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The distribution function is $F(x)=x^2/{\theta }^2$ on $[0,\theta ]$, so its median is $\theta /\sqrt{2}$. The likelihood function iswhich decreases over its admissible range. Hence $\hat{\theta }=X_{(n)}={\max }_i{X}_i$. By the invariance property of maximum likelihood estimation,

By the invariance property of maximum likelihood estimation,The sampling distribution is $\hat{\theta }\sim N(\theta ,{\sigma }_{\hat{\theta }}^2)$, where$\hat{h}$ has a log-normal distribution withWriting $v={\sigma }_{\hat{\theta }}^2/{\alpha }^2$, its exact fractional bias and variance areHence, to leading order,

The maximum-likelihood estimators are the success proportionsThe invariance property of maximum likelihood estimation therefore gives

The invariance property of maximum likelihood estimation givesSince $\hat{\theta }-\theta \sim N(0,{\sigma }_{\theta }^2)$, the ratio $\hat{h}/h$ has a log-normal distribution. Its exact variance isand the lowest-order delta method approximation isFor fixed ${\sigma }_{\mathrm{tot}}$, case (i) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2$, whereas case (ii) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2/K$. Because $K\ge 2$ 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.

With independent Gaussian errors, maximizing the likelihood over $\alpha ,\beta$ minimizes the residual sum of squares, so the maximum-likelihood estimators are the same as least squares. The invariance property of maximum likelihood estimation givesThe two estimates have zero covariance by part (ii). ThereforeThis is the uncertainty in the estimated expected response. Estimating an additional noisy observation would add its own error variance, but the quantity requested is the mean response $E_t$.