# Lower bound Cramer rao for a Gamma with parameter θ=θ^2

I have this Gamma density function:

$$\frac{1}{6\theta^8} x^3 \exp\Big[\frac{-1}{\theta^2}x\Big]$$

If I calculate the MSE I have that:

$$\hat\theta=\frac{1}{2}\sqrt{\frac{1}{n}\sum{x_i}}$$

Now, if I calculate the Cramer Rao lower bound for this distribution, I have:

\begin{align} \frac{\partial^2\log(f(x))}{\partial\theta^2} &= \frac{8\theta^2-6x}{\theta^4} \\[5pt] -nE\bigg[\frac{\partial^2}{\partial\theta^2}\log(f(x))\bigg] &= \frac{16n}{\theta^2} \end{align}

Because $E(x)=4\theta^2$ then, Cramer Rao lower bound is equal to $\frac{\theta^2}{16n}$, but if I calculate the variance of $\hat\theta$ I have that as equal to $\frac{1}{4}\sqrt{\frac{4\theta^2}{n}}=\frac{\theta}{2\sqrt{n}}$. I assumed that the variance of sample mean is ${\rm Var}(\bar{x})= \frac{\sigma^2}{n}$, where $\sigma^2$ indicates the variance of the population that is distributed as a $\Gamma(\theta^2,4)$. So ${\rm Var}(\frac{1}{2}\sqrt{\bar{x}})=(\frac{1}{2})^2\sqrt{\frac{4\theta^4}{n}}= \frac{2\theta^2}{4\sqrt{n}}=\frac{\theta^2}{2\sqrt{n}}$, which is different from the Cramer Rao lower bound $\frac{\theta^2}{ 16n}$

But if I write the density function in the form of the exponential family, I have:

\begin{align} \frac{1}{6\theta^8} x^3 \exp\Big[\frac{-1}{\theta^2}x\Big] &= \frac{1}{6}x^3 \exp\big[-\frac{1}{\theta^2}x-8\log\theta\big] \\[5pt] a(\theta) &= -\frac{1}{\theta^2} \\[5pt] t(x) &= x \\[5pt] c(\theta) &= -8\log\theta \\[5pt] -\frac{c'(\theta)}{a'(\theta} &= 4\theta^2 \end{align}

And I find that $\frac{1}{n}\sum{x_i}$ is the UMVUE estimator for $4\theta^2$.

And so I have that $\frac{1}{2}\sqrt{\frac{1}{n}\sum{x_i}}$ is UMVUE estimator for $\theta$.

But how can it be possible if the variance of this estimator is different from the Cramer Rao lower bound?

• Could you show how you compute the variance of $\hat\theta$? – whuber Jun 1 '17 at 17:54
• Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. – gung Jun 1 '17 at 20:01
• What justifies your assumption about the variance of the sample mean and what is "$\sigma$"? Moreover, $\hat \theta$ is not the sample mean--it's proportional to its square root. It begins to look like your question might be rooted in misconceptions concerning the definitions and calculations of moments, rather than in any paradoxical behavior of the Carmer-Rao bound. – whuber Jun 1 '17 at 20:42