Optimal Sample Allocation Across Independent Strata

Why Sample Allocation Matters

In stratified sampling, observations are collected separately within groups, or strata, and then combined to estimate an overall mean. A fixed budget forces a choice: how many observations should be taken from each stratum? Sampling more in a stratum improves the precision of its estimated mean, but the value of that improvement depends on the stratum's weight, its variability, and the cost of collecting an observation there.

The goal here is to minimize the variance of the weighted mean subject to a fixed total sampling cost. The resulting allocation is a continuous optimum; actual sample sizes must be integers, and practical constraints such as minimum sample sizes can alter the solution.

Model and Variance

Suppose there are $H$ strata. In stratum $h$, observations $X_{h1},\ldots,X_{hn_h}$ are independent draws with mean $\mu_h$ and variance $\sigma_h^2$. Assume samples from distinct strata are independent as well. The target is the weighted mean

\mu=\sum_{h=1}^H W_h\mu_h,\qquad W_h>0,\qquad \sum_{h=1}^H W_h=1.

The weights $W_h$ are fixed and known, such as population proportions. Estimate each stratum mean by $\bar X_h=n_h^{-1}\sum_{i=1}^{n_h}X_{hi}$, and estimate the overall mean by $\hat\mu=\sum_{h=1}^H W_h\bar X_h$.

Definition: For independent sampling within and across strata, the variance of the weighted mean is $$ \operatorname{Var}(\hat\mu)=\sum_{h=1}^H \frac{W_h^2\sigma_h^2}{n_h}. $$ The sampling cost is $\sum_{h=1}^H c_hn_h$, where $c_h>0$ is the cost of one observation in stratum $h$.

The variance expression follows because $\operatorname{Var}(\bar X_h)=\sigma_h^2/n_h$, and independence makes the covariance terms between distinct stratum means zero. It is exact under the stated sampling model. Under finite-variance conditions, if all relevant $n_h$ grow, it is also the variance of the leading normal approximation to $\hat\mu$. The allocation result below minimizes this variance, or equivalently its first-order large-sample approximation when the sample sizes are interpreted asymptotically.

Optimal Allocation With Unequal Costs

Write $a_h=W_h\sigma_h$. The optimization problem with total budget $C$ is

\min_{n_1,\ldots,n_H>0}\ \sum_{h=1}^H \frac{a_h^2}{n_h} \quad\text{subject to}\quad \sum_{h=1}^H c_hn_h=C.

Assume $a_h>0$ and $c_h>0$ for every stratum. The budget must be large enough to permit positive sample sizes. The following inequality proves the minimum directly.

Theorem: Under the preceding assumptions, the unique continuous allocation minimizing $\operatorname{Var}(\hat\mu)$ subject to $\sum_h c_hn_h=C$ is $$ n_h=\frac{C\,W_h\sigma_h}{\sqrt{c_h}\sum_{j=1}^H W_j\sigma_j\sqrt{c_j}}, \qquad h=1,\ldots,H. $$ The minimum variance is $$ \frac{\left(\sum_{h=1}^H W_h\sigma_h\sqrt{c_h}\right)^2}{C}. $$

Proof. Apply the Cauchy–Schwarz inequality to the vectors with entries $a_h/\sqrt{n_h}$ and $\sqrt{c_hn_h}$. It gives

\left(\sum_{h=1}^H a_h\sqrt{c_h}\right)^2 = \left(\sum_{h=1}^H \frac{a_h}{\sqrt{n_h}}\sqrt{c_hn_h}\right)^2 \leq \left(\sum_{h=1}^H\frac{a_h^2}{n_h}\right) \left(\sum_{h=1}^Hc_hn_h\right).

Since the cost is $C$, every feasible allocation has variance at least $(\sum_h a_h\sqrt{c_h})^2/C$. Equality in Cauchy–Schwarz holds precisely when the two vectors are proportional: for some $t>0$, $a_h/\sqrt{n_h}=t\sqrt{c_hn_h}$ for every $h$. Rearranging yields $n_h=a_h/(t\sqrt{c_h})$. Substituting this form into the budget constraint gives $t=(\sum_j a_j\sqrt{c_j})/C$, which produces the stated allocation and minimum variance. Equality conditions also show that no other positive allocation attains the minimum. $\square$

The rule allocates more observations to a stratum with a larger weight or standard deviation, but fewer to a stratum with a larger cost. The square root of the cost appears because a more expensive observation affects both the budget constraint and the marginal reduction in variance.

