Suppose population size \(N\) is exactly divisible by sample size \(n\), so \(N = nk\) where \(k\) is the sampling interval. Choose a random integer \(r\) (called the random start) between 1 and \(k\), then select units numbered
\[ r,\; r + k,\; r + 2k,\; \ldots,\; r + (n-1)k. \]This gives a sample of size \(n\). Notation: SYS\((N, n, k)\).
Example. A company with \(N = 5000\) customers wants a sample of \(n = 100\). The sampling ratio is \(n/N = 100/5000 = 1/50\), so the interval is \(k = 50\): one customer in every 50. A random start is drawn from 1 to 50, say 11, and the sample is customers 11, 61, 111, 161, …, up to \(11 + 99 \times 50 = 4961\). Every unit after the first is fixed by the interval.
Notation. Write \(y_{ij}\) for the \(j\)th unit of the \(i\)th systematic sample (\(i = 1, \ldots, k\) is the random start, \(j = 1, \ldots, n\)). Then
\[ \bar y_{i.} = \frac1n\sum_{j=1}^n y_{ij}, \qquad \bar y_{..} = \frac{1}{nk}\sum_{i=1}^k\sum_{j=1}^n y_{ij} = \frac1k\sum_{i=1}^k \bar y_{i.}, \] \[ S^2 = \frac{1}{nk-1}\sum_{i=1}^k\sum_{j=1}^n (y_{ij} - \bar y_{..})^2 . \]The population mean \(\bar Y\) is written \(\bar y_{..}\) here, and \(\bar y_{sys}\) is the \(\bar y_{i.}\) of whichever start was drawn.
| Random start | Units in the sample | Probability | Mean |
|---|---|---|---|
| 1 | \(1,\; 1+k,\; \ldots,\; 1+(n-1)k\) | \(1/k\) | \(\bar y_{1.}\) |
| 2 | \(2,\; 2+k,\; \ldots,\; 2+(n-1)k\) | \(1/k\) | \(\bar y_{2.}\) |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| \(i\) | \(i,\; i+k,\; \ldots,\; i+(n-1)k\) | \(1/k\) | \(\bar y_{i.}\) |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| \(k\) | \(k,\; 2k,\; \ldots,\; nk\) | \(1/k\) | \(\bar y_{k.}\) |
Each row is one possible sample, and each of the \(N\) units lies in exactly one row, so every unit has the same chance \(1/k = n/N\) of selection. Each column is a block of \(k\) consecutive units, so the columns form \(n\) strata. Systematic sampling takes the unit in the same position from every stratum; stratified sampling with one unit per stratum would choose the position afresh in each stratum. This picture is behind every comparison below.
The textbook says systematic sampling “may be more efficient than simple random sampling provided the sampling frame is arranged totally at random”, for example an alphabetical list. A list in random order makes systematic sampling equivalent to SRSWOR on average, not better: averaged over all \(6! = 720\) orderings of the values 1, 2, 4, 7, 11, 16 (\(n = 2\), \(k = 3\)), \(\text{Var}(\bar y_{sys})\) is exactly \(11.12\), the SRSWOR variance. The gain comes when the list is ordered by something related to \(y\) (a trend, §1.3). A periodic order can make it much worse: for the values 10, 20, 30, 20 repeated three times (\(N = 12\), \(k = 4\)), every systematic sample consists of one value repeated, and \(\text{Var}(\bar y_{sys}) = 50\) against \(13.64\) for SRSWOR.
Systematic sampling is also not a simple random sample in the strict sense: only the first unit is random, and most of the \(\binom Nn\) subsets can never be drawn.
The systematic sample mean
It is unbiased: \(E(\bar y_{sys}) = \bar Y\).
The random start \(i\) takes each value \(1, \ldots, k\) with probability \(1/k\), so
\[ E(\bar y_{sys}) = \frac1k\,\bar y_{1.} + \frac1k\,\bar y_{2.} + \cdots + \frac1k\,\bar y_{k.} = \frac1k\sum_{i=1}^k \bar y_{i.} = \bar y_{..} = \bar Y . \]This needs \(N = nk\): then all \(k\) samples have \(n\) units and the average of their means is the population mean.
