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I am just asking for a hint. I need help with this example:

Let $ {X} $ is a random variable with density

$ f (x, \theta) = (\frac{2}{\pi})^{\frac{1}{2}} \theta^{-1} e^{\frac{-x^{2}}{2\sigma^{2}}}, x > 0, \theta > 0, $

where $\theta$ is unknown parameter. Prove that $ T_{n} = (\frac{\pi}{2})^{\frac{1}{2}} \cdot \bar{X} $ is unbiased and consistent estimator.

I know that if estimator is unbiased then $ E(T_{n}) = \theta $ and if $ \lim_{n\to\infty} E(T_{n}) = \theta < \infty $ and $ \lim_{n\to\infty} D(T_{n}) = 0 $ then estimator is consistent, but in this example i do not know how to prove it. I think it is incorrect description of example, beacuse of that part . $ \bar{X} $. But if you have some hints, it will be great. Thanks in advance.

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  • $\begingroup$ If $ \lim_{n\to\infty} E(T_{n}) = \theta < \infty $ and $ \lim_{n\to\infty} D(T_{n}) = 0 $, then $T_n$ is consistent. Not the other way around. $\endgroup$ – Shanks May 20 '18 at 21:14
  • $\begingroup$ @Shanks yes you are right, sorry. The question is edited due this good catch $\endgroup$ – Bopinko May 20 '18 at 21:17
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    $\begingroup$ Can you verify that the density written above is correct? I'm showing that $T_n$ is a biased estimator of $1/\theta$, that is, $E[T_n]=\sigma^2\cdot 1/\theta$ $\endgroup$ – user1993951 May 20 '18 at 22:23
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    $\begingroup$ @user1993951 - the density can't be correct; note that the kernel (the part that's not a constant function of $x$) is just that of a Normal distribution with mean 0 and variance $\sigma^2$, so $\theta$ is hardly unknown if $\sigma^2$ is not. Furthermore, if you work through the algebra, $\theta = 2\sigma$, and obviously $\mathbb{E}\bar{x} = 0$, therefore $\mathbb{E}T_n = 0$ as well, which is not consistent (ha ha) with the ideas that $\theta = 2\sigma$ and $\mathbb{E}T_n = \theta$. $\endgroup$ – jbowman May 20 '18 at 23:58
  • $\begingroup$ Presumably "$\sigma$" is a typographical error for some constant multiple of "$\theta$" (and there may be other errors in the normalizing constant). $\endgroup$ – whuber May 21 '18 at 13:12

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