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Jun 8, 2012 at 12:25 history edited Macro CC BY-SA 3.0
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Aug 2, 2010 at 3:10 history edited Rob Hyndman CC BY-SA 2.5
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Aug 1, 2010 at 15:58 comment added user88 Aww, those comments locks; $\frac{\hat{\sigma}}{\sqrt{2n}}$. At least this one gives the result in agreement with bootstrap.
Aug 1, 2010 at 15:34 comment added user88 What Srikant found (and what seems confirmed at PhysicsForums) there should be $\sqrt{2}$, so rather $\hat{\sigma}\frac{\sqrt{2}}{2n}$.
Aug 1, 2010 at 14:27 history edited robin girard CC BY-SA 2.5
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Jul 26, 2010 at 17:09 comment added robin girard ok, then you will possibly estimate the square root of the variance of the estimation of the square root of the variance... right :) should be something like $\hat{\sigma}/n$ ?
Jul 26, 2010 at 16:45 comment added user88 Isn't that a function of estimator is still an estimator? I still don't know \sigma, only X_i.
Jul 26, 2010 at 16:23 history answered robin girard CC BY-SA 2.5