The <maximum-likelihood estimators> are the success proportions
$$
\widehat\phi_0=\frac{120}{200}=\frac35,
\qquad
\widehat\phi_1=\frac{110}{220}=\frac12.
$$
The <invariance property of maximum likelihood estimation> therefore gives
$$
\widehat\alpha=\widehat\phi_1-\widehat\phi_0
=\frac12-\frac35=-\frac1{10}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 2 / 2 / Solution

The <invariance property of maximum likelihood estimation> gives
$$
\widehat h=10^{\widehat\theta/5}
=\exp(\widehat\theta/\alpha).
$$
Since $\widehat\theta-\theta\sim N(0,\sigma_\theta^2)$, the ratio $\widehat h/h$ has a <log-normal distribution>. Its exact variance is
$$
\operatorname{Var}\!\left(\frac{\widehat h}{h}\right)
=e^{\sigma_\theta^2/\alpha^2}
\left(e^{\sigma_\theta^2/\alpha^2}-1\right),
$$
and the lowest-order <delta method> approximation is
$$
\operatorname{Var}\!\left(\frac{\widehat h}{h}\right)
=\frac{\sigma_\theta^2}{\alpha^2}+O(\sigma_\theta^4).
$$
For fixed $\sigma_{\rm tot}$, case (i) has $A=\sigma_\mu^2+\sigma_{\rm tot}^2$, whereas case (ii) has $A=\sigma_\mu^2+\sigma_{\rm tot}^2/K$. Because $K\geq2$ and $B$ is the same in both cases, independent supernova-level variation in case (ii) gives the smaller fractional variance: averaging reduces it, while a shared galaxy fluctuation does not average away.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 / 1 / d / Solution

If $\widehat\theta$ maximizes a likelihood and the target parameter is $\eta=g(\theta)$, a maximum-likelihood estimate of $\eta$ is $g(\widehat\theta)$, with the usual interpretation when $g$ is not one-to-one.


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 Incoming links: Invariance property of maximum likelihood estimation