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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.