I'm not even sure if the title of the question makes sense...
I have the solution to the following question in front of me, but can not make sense of it:
Let $X_1, \ldots, X_n$ be a random sample from a distribution $\mathrm{N}(0, \theta)$.
Is $\hat{\theta}_{MLE}$ a UMVUE of $\theta$?(HINT: $\frac{X^2}{\theta} \sim \chi_{1}^2$, and $\mathrm{E}(\frac{X^2}{\theta})=1$, $\mathrm{Var}(\chi_{k}^2) = 2k$.)
I fairly easily found $\hat{\theta}_{MLE} = \frac{1}{n} \sum X_i^2$; and further that this is unbiased, since $\mathrm{E}(\hat{\theta}_{MLE})=\mathrm{E}(X_{1}^{2})=\theta$. Now, upon computing the relevant derivative to find the CRLB I have arrived at the following expectation: $$ \mathrm{E} \left[ \frac{\partial}{\partial\theta} \mathrm{ln} \ f(X; \theta) \right]^2 = \mathrm{E} \left[- \frac{1}{2\theta} + \frac{\mathrm{X}^2}{2\theta^2} \right]^2 = \frac{1}{4\theta^2} \times \mathrm{E} \left[ \frac{\mathrm{X}^2}{\theta} - 1 \right]^2 $$
I am certain that it is here I need to apply the hint given to me in the question, but I have lost my way and would greatly appreciate a pointer in the right direction.