Solution
= Solution
{c}
The <recurrence relation> at $k=3$ gives $I_5=2\theta I_3=2\theta^2$. Consequently
$$
\operatorname{var}(X_i^2)=\mathbb E[X_i^4]-\bigl(\mathbb E[X_i^2]\bigr)^2=I_5-I_3^2=2\theta^2-\theta^2=\theta^2.
$$
By <variance additivity for independent random variables>, the <variance> of their <sample mean> is
$$
\boxed{\operatorname{var}(\widehat\theta)=\frac{n\theta^2}{n^2}=\frac{\theta^2}{n}.}
$$