Talk:Cramér-Rao inequality
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[edit] Lower/Upper bound
"Upper bound" is correct in the first paragraph, where we speak of "accuracy". The Cramer-Rao inequality gives the maximum accuracy that can be achieved. Later, we speak about variance and there it is in fact a lower bound. High variance means low accuracy and vice versa.
This was changed recently, I have changed it back. -BB
- I didn't read carefully the first time, if we're talking about variance then the correct concept is precision not accuracy. Accuracy ties in with unbiasness. I feel we should not bring in accuracy or precision here since traditionally the CRLB is used in direct relation to the variance.
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- I like it as you have put it now, using precision instead of accuracy. The first sentence now expresses well the basic message of the CRB: It tells you how good any estimator can be, thus limiting the "goodness" by giving its maximum value. -BB
[edit] Too Technical
I have a science/engineering background, but can't begin to understand this. If I could suggest a change, I would. —BenFrantzDale
- Hi BenFranz. I'm happy to improve the article. What bit don't you understand? Robinh 14:32, 6 January 2006 (UTC)
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- For starters, a sentence or two on how this inequality is used (i.e., in what field does it come up) would be helpful. I don't know how specialized this topic is, but I like to think I have most of the background needed to get a rough understanding of it. A list of prerequisites, as described on Wikipedia:Make technical articles accessible would be helpful. —BenFrantzDale 16:25, 6 January 2006 (UTC)
It's used where statistical estimation is used. Read the article on Fisher information. Michael Hardy 22:54, 9 January 2006 (UTC)
[edit] a correction for the example for the cramer-rao ineqality
My name is Roey and I am a student in my third year for industrial engineering in T.A.U, I have a correction for the example given here for cramer rao inequality: the normal distibution formula is incorrect and for some reason I can`t insert it here, in the example the formula has not divided the (x-m)^2 by 2*teta, instead, it is devided by teta. thus the final result is incorrect. The right result should be: 1/2*teta^2 (and not 3 times this number) I have continued this example correctly by I can`t get it to here.... Have a good day and tell me how can I put the correct answer here please...
Roey
[edit] Error in Example
As Roey I found that the example for the Gaussian case was mistaken. I have corrected it. Indeed this example achieves the CR bound. Anyway I'm not too much familiar with this math. I would acknowledge if the original author or other reader will confirm the change.
Viroslav
