Differentiating the log-likelihood gives the unique stationary point
$$
\widehat M_0^*=s_1,
\qquad
\widehat{\Delta M}=\frac1G\sum_gs_{2g},
\qquad
\widehat{\mathcal M}=\frac1{N_{\mathrm{SN}}}\sum_is_{3i}.
$$
Its Hessian is diagonal with entries $-1/V_1$, $-G/V_2$, and $-N_{\mathrm{SN}}/\sigma_{\mathrm{SN}}^2$, so it is the unique maximum. The estimators are unbiased and have variances
$$
V_1,
\qquad
\frac{V_2}{G},
\qquad
\frac{\sigma_{\mathrm{SN}}^2}{N_{\mathrm{SN}}},
$$
which equal their <Cramér-Rao lower bounds> because the normal location statistics are efficient.

Since $\theta=M_0^*+\Delta M-\mathcal M$, invariance of the <maximum-likelihood estimator> gives
$$
\widehat\theta=\widehat M_0^*+\widehat{\Delta M}-\widehat{\mathcal M}.
$$
It is unbiased and normal with variance
$$
\sigma_\theta^2
=V_1+\frac{V_2}{G}
+\frac{\sigma_{\mathrm{SN}}^2}{N_{\mathrm{SN}}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 219 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-219/1/e/solution