Timeline for Are inconsistent estimators ever preferable? A twist
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| Jun 11, 2020 at 14:32 | history | edited | CommunityBot |
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| May 19, 2020 at 9:45 | comment | added | Richard Hardy | @SextusEmpiricus, thank you! I am not sure trying to phrase things my way is helpful (redefining well-known terms in a new way rarely is), but I noticed immediately how you used non-consistent there, and it was not in the spirit I meant. Of course, the parenthetical explication makes it unambiguous, so no problem; just a note. Interesting questions you are referring to; I am aware of and like both of them. | |
| May 19, 2020 at 7:29 | comment | added | Sextus Empiricus | @RichardHardy I have updated the answer again. This question now makes me think of stats.stackexchange.com/questions/311417/… and stats.stackexchange.com/questions/386337/… | |
| May 19, 2020 at 7:23 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 19, 2020 at 5:12 | comment | added | Richard Hardy | Excellent, thank you very much! | |
| May 18, 2020 at 22:11 | comment | added | Sextus Empiricus | @RichardHardy I have added graphs. The unbiased estimator will be half the time negative (cost = 1) and half the time positive (cost ~ 0), which will result in an expectation value of about 0.5 (if is half-half positive/negative because for finite samplesize the zero value almost never occurs). The biased estimator will be closer and closer to the value $d$, as the sample size grows, for which the cost is $d^2$. | |
| May 18, 2020 at 22:07 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 16, 2020 at 10:40 | vote | accept | Richard Hardy | ||
| May 16, 2020 at 10:39 | comment | added | Richard Hardy | Disregard my last comment. I now see the loss function in the example I was puzzled by is discontinuous! The only remaining questions are how you get $0.5$ and $d^2$ (but they are not so important). Once again, a sincere thanks for your great answer! | |
| May 16, 2020 at 10:36 | comment | added | Richard Hardy | @SextusEmpiricus, thank you very much for you answer! It is enlightening, but I am still puzzled. You say that in the case of unique minimum, the pathological cases should not remain (barring what you note in the last paragraph in parentheses). This is intuitive for me. However, is the example following Edit: The question has changed not contradicting this? If so, could you please make it a little more explicit / "for dummies"? How do you get the expected losses $0.5$ and $d^2$? | |
| May 5, 2020 at 20:09 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 5, 2020 at 15:27 | comment | added | Sextus Empiricus | But wouldn't in that case, if you smooth a bit, the consistent estimator 'get the cost as close' to $L=0$ as you want (if you increase $n$ sufficiently) and become better than the inconsistent estimator? In any case, the example given in this answer won't work anymore. | |
| May 5, 2020 at 15:23 | comment | added | whuber♦ | Continuity doesn't matter. Little changes, as far as I can tell, if you smooth the cost function a tiny bit. | |
| May 5, 2020 at 13:33 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 5, 2020 at 13:16 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 5, 2020 at 12:57 | comment | added | Sextus Empiricus | In the current formulation of Richard Hardy's problem we can easily make a counterexample by using a discontinuous cost function. $$L = \begin{cases} 2 & if \quad t < \mu \\ 0 & if \quad t = \mu \\ 0.5 & if \quad \mu < t \leq \mu + 1 \\ 2 & if\quad \mu + 1 < t \end{cases}$$ In this example the discontinuity is the driving force that makes it work. I believe that continuity of the cost function should be one of the regularity conditions for the idea from Richard Hardy, about inconsistent estimators not being able to dominate a consistent estimator for all $n$. | |
| May 5, 2020 at 12:39 | comment | added | Sextus Empiricus | It may be oversimplified but it is intentionally stripped away from technical details such that it is intuitive. The simple story is that if the cost function is such that it's value is minimum if the estimate is close/near to the true value (whether it is admissible or not) then the consistent estimator will be the only estimator that has the lowest possible value of the cost function as a limit and it should for sufficiently large $n$ dominate inconsistent estimators. For this limit it is important that the cost function is continuous so I believe that it is part of the problem. | |
| May 5, 2020 at 11:32 | comment | added | whuber♦ | As far as I can see, discontinuity of $L$ is irrelevant. I think you lose your way in this analysis by oversimplifying the loss, which is a function of two variables: the statistic and the parameter. Moreover, there is a logical flaw: the limit does not imply the sample mean is admissible. | |
| May 4, 2020 at 21:35 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 4, 2020 at 21:30 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| May 4, 2020 at 21:24 | history | answered | Sextus Empiricus | CC BY-SA 4.0 |