I have a random sample $X_{i}, \ldots, X_{n} \sim \mathcal{N}(\theta,1)$. I would like to find the Cramer-Rao lower bound for the variance of unbiased estimators of $\theta^{2}$.
Here's what I have:
The Cramer-Rao bound is in general given by $\frac{1}{I(\theta)}$ where $I(\theta)$ denotes the Fisher information and $\theta$ is the parameter of interest. In my case the parameter of interest is $\theta^{2}$.
My approach is to reparameterize the log-likelihood to find the Fisher information for $\theta^{2}$. So I for the log-likelihood of my sample I have:
$$l(a, \boldsymbol{x}) = -nln(\sqrt{2\pi})-\sum_{i=1}^{n}(x_{i}-a^{\frac{1}{2}})^{2}$$ with $a := \theta^{2}$.
I get
$$\frac{\partial^{2} l}{\partial a^{2}} = -\frac{1}{2}a^{-\frac{3}{2}}\sum_{i=1}^{n}x_{i}$$
For the Fisher Information I get
$$I(a) = -E\left[\frac{\partial^{2} l}{\partial a^{2}} \right] = \frac{1}{2}na^{-1}$$ where I used the fact that the expectation of $x_{i}$ is given by $a^{\frac{1}{2}} (=\theta)$.
Therefore the Cramer-Rao lower bound is given by
$$CRLB = \frac{1}{I(a)} = \frac{2}{n}a = \frac{2}{n}\theta^{2}$$.
Is this correct?