By the <invariance property of maximum likelihood estimation>,
$$
\boxed{\widehat h=10^{\widehat\theta/5}=e^{\widehat\theta/\alpha}.}
$$
The <sampling distribution> is $\widehat\theta\sim N(\theta,\sigma_{\widehat\theta}^2)$, where
$$
\sigma_{\widehat\theta}^2=C/K+H/N,
$$
$\widehat h$ has a <log-normal distribution> with
$$
\log\widehat h\sim N\!\left(\log h,
\frac{\sigma_{\widehat\theta}^2}{\alpha^2}\right).
$$
Writing $v=\sigma_{\widehat\theta}^2/\alpha^2$, its exact fractional bias and variance are
$$
\frac{\mathbb E\widehat h-h}{h}=e^{v/2}-1,
\qquad
\frac{\operatorname{Var}(\widehat h)}{h^2}=e^v(e^v-1).
$$
Hence, to leading order,
$$
\boxed{\frac{\mathbb E\widehat h-h}{h}\simeq
\frac{\sigma_{\widehat\theta}^2}{2\alpha^2},
\qquad
\frac{\operatorname{Var}(\widehat h)}{h^2}\simeq
\frac{\sigma_{\widehat\theta}^2}{\alpha^2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 219 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-219/1/d/solution