Find the Cramer-Rao bound for an unbiased estimator of $b^2$ $X$ is a RV with pdf $f(x,b) = \frac{x}{b^2} \exp \{-\frac{x^2}{2b^2} \}$
I've got two different estimates: $\hat{b^2} =\frac{2}{\pi} (\frac{1}{n} \sum_{i=1}^n X_i)^2 $ using MME, and $\hat{b^2} = \frac{1}{n} \sum_{i=1}^n X_i^2$ using MLE. MLE is generally considered unbiased, right? So I'll go with that. 
The Cramer-Rao bound is defined as $\mathbb{E}[T(X)-g(b)]^2 \ge \frac{(g'(b))^2}{J(b)}$ where $J(b)$ is the Fisher information and $g(b) = \mathbb{E}T(X) =\int_0^{\infty} T(x) \frac{x}{b^2} \exp \{-\frac{x^2}{2b^2} \} dx$. 
Of course, $T(X)$ is the estimate : $T(X)=\hat{b^2} = \frac{1}{n} \sum_{i=1}^n X_i^2$.
I've got $J(b) = \frac{4}{b^2}$ for the Fisher info (God, the calculations are a pain!) and now I'm a little unsure whether I'm proceeding correctly to compute the $g'(b)$. 
\begin{align}
g'(b) &=\frac{\partial}{\partial b}\int_0^{\infty} \frac{1}{n} \sum_{i=1}^n x_i^2 \frac{x_i}{b^2} \exp \{-\frac{x_i^2}{2b^2} \} dx\\
&=\frac{1}{n} \sum_{i=1}^n \int_0^{\infty}\Bigg(\frac{-2x^3}{b^3}+\frac{x^5}{b^5} \Bigg) e^{-\frac{x_i^2}{2b^2} } dx\\
&= 4b\\
\end{align}
is the setup correct?
This way my ultimate answer for the CRB would be $\mathbb{E}[T(X)-g(b)]^2 \ge 4b^4$.
Thanks for any help. Also : unless you really love calculus, I suggest you use Wolframalpha ;-)
 A: One can check one's work on such things with a computer algebra system. For this problem, you have a random variable $X$ ~ Rayleigh($b$) with pdf $f(x)$:

(source: tri.org.au) 
Then, for random samples of size $n$ drawn on $X$, the CRLB for all unbiased estimators of $b^2$ is:
$$\text{CRLB}=\frac{1}{n *\text{FisherInformation}\left[b ^2,f\right]}$$
where FisherInformation is a mathStatica funkeh monkeh, and $f$ is the pdf. Here we go:

(source: tri.org.au) 

Alternatively, since you have correctly derived yourself that the Fisher Information on $b$ (rather than $b^2$) is: 
$$J(b) = \frac{4}{b^2}$$
and if $g(b)$ is some differentiable function of $b$ that we are interested in estimating, then, subject to some regularity conditions, the CRLB on $g(b)$ is: 
$$\left(\frac{\partial g(b) }{\partial b }\right)^2 \frac{1}{n J(b)} $$
In your case, $g(b)=b^2$ ... so after differentiating, you get the same result $\frac{b^4}{n}$ .

P.S. Your title should be:  CRLB for an unbiased estimator of $b^2$  ...not estimate
