Stratified estimator unbiasedness, variance, and proportional allocation

For each $h$, [F1] gives ${\int }_{\mathsf{X}}{f}^{2}d{P}_h=p_h^{-1}{\int }_{A_h}{f}^{2}dP<\mathrm{\infty }$. Since $\mathrm{\mid }f\mathrm{\mid }\le 1+f^2$, $f$ is integrable under $P$ and every $P_h$. Thus [F3] gives ${\mu }_h=p_h^{-1}{\int }_{A_h}fdP$, and [F4] plus the finite partition identity yield $E{\widehat{\mu }}_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}}={\sum }_h{p}_h{\mu }_h={\sum }_h{\int }_{A_h}fdP=\int fdP=\mu$.

For fixed $h$, expand the centered square of ${\bar{f}}_h$. The diagonal terms give ${\sigma }_h^2\mathrm{/}{n}_h$; for distinct draws the centered product expectation is zero by independence and [F5]. Thus $\mathrm{Var}({\bar{f}}_h)={\sigma }_h^2\mathrm{/}{n}_h$.

Splitting $\int (f-\mu {)}^{2}dP$ over the partition and writing $f-\mu =(f-{\mu }_h)+({\mu }_h-\mu )$ on $A_h$, the cross term integrates to zero since ${\int }_{A_h}(f-{\mu }_h)dP=p_h{E}_{P_h}(f-{\mu }_h)=0$. Therefore ${\mathrm{Var}}_P(f)={\sum }_h{p}_h{\sigma }_h^2+{\sum }_h{p}_h({\mu }_h-\mu {)}^2$.

By [F1] and [F7], the sigma-algebras generated by all individual draws are independent. Grouping them by stratum and applying [F8] shows that the sigma-algebras generated by the disjoint stratum blocks are independent. Each ${\bar{f}}_h$ is measurable with respect to its block sigma-algebra because it is a finite sum of measurable values of $f$, using [F9]; hence the stratum averages are independent. Applying [F5] to their centered products and expanding the variance of ${\sum }_h{p}_h{\bar{f}}_h$ gives ${\sum }_h{p}_h^2{\sigma }_h^2\mathrm{/}{n}_h$.

Under $n_h=n{p}_h$, step 3.1 becomes $n^{-1}{\sum }_h{p}_h{\sigma }_h^2$. The ordinary mean of $n$ IID $P$-draws has variance ${\mathrm{Var}}_P(f)\mathrm{/}n$, by the same centered-square expansion as step 2.1. Step 2.2 shows the difference is $n^{-1}{\sum }_h{p}_h({\mu }_h-\mu {)}^2\ge 0$, proving the comparison.