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I would like to know:

  1. Is $\beta$ a MLE?

  2. If yes what is the mean of it: $E(\beta)=$?

given:

$x$ is a random variable

$f(x)=\sqrt{\frac{2}{\pi \theta^2}}\exp \left(-\frac{x^2}{2\theta^2}\right)$ with $\theta,x>0$

I got $\beta = \sqrt{\sum x_i^2/n}$

Doing the likelihood function and the first derivative however when I check the 2nd derivative it was not evident to be negative.

If anyone could help I would appreciate a lot!

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    I cant prove it,it $s$upposed to be but now I am not so sure2012-12-29

2 Answers 2

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I will not get into much detail. As pointed out the first order condition leads to the estimator $ \hat{\theta} = \sqrt{\frac{\sum_{i = 1}^n x_i^2}{n}} $ The second order condition leads to $ \frac{n}{\theta^2} - \frac{3}{\theta^4} \sum_{i = 1}^n x_i^2 $ Upon substituion of $\hat{\theta}$, this becomes $ - \frac{2 n^2}{\sum_{i = 1}^n x_i^2} < 0 $ which is what needed to be verified.

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    I need to do on paper but I almost finish with this exercise thank you @learner again !!!2012-12-30
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It's the MLE for $\theta$. We it a normal distribution it would be $\sigma$, by symmetry it is the same for your truncated normal. The truncation affects the mean, but as $x^2 = (-x)^2$, not the estimate of $\theta$.

The bias isn't that simple, but is the same as for a normal: http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation