Prove the MLE is an efficient estimator for $\theta$ in the context of Normal distribution

Let $$X_{1},X_{2},\ldots,X_{n}$$ represent a random sample whose distribution is given by $$X\sim\mathcal{N}(0,\theta)$$. Find the MLE of $$\theta$$. Then obtain the score function, find the Fischer information and prove the MLE is an efficient estimator for $$\theta$$.

MY ATTEMPT

Firstly, let us determine the MLE estimator \begin{align*} & p(x|\theta) = \frac{1}{\sqrt{2\pi\theta}}\exp\left\{-\frac{x^{2}}{2\theta}\right\} \Rightarrow\\ & p(\textbf{x}|\theta) = \left(2\pi\theta\right)^{-n/2}\exp\left\{-\frac{1}{2\theta}\sum_{i=1}^{n}x^{2}_{i}\right\}\Rightarrow\\\\ & \ln p(\textbf{x}|\theta) = -\frac{n}{2}\ln(2\pi) - \frac{n}{2}\ln(\theta) - \frac{1}{2\theta}\sum_{i=1}^{n}x^{2}_{i} \Rightarrow\\ &\frac{\partial\ln p(\textbf{x}|\theta)}{\partial\theta} = -\frac{n}{2\theta} + \frac{1}{2\theta^{2}}\sum_{i=1}^{n}x^{2}_{i} = 0 \Rightarrow \hat{\theta} = \frac{1}{n}\sum_{i=1}^{n}x^{2}_{i} \end{align*}

From then on I get stuck. This is what I have tried: \begin{align*} \frac{\partial^{2}\ln p(\textbf{x}|\theta)}{\partial\theta^{2}} = \frac{n}{2\theta^{2}} - \frac{1}{\theta^{3}}\sum_{i=1}^{n}x^{2}_{i} \Rightarrow I(\theta) = -\frac{n}{2\theta^{2}} + \frac{1}{\theta^{3}}\sum_{i=1}^{n}\textbf{E}(X^{2}_{i}) \end{align*}

Unfortunately, this leads to nowhere, even though I know that $$\textbf{E}(X^{2}_{i}) = \theta$$. Can someone help me out with the Fischer Information and $$\textbf{Var}(X^{2}_{i})$$?

• For fisher information $\sum_{i=1}^{n}\textbf{E}(X^{2}_{i}) = n\theta$. For $\textbf{Var}(X^{2}_{i})$, find $\textbf{Var}(X^{4}_{i})$ by searching normal distribution. – user158565 May 30 at 4:26
• $Var(X_i^4)$ should be $E(X_i^4)$. – user158565 May 30 at 4:42

To establish efficiency, you need to compare the variance of your estimator with the Cramér-Rao bound. You have already derived an expression for the Fisher information, so now you just need to simplify it. Continuing what you have already found, you have $$\mathbb{E}(X_i^2) = \theta$$ which implies that the estimator is unbiased. Thus, using your existing expression, the Fisher information can be written as:
\begin{equation} \begin{aligned} \mathcal{I}(\theta) &= \mathbb{E} \bigg( \frac{n}{\theta^3} \Bigg[ \frac{1}{n} \sum_{i=1}^n X_i^2 - \frac{\theta}{2} \Bigg] \bigg) \\[6pt] &= \mathbb{E} \bigg( \frac{n}{\theta^3} \Bigg[ \hat{\theta} - \frac{\theta}{2} \Bigg] \bigg) \\[6pt] &= \frac{n}{\theta^3} \Bigg[ \mathbb{E}(\hat{\theta}) - \frac{\theta}{2} \Bigg] \\[6pt] &= \frac{n}{\theta^3} \Bigg[ \theta - \frac{\theta}{2} \Bigg] \\[6pt] &= \frac{n}{2 \theta^2}. \\[6pt] \end{aligned} \end{equation}
Since $$\hat{\theta}$$ is an unbiased estimator of $$\theta$$, the Cramér-Rao bound is:
$$\mathbb{V}(\hat{\theta}) \geqslant \mathcal{I}(\theta)^{-1} = \frac{2}{n} \cdot \theta^2.$$