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?
Best Answer
Remember an unbiased estimator that reaches the Cramer-Rao lower-bound is UMVUE, but if an estimator is UMVUE it does not necessarily reach the Cramer-Rao lower bound.