Continuous differentiability gives the first-order expansion
$$
\Phi(\widehat\theta_n)-\Phi(\theta_0)
=\nabla\Phi(\theta_0)^T(\widehat\theta_n-\theta_0)
+o(\|\widehat\theta_n-\theta_0\|).
$$
The assumed convergence in distribution implies $\sqrt n(\widehat\theta_n-\theta_0)=O_p(1)$, so after multiplication by $\sqrt n$ the remainder is $o_p(1)$. The <Slutsky theorem> gives the multivariate <delta method>
$$
\boxed{\sqrt n\bigl(\Phi(\widehat\theta_n)-\Phi(\theta_0)\bigr)
\xrightarrow d\nabla\Phi(\theta_0)^TZ}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 29K / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/29k/a/solution