Let \(S_{wsy}^2\) be the variance within systematic samples. Then
Equivalently, variance can be expressed via intra-class correlation:
\[ \text{Var}(\bar y_{sys}) \;=\; \dfrac{N-1}{N}\cdot\dfrac{S^2}{n}\bigl[1 + (n-1)\rho\bigr], \]where \(\rho\) is the intra-class correlation among units of the same systematic sample.
Since \(\bar y_{sys}\) takes the value \(\bar y_{i.}\) with probability \(1/k\),
\[ \text{Var}(\bar y_{sys}) = E(\bar y_{i.} - \bar y_{..})^2 = \frac1k\sum_{i=1}^k (\bar y_{i.} - \bar y_{..})^2 . \qquad (1) \]Split each deviation as \(y_{ij} - \bar y_{..} = (y_{ij} - \bar y_{i.}) + (\bar y_{i.} - \bar y_{..})\). The cross term vanishes, because \(\sum_j (y_{ij} - \bar y_{i.}) = n\bar y_{i.} - n\bar y_{i.} = 0\) for each \(i\). Hence
\[ (N-1)S^2 = \sum_{i}\sum_{j}(y_{ij} - \bar y_{i.})^2 + n\sum_i (\bar y_{i.} - \bar y_{..})^2 = k(n-1)S_{wsy}^2 + nk\,\text{Var}(\bar y_{sys}), \]where \(S_{wsy}^2 = \frac{1}{k(n-1)}\sum_i\sum_j (y_{ij} - \bar y_{i.})^2\) is the mean square within systematic samples. Dividing by \(N = nk\),
\[ \text{Var}(\bar y_{sys}) = \frac{N-1}{N}\,S^2 - \frac{k(n-1)}{N}\,S_{wsy}^2 . \]The \(S^2\) terms combine as \(\frac{N - n - Nn + n}{Nn}S^2 = -\frac{n-1}{n}S^2\), and \(\frac{k(n-1)}{N} = \frac{n-1}{n}\), so
\[ \text{Var}_{SRS} - \text{Var}_{sys} = \frac{n-1}{n}\big(S_{wsy}^2 - S^2\big) . \]Systematic sampling is more precise than SRSWOR exactly when \(S_{wsy}^2 > S^2\): when the units within a systematic sample are more varied than the population as a whole. Spreading the sample across the list is what achieves this.
Let \(\rho\) be the correlation between pairs of units in the same systematic sample,
\[ \rho = \frac{\sum_i \sum_{j \ne j'} (y_{ij} - \bar y_{..})(y_{ij'} - \bar y_{..})}{(n-1)(nk-1)S^2} . \]From (1), \(n^2k\,\text{Var}(\bar y_{sys}) = \sum_i \big[\sum_j (y_{ij} - \bar y_{..})\big]^2\). Squaring the inner sum gives the squares, which add to \((nk-1)S^2\), and the cross products, which add to \((n-1)(nk-1)\rho S^2\). Hence
\[ \text{Var}(\bar y_{sys}) = \frac{nk-1}{nk}\cdot\frac{S^2}{n}\big[1 + (n-1)\rho\big] . \]The short form \(\frac{S^2}{n}[1 + (n-1)\rho]\), often quoted, drops the factor \(\frac{N-1}{N} = \frac{nk-1}{nk}\); it is the large-\(N\) approximation.
With \(\text{Var}_{SRS} = \frac{N-n}{N}\cdot\frac{S^2}{n} = \frac{k-1}{k}\cdot\frac{S^2}{n}\),
\[ E = \frac{\text{Var}_{SRS}}{\text{Var}_{sys}} = \frac{n(k-1)}{(nk-1)\big[1 + (n-1)\rho\big]} . \]Systematic sampling is the more efficient when \(E > 1\), that is, \(n(k-1) > (nk-1) + (n-1)(nk-1)\rho\), which simplifies to \((n-1)\big[1 + (nk-1)\rho\big] < 0\):
\[ E > 1 \iff \rho < -\frac{1}{nk-1}, \qquad E < 1 \iff \rho > -\frac{1}{nk-1} . \]A negative correlation within samples is what helps: each sample then mixes high and low values. Two special cases: if \(\rho = 1\) (every sample constant), \(E = \frac{k-1}{nk-1}\), the worst possible; and since \(\text{Var}(\bar y_{sys}) \ge 0\), \(\rho\) can be no smaller than \(-\frac{1}{n-1}\), at which value \(\text{Var}(\bar y_{sys}) = 0\) and \(E\) is unbounded.
