ||
| S Apr 22, 2021 at 5:59 | history | bounty ended | The Pointer | ||
| S Apr 22, 2021 at 5:59 | history | notice removed | The Pointer | ||
| Apr 22, 2021 at 5:58 | comment | added | The Pointer | @SextusEmpiricus Oh my goodness, you're right. It seems that I completely screwed it up. The problem is that this was all based on memory, so I was trying to recall the idea and then ask. And it seems that, in doing so, I confused myself and wrote the wrong thing! My apologies! Given what I have written, it seems that what Thomas wrote is absolutely correct. Again, I apologise for any confusion. | |
| Apr 22, 2021 at 5:50 | comment | added | Sextus Empiricus | Your phrase should be "if we define, say, a statistic $T_1(\mathbf{X})$ and a sufficient (complete) statistic $T_2(\mathbf{X})$ for some parameter $\varphi$ then, under some conditions, we can say that $\text{Var}(T_3(\mathbf{X})) \leq \text{Var}(T_1(\mathbf{X}))$, where $T_3(\mathbf{X})=E[T_1(\mathbf{X})\, | \,T_2(\mathbf{X})]$. " 1: This $T_3$ is a different estimator from $T_1$ and $T_2$, but it is related. 2: $T_1$ need not be sufficient. 3: $T_2$ doesn't need to be complete (for RB theorem and the inequality can be equality as well). | |
| Apr 19, 2021 at 1:05 | answer | added | Thomas Lumley | timeline score: 7 | |
| Apr 18, 2021 at 22:42 | history | edited | The Pointer | CC BY-SA 4.0 |
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| S Apr 18, 2021 at 22:17 | history | bounty started | The Pointer | ||
| S Apr 18, 2021 at 22:17 | history | notice added | The Pointer | Draw attention | |
| Apr 15, 2021 at 3:13 | history | tweeted | twitter.com/StackStats/status/1382532483439288320 | ||
| Apr 14, 2021 at 18:41 | comment | added | The Pointer | If $S^2$ and $\bar{X}$ are unbiased estimators for $\lambda$, then could it mean that $T_1(\mathbf{X})$ and $T_2(\mathbf{X})$ are $S^2$ and $\bar{X}$? We can say that, if $S^2$ is an unbiased estimator of $\theta$ and $\bar{X}$ is a sufficient statistic for $\theta$, then, since we know that $E[S^2|\bar{X}] = \bar{X}$, (1) $\bar{X}$ is an unbiased estimator of $\theta$ (that is, $E(\bar{X}) = \theta$) and (2) $\text{Var}(\bar{X}) \le \text{Var}(S^2)$. (1) and (2) are just the Rao-Blackwell theorem, right? | |
| Apr 14, 2021 at 18:16 | comment | added | AdamO | @Xi'an I thought one would distinguish $T$ and $(T,T)$ in that $T$ is minimally complete & sufficient, whereas $(T,T)$ would simply be complete, sufficient. Is that not correct? | |
| Apr 14, 2021 at 18:01 | comment | added | The Pointer | @Xi'an Sorry about that. I will not link anymore documents. Is my question ok? As I said, I don't really understand what it's saying myself, so I hope I managed to get the message across. | |
| Apr 14, 2021 at 17:55 | comment | added | Xi'an | Please avoid linking to documents without providing the context as it is unrealistic to expect people to read these before addressing the question. | |
| Apr 14, 2021 at 17:54 | history | edited | The Pointer | CC BY-SA 4.0 |
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| Apr 14, 2021 at 17:48 | history | edited | The Pointer | CC BY-SA 4.0 |
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| Apr 14, 2021 at 17:31 | comment | added | The Pointer | @AdamO Sorry about that. I've been trying to read from multiple documents explaining the same thing, so I got confused when typing out my question. Is that better? | |
| Apr 14, 2021 at 17:30 | history | edited | The Pointer | CC BY-SA 4.0 |
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| Apr 14, 2021 at 17:24 | history | edited | The Pointer | CC BY-SA 4.0 |
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| Apr 14, 2021 at 17:23 | comment | added | AdamO | For a Poisson model, the variance = mean. You have a normal model here! | |
| Apr 14, 2021 at 17:21 | comment | added | The Pointer | @AdamO My research turned up this math.stackexchange.com/q/2881768/356308 | |
| Apr 14, 2021 at 17:21 | comment | added | AdamO | I think the Hodge's Superefficient estimator is a simple counter example that actually helps understand how ill-formed some of this early variance bound research was: en.wikipedia.org/wiki/Hodges%27_estimator | |
| Apr 14, 2021 at 17:16 | history | rollback | The Pointer |
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| Apr 14, 2021 at 17:15 | comment | added | AdamO | I can't think of an example of an incomplete and sufficient statistic and another complete and sufficient statistic for the same probability model. R-B says if you take an unbiased statistic and condition on the sufficient statistic, you will get a new estimator with lower variance but there may be other unbiased estimators with even lower variance. L-S says if it's complete and sufficient, you get the UMVUE. | |
| Apr 14, 2021 at 17:13 | history | edited | The Pointer | CC BY-SA 4.0 |
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| Apr 14, 2021 at 17:10 | comment | added | AdamO | $E[S^2|\bar{X}] = \bar{X}$??? | |
| Apr 14, 2021 at 17:07 | history | asked | The Pointer | CC BY-SA 4.0 |