# Mean of a possible MLE ,cant figure out the distribution of the sum of $x^{2}$

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

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
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

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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