The textbook states that at \(\rho = -\frac{1}{nk-1}\), \(\text{Var}(\bar y_{sys}) = 0\) and \(E \to \infty\). That value is the boundary just derived, where \(E = 1\) exactly: the two designs are equally precise (with \(n = 3\), \(k = 5\): \(\rho = -1/14\), \(1 + 2\rho = 6/7\), and \(E = 1\)). The variance vanishes only at \(\rho = -\frac{1}{n-1}\), where \(1 + (n-1)\rho = 0\).
Take the \(n\) columns of the table as strata of \(k\) units each, with means \(\bar y_{.j} = \frac1k\sum_i y_{ij}\), mean squares \(S_j^2 = \frac{1}{k-1}\sum_i (y_{ij} - \bar y_{.j})^2\), and pooled within-stratum mean square \(S_{wst}^2 = \frac{1}{n(k-1)}\sum_i\sum_j (y_{ij} - \bar y_{.j})^2\). Stratified sampling with one unit per stratum has \(N_j = k\), \(n_j = 1\), \(W_j = 1/n\), so
\[ \text{Var}(\bar y_{st}) = \sum_{j=1}^n \Big(1 - \frac1k\Big)\frac{S_j^2}{n^2} = \frac{k-1}{n^2k}\sum_j S_j^2 = \frac{k-1}{nk}\,S_{wst}^2 . \]For the systematic sample, \(\bar y_{i.} - \bar y_{..} = \frac1n\sum_j (y_{ij} - \bar y_{.j})\), because \(\bar y_{..}\) is also the average of the column means. Squaring and summing as in Theorem 3, with \(\rho_{wst}\) the correlation between deviations from the stratum means within the same systematic sample,
\[ \rho_{wst} = \frac{\sum_i\sum_{j \ne j'} (y_{ij} - \bar y_{.j})(y_{ij'} - \bar y_{.j'})}{(n-1)\,n(k-1)\,S_{wst}^2}, \] \[ \text{Var}(\bar y_{sys}) = \frac{k-1}{nk}\,S_{wst}^2\big[1 + (n-1)\rho_{wst}\big], \qquad \frac{\text{Var}(\bar y_{st})}{\text{Var}(\bar y_{sys})} = \frac{1}{1 + (n-1)\rho_{wst}} . \]So: \(\rho_{wst} > 0\) favours stratified sampling, \(\rho_{wst} = 0\) makes the two equal, and \(\rho_{wst} < 0\) favours systematic sampling. (The textbook states the first two cases only.)
If the population values follow a linear trend \(Y_i = a + bi\), the variances simplify to:
Variance ordering for linear trend:
Stratified (one unit per stratum) is best for a linear trend and SRSWOR is worst; systematic sits in between. In fact \(\text{Var}_{SYS} = n\,\text{Var}_{StRS,prop}\) and \(\text{Var}_{SRSWOR} = \tfrac{N+1}{k+1}\text{Var}_{SYS}\), so systematic sampling always beats SRSWOR here — spreading the sample evenly across the ordered trend captures its full range better than a purely random draw.
Since populations rarely follow an exact linear trend, in practice systematic ≈ stratified for many real data sets. (Systematic is only poor when the population is periodic with period equal to \(k\).)
Take \(Y_i = i\), \(i = 1, \ldots, N\) (any trend \(a + bi\) multiplies every variance by \(b^2\)). Then \(\bar Y = \frac{N+1}{2}\) and
\[ S^2 = \frac{1}{N-1}\Big[\frac{N(N+1)(2N+1)}{6} - N\Big(\frac{N+1}{2}\Big)^2\Big] \] \[ = \frac{N(N+1)}{N-1}\cdot\frac{N-1}{12} = \frac{N(N+1)}{12} . \]SRSWOR. \(\text{Var}_{SRS} = \frac{N-n}{Nn}S^2 = \frac{(k-1)(nk+1)}{12}\).
Stratified. Each stratum is \(k\) consecutive integers, so \(S_j^2 = \frac{k(k+1)}{12}\), and \(\text{Var}_{st} = \frac{k-1}{n^2k}\cdot n\,\frac{k(k+1)}{12} = \frac{k^2-1}{12n}\).
