Timeline for An alternative proof of Bayesian Cramer-Rao
Current License: CC BY-SA 4.0
20 events
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| Jan 21 at 21:57 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Nov 10, 2017 at 12:50 | history | edited | Iosif Pinelis |
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| Nov 10, 2017 at 3:17 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Oct 24, 2017 at 22:04 | comment | added | Deane Yang | What I meant to say is that proving that Jensen implies Cauchy-Schwarz is not a research level question. | |
| Oct 24, 2017 at 18:36 | comment | added | Boby | @DeaneYang The question is about an alternative proof of CR bound. It didn't get flagged for being irrelevant for this site. Now, you are saying it can be done with Jensen's instead of Cauchy-Swartz. To me, this seems like you have an alternative proof. I don't see why you would not post it, but it is up to you. | |
| Oct 24, 2017 at 18:12 | comment | added | Deane Yang | Sorry but that's not a research level question. You should try to figure it out yourself. It might take a while. It always does the first few times. | |
| Oct 24, 2017 at 18:08 | comment | added | Boby | @DeaneYang But how would you do that with a product of two terms. How would you separate the terms? Can you post this? | |
| Oct 24, 2017 at 16:34 | comment | added | Deane Yang | In the step where you use Cauchy-Schwarz, just use Jensen instead. | |
| Oct 24, 2017 at 16:34 | comment | added | Deane Yang | If it implies Cauchy-Schwarz, it has to be sufficient. | |
| Oct 24, 2017 at 13:55 | comment | added | Boby | @DeaneYang Sure. But do you think Jensen's inequality is sufficient to show CR bound? | |
| Oct 24, 2017 at 13:52 | comment | added | Deane Yang | Yes. It's a more general inequality. Cauchy-Schwartz is a special case of Jensen. | |
| Oct 24, 2017 at 13:42 | comment | added | Boby | @DeaneYang So, how about Jensen's inequality, can we use it to show Cramer-Rao inequality? | |
| Oct 23, 2017 at 22:15 | comment | added | Deane Yang | The Cauchy-Schwarz inequality is equivalent to saying that the function $x \mapsto x^2$ is convex. Any proof of the Cramer-Rao inequality has to use a convexity inequality like that, and the Cauchy-Schwarz inequality is the simplest possible one, since $x\mapsto x^2$ is the simplest convex function. | |
| Oct 23, 2017 at 18:23 | history | edited | Boby | CC BY-SA 3.0 |
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| Oct 23, 2017 at 16:35 | history | edited | Boby | CC BY-SA 3.0 |
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| Oct 23, 2017 at 14:47 | answer | added | Tom Leinster | timeline score: 4 | |
| Oct 23, 2017 at 14:36 | comment | added | Boby | @DeaneYang I don't see how convexity comes in. Could you show this proof? Another question. Why do you think there is no proof that avoids Cauchy-Schwartz but uses some other inequality? | |
| Oct 23, 2017 at 14:27 | comment | added | Deane Yang | It is possible to write this proof a bit differently in a way you find more natural. But the key step uses convexity, so there is no way to avoid Cauchy-Schwarz or an equivalent inequality. | |
| Oct 23, 2017 at 14:11 | history | edited | Boby | CC BY-SA 3.0 |
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| Oct 23, 2017 at 13:12 | history | asked | Boby | CC BY-SA 3.0 |