How does stratified sampling work?

Question

I don't understand stratified sampling as it is described on these slides on p. 4. They consider a simple example of $$\int_{[0,\:1)}f\:{\rm d}\lambda,$$ where $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R)$. Now they divide $I=[0,1)$ into strata. Say (, more generally than in the paper,) $$I=\biguplus_{j=1}^mI_j.$$ Now they take $n_j$ (although slightly more general) samples $U_{ij}$ from the uniform distribution $\mathcal U_{I_j}$ on $I_j$ and consider $$F_j:=\frac1{n_j}\sum_{i=1}^{n_j}f(U_{ij}$$ which almost surely tends to $\int_{I_j}f\:{\rm d}\lambda$ as $n_j\to\infty$. So, the estimator of $\int_If\:{\rm d}\lambda$ should be $$F:=\sum_{j=1}^mF_j.$$ However, they divide the sum on the right-hand side by the number of strata $m$. But why? This would only be correct if all the $F_j$ would be estimators for the same quantity (here $\int_If\:{\rm d}\lambda$) or am I missing something?

Answer 1

That's because $$ F_j = \frac{1}{n_j}\sum_{i=1}^{n_j} f(U_{ij}) $$ tends to $$ F_j \to \frac{1}{|I_j|}\int_{I_j} f \,\mathrm{d}\lambda $$ as $n_j \to \infty$. Indeed, the density of a uniform r.v. on $I_j$ is equal to $|I_j|^{-1}\mathbb{I}_{I_j}(x)$. For $m$ strata of equal length you have $|I_j|=1/m$ so that $$ F_j \to m\int_{I_j}f \,\mathrm{d}\lambda. $$ Reassembling the estimator over the whole interval $I=(0,1)$, you obtain $$ \sum_{j=1}^m F_j\to m \int_I f \, \mathrm{d}\lambda. $$ Hence the division by $m$ in the overall estimator.