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Got a distribution of $f_X(x;\theta) = (x/\theta^2) \exp(-x^2/2\theta^2)$ for $x \ge 0$

where the MLE is calculated as $\theta_{MLE} = \sqrt{(\sum_{i=1}^{n}x^2_i)/2n}$

So now need to find if it's unbiased by taking the expected value of the beast. How would this be approached?

JakeM
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  • You should use different notations for the parameter $\theta$ on one hand and the MLE estimator on the other. – Learner Mar 16 '13 at 07:56
  • Also, $f_x$ is not a density in your question. – Learner Mar 16 '13 at 07:57
  • I'd write $f_X(x)$, being careful about which is capital $X$ (the random variable) and which is lower-case $x$ (the argument to the (alleged) density function). That way it's possible to understand expressions like $f_X(3)$, or like $\Pr(X\le x)$, etc. – Michael Hardy Mar 16 '13 at 12:57
  • The whole point of the $\exp$ notation is to make it unnecessary to write some expressions as superscripts. I've edited the question accordingly. – Michael Hardy Mar 16 '13 at 13:01
  • New to the tex syntax. Thanks. – JakeM Mar 16 '13 at 20:30

1 Answers1

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Since $\mathbb E_\theta(X^2)=2\theta^2$, the random variable $\Theta_n=\sqrt{\frac1{2n}\sum\limits_{k=1}^nX_k^2}$ is such that $\mathbb E_\theta(\Theta_n^2)=\theta^2$. Cauchy-Schwarz inequality yields $\mathbb E_\theta(\Theta_n)\lt\sqrt{\mathbb E_\theta(\Theta_n^2)}$ hence $\mathbb E_\theta(\Theta_n)\lt\theta$. Thus, $\Theta_n$ is a biased estimator of $\theta$ for every $n\geqslant1$. For example $\mathbb E_\theta(\Theta_1)=\frac1{\sqrt2}\theta$, but $\mathbb E_\theta(\Theta_n)\to\theta$ when $n\to\infty$.

Did
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