# Unbiased and consistent estimate

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.

• 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. – Shanks May 20 '18 at 21:14
• @Shanks yes you are right, sorry. The question is edited due this good catch – Bopinko May 20 '18 at 21:17
• 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, $E[T_n]=\sigma^2\cdot 1/\theta$ – user1993951 May 20 '18 at 22:23
• @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 $\mathbb{E}T_n = \theta$. – jbowman May 20 '18 at 23:58
• Presumably "$\sigma$" is a typographical error for some constant multiple of "$\theta$" (and there may be other errors in the normalizing constant). – whuber May 21 '18 at 13:12