Worked Example: Three Strata With Equal Costs

Suppose three strata have weights $W=(0.5,0.3,0.2)$, standard deviations $\sigma=(2,3,1)$, equal unit costs $c=(1,1,1)$, and budget $C=210$. Calculate the continuous optimum and its variance.

The products $a_h=W_h\sigma_h$ are $1$, $0.9$, and $0.2$, so $\sum_h a_h\sqrt{c_h}=2.1$. Because the costs are equal, the allocation is proportional to these products:

(n_1,n_2,n_3) =210\left(\frac{1}{2.1},\frac{0.9}{2.1},\frac{0.2}{2.1}\right) =(100,90,20).

The cost is $100+90+20=210$. The variance is

\frac{1^2}{100}+\frac{0.9^2}{90}+\frac{0.2^2}{20} =0.010+0.009+0.002=0.021.

For comparison, equal allocation would use 70 observations per stratum and have variance $1/70+0.81/70+0.04/70\approx0.02643$, which is larger. The comparison confirms that, with equal costs, more observations go to strata with larger $W_h\sigma_h$.

Worked Example: Two Strata With Unequal Costs

Take $W=(0.6,0.4)$, $\sigma=(2,1)$, $c=(4,1)$, and $C=100$. Then $a=(1.2,0.4)$, and the denominator in the allocation formula is

\sum_h a_h\sqrt{c_h}=1.2(2)+0.4(1)=2.8.

Thus

n_1=\frac{100(1.2)}{2(2.8)}=\frac{150}{14}\approx21.429, \qquad n_2=\frac{100(0.4)}{2.8}=\frac{100}{7}\approx14.286.

The budget check is $4(150/14)+100/7=600/14+200/14=800/14$, which is not 100. That would indicate a calculation error, so check the formula carefully: the denominator for $n_h$ is $\sum_j a_j\sqrt{c_j}$, and the numerator is $C a_h/\sqrt{c_h}$. Therefore the correct calculations are

n_1=\frac{100(1.2/2)}{2.8}=\frac{300}{14}\approx21.429, \qquad n_2=\frac{100(0.4)}{2.8}=\frac{100}{7}\approx14.286.

The apparent discrepancy is resolved by noting that $n_1=300/14=150/7$, not $150/14$. The cost is $4(150/7)+100/7=700/7=100$. The variance is

\frac{1.2^2}{150/7}+\frac{0.4^2}{100/7} =\frac{1.44(7)}{150}+\frac{0.16(7)}{100} =0.0672+0.0112=0.0784.

The more expensive stratum receives more observations because its weighted variability is larger, but its allocation is moderated by its higher cost. The formula does not simply allocate in proportion to $W_h\sigma_h/\sqrt{c_h}$; that quantity is proportional to $n_h\sqrt{c_h}$, not to $n_h$ itself.

Minimum Sample Sizes

Some designs require at least $m_h$ observations in each stratum, for example to ensure adequate representation. The constrained problem is to minimize the same variance subject to $n_h\geq m_h$ and $\sum_h c_hn_h=C$. It is feasible only if $C\geq\sum_h c_hm_h$.

Theorem: For positive $a_h,c_h$, the continuous optimum with lower bounds $n_h\geq m_h$ has the form $$ n_h=\max\left\{m_h,\frac{k\,a_h}{\sqrt{c_h}}\right\}, $$ where $k>0$ is chosen so that $\sum_h c_hn_h=C$.

To justify this form, consider the Lagrangian

\mathcal L(n,\lambda,\nu) =\sum_h\frac{a_h^2}{n_h} +\lambda\left(\sum_hc_hn_h-C\right) +\sum_h\nu_h(m_h-n_h),

where $\nu_h\geq0$ are the multipliers for the lower bounds. At an unconstrained stratum, $n_h>m_h$, so $\nu_h=0$. Stationarity gives $-a_h^2/n_h^2+\lambda c_h=0$, hence $n_h=a_h/(\sqrt{\lambda}\sqrt{c_h})$. At a stratum fixed at its minimum, $n_h=m_h$, and the one-sided optimality condition is $-a_h^2/m_h^2+\lambda c_h\geq0$. This is equivalent to $m_h\geq a_h/(\sqrt{\lambda}\sqrt{c_h})$. Taking $k=1/\sqrt{\lambda}$ gives the maximum formula. The value of $k$ is determined by the budget; the left side of the budget equation is continuous and nondecreasing in $k$, and increases once any stratum is above its minimum. Thus the budget can be met whenever it exceeds the minimum required cost. $\square$

In practice, one can increase $k$ until the budget is met, fixing at their minima the strata whose unconstrained allocations fall below those minima and reallocating among the others. The minima are not handled by adding an unconstrained allocation of the leftover budget to every minimum: that procedure can over-allocate a stratum already at its lower bound.

