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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 supposed to be but now I am not so sure –  Gmath Dec 29 '12 at 22:42
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2 Answers

up vote 2 down vote accepted

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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Ok it is a MLE but how can I know the Mean of it ?? –  Gmath Dec 30 '12 at 5:32
    
Do you know about Jensen's inequality? –  Learner Dec 30 '12 at 5:37
    
Yes, I will try it thanks :) –  Gmath Dec 30 '12 at 5:49
    
but the equality is only for linear functions so I would only get some kind of bound,how should I use it ? –  Gmath Dec 30 '12 at 6:48
1  
You could use it to quickly prove that the estimator is biased (here different from $\theta/\sqrt{n}$). Its square is unbiased though. –  Learner Dec 30 '12 at 6:52
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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

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