Timeline for How to show unbiased estimator of combination of bernoulli and normal variables?
Current License: CC BY-SA 3.0
11 events
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Jun 13, 2012 at 23:29 | vote | accept | David | ||
Feb 16, 2012 at 15:09 | comment | added | cardinal | @David: If this answer has been helpful, you should consider formally accepting it. :) | |
Feb 14, 2012 at 2:59 | comment | added | David | Thank you so much all. Its really good discussion. Finally, we got the estimator $\hat\theta$ is not unbaised. | |
Feb 14, 2012 at 1:36 | history | edited | Macro | CC BY-SA 3.0 |
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Feb 14, 2012 at 1:29 | history | edited | Macro | CC BY-SA 3.0 |
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Feb 14, 2012 at 1:01 | history | edited | Macro | CC BY-SA 3.0 |
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Feb 14, 2012 at 0:52 | comment | added | cardinal | I think that would be my answer, too. Honestly, it makes me suspicious that either (a) the OP hasn't quite stated the problem as intended or (b) the question was poorly thought out in the first place. | |
Feb 14, 2012 at 0:01 | comment | added | Macro | Maybe the answer is that $\hat{\theta}$ is not unbiased because $E(\hat{\theta})$ doesn't exist? | |
Feb 13, 2012 at 23:54 | comment | added | Macro | Good point, cardinal. I was thinking of this in the case where $X$ has an arbitrary distribution but didn't think of the case where there is mass at 0. So, of course, the logic above only applies when $E(1/\overline{X})$ exists. | |
Feb 13, 2012 at 14:44 | comment | added | cardinal | This seems to be a bit of a weird problem since $\mathbb P(\bar X_n = 0) >0$ for all $n$. I've only thought about this for a moment, but navigating around this fact looks difficult at the moment. | |
Feb 13, 2012 at 4:01 | history | answered | Macro | CC BY-SA 3.0 |