Worked Example: Checking a Minimum-Size Constraint

Let $a_1=a_2=c_1=c_2=1$, with minimum sizes $m=(9,1)$ and budget $C=20$. Without the minima, equal costs and equal $a_h$ give $n_1=n_2=10$. This allocation already satisfies both lower bounds, so it remains optimal and has variance

\frac{1}{10}+\frac{1}{10}=0.2.

A tempting but incorrect procedure would first reserve the minimums, leaving $20-(9+1)=10$, and then divide those 10 observations equally on top of the minimums. It yields $(14,6)$, with variance $1/14+1/6=10/42\approx0.2381$, which is worse than $0.2$. The correct maximum formula instead gives $(10,10)$, since the unconstrained allocation is above both minima.

If the first minimum were raised to 12 while the budget remained 20, the unconstrained allocation $(10,10)$ would violate that bound. The constrained optimum would be $(12,8)$: the first stratum is fixed at its minimum, and the remaining budget goes to the second. Its variance is $1/12+1/8=5/24\approx0.2083$. This illustrates why binding minima must be incorporated in the optimization rather than added to a fresh unconstrained allocation.

Continuous Solutions and Integer Designs

The formulas above solve a continuous optimization problem, so they can produce noninteger sample sizes. A real study requires integer $n_h$, and may also require that the budget be met exactly or not exceeded. Rounding each continuous allocation independently can violate the budget, while a rounded allocation need not be the best feasible integer allocation.

For equal unit costs, a useful exact discrete perspective is the variance reduction from adding one observation to stratum $h$ when its current size is $n_h$:

\frac{a_h^2}{n_h}-\frac{a_h^2}{n_h+1} =\frac{a_h^2}{n_h(n_h+1)}.

With equal costs and a fixed number of observations, repeatedly adding an observation where this reduction is largest gives an optimal integer allocation, subject to any minimum sizes. With unequal costs, each addition has a different cost, and choosing the largest reduction per unit cost is not in general guaranteed to solve the exact discrete budget problem. Integer designs can instead be compared over feasible allocations or solved as a discrete optimization problem. The continuous optimum remains a useful guide, not a substitute for checking feasibility.

Why the Allocation Rule Matters

Optimal allocation makes explicit the trade-off hidden in a fixed sampling budget. A large population weight raises the contribution of a stratum's estimation error to the overall mean; high within-stratum variability makes that error harder to control; and high sampling cost makes additional precision more expensive. The rule balances all three.

With equal costs, allocation is proportional to $W_h\sigma_h$.
With unequal costs, the continuous allocation is proportional to $W_h\sigma_h/\sqrt{c_h}$, with the common scaling fixed by the total cost.
Minimum sample sizes change the solution only for strata whose unconstrained allocations fall below their required minima; the remaining strata must then absorb the budget consistently with the cost constraint.

A common mistake is to use $W_h\sigma_h/\sqrt{c_h}$ as the full formula without checking the cost constraint, or to add a proportional allocation to minimum sample sizes. The first can spend the wrong amount because allocations must be scaled with the costs $c_h$; the second can assign too many observations to a stratum whose minimum already binds. Substitute the proposed allocation into the budget and verify all lower bounds before treating it as a design.

Key takeaway: For independent strata, minimizing the variance of a weighted mean under a fixed cost budget gives $n_h$ proportional to $W_h\sigma_h/\sqrt{c_h}$. Under minimum-size constraints, use the maximum of the minimum and this scaled allocation, with the scale chosen to satisfy the budget.

Check Your Understanding

Work through these questions by identifying the variance contribution and the budget constraint in each case.

Why does the variance of a weighted mean contain $W_h^2\sigma_h^2/n_h$ for stratum $h$? State the role of independence.
With equal costs, which stratum receives more observations: one with larger $W_h\sigma_h$ or one with smaller $W_h\sigma_h$? Explain.
For unequal costs, why is the continuous allocation not simply proportional to $W_h\sigma_h/\sqrt{c_h}$ without an additional scaling factor?
Suppose a continuous allocation violates one minimum sample size. What must be adjusted, and how is the scale $k$ determined?
Why can independently rounding all continuous sample sizes produce an infeasible design?