Systematic. The \(i\)th sample is \(i, i+k, \ldots, i+(n-1)k\), so
\[ \bar y_{i.} = \frac1n\big[ni + (1 + 2 + \cdots + (n-1))k\big] = i + \frac{(n-1)k}{2}, \]and \(\bar y_{..} = \frac{nk+1}{2}\), so \(\bar y_{i.} - \bar y_{..} = i - \frac{k+1}{2}\): the \(k\) sample means are equally spaced, one unit apart. Their variance is that of \(1, \ldots, k\):
\[ \text{Var}_{sys} = \frac1k\sum_{i=1}^k\Big(i - \frac{k+1}{2}\Big)^2 = \frac{k^2-1}{12} . \]Dividing all three by \(\frac{k-1}{12}\):
\[ \text{Var}_{st} : \text{Var}_{sys} : \text{Var}_{SRS} = \frac{k+1}{n} : (k+1) : (nk+1), \]which for large \(k\) is about \(\frac1n : 1 : n\). Stratified sampling removes the trend best, and systematic sampling comes next. (Check with \(N = 15\), \(n = 3\), \(k = 5\): the three variances are \(\frac23\), \(2\) and \(\frac{16}{3}\), in the ratio \(2 : 6 : 16 = \frac63 : 6 : 16\).)
| Aspect | SRSWOR | Stratified | Systematic |
|---|---|---|---|
| Sample selection | Random | Random in each stratum | One random start, every k-th |
| Variance for homogeneous pop. | Same | Same | Same |
| Variance with linear trend | Worst | Best | Intermediate (beats SRSWOR) |
| Variance with periodic pop. | Same | Same | Can be very bad |
| Operational ease | Moderate | Hard (frame, allocation) | Easiest |
From \(N = 50\) employees, draw a sample of 10 by systematic sampling. \(k = 5\); random start \(r = 3\); sample = {3, 8, 13, 18, 23, 28, 33, 38, 43, 48}.
For population values \(1, 2, 3, \ldots, 12\) (\(N = 12, n = 4, k = 3, b = 1\)):
\(\text{Var}_{SYS} = (3^2 - 1)/12 = 8/12 = 0.667\).
\(\text{Var}_{SRSWOR} = (12-4)(1)(13)/(12 \cdot 4) = 104/48 = 2.167\).
\(\text{Var}_{StRS,prop} = 8/(12 \cdot 4) = 0.167\).
So Stratified < Systematic < SRSWOR for this linear trend (note systematic is BETTER than SRSWOR here because the linear trend's variance reduction within systematic samples dominates).
The textbook's problem, worked by both routes (the formulas of Theorems 2 and 4, and directly from the \(k\) possible samples), then its exercise with the answers checked.
Source note. Every figure was recomputed exactly. Worked Problem 1 agrees with the textbook up to rounding; in the exercise, the textbook's systematic variance is wrong and is corrected below.
A population of \(N = 15\) units forms \(k = 5\) systematic samples of \(n = 3\) units (rows); the columns are the 3 strata:
| Sample \(i\) | Stratum 1 | Stratum 2 | Stratum 3 | Total | \(\bar y_{i.}\) | \(\sum_j y_{ij}^2\) |
|---|---|---|---|---|---|---|
| 1 | 1 | 4 | 8 | 13 | 4.33 | 81 |
| 2 | 6 | 12 | 20 | 38 | 12.67 | 580 |
| 3 | 4 | 8 | 16 | 28 | 9.33 | 336 |
| 4 | 3 | 10 | 18 | 31 | 10.33 | 433 |
| 5 | 4 | 12 | 24 | 40 | 13.33 | 736 |
| Column total | 18 | 46 | 86 | 150 | 2166 | |
| \(\bar y_{.j}\) | 3.6 | 9.2 | 17.2 |
Find the variance of the sample mean under SRSWOR, stratified sampling (one unit per stratum) and systematic sampling, and compare their efficiencies.
Population constants. \(\bar y_{..} = 150/15 = 10\), and
\[ S^2 = \frac{1}{nk-1}\Big[\sum\sum y_{ij}^2 - nk\,\bar y_{..}^2\Big] = \frac{2166 - 15(10)^2}{14} = \frac{666}{14} = 47.57 . \]SRSWOR. \(\text{Var}_{SRS} = \frac{N-n}{N}\cdot\frac{S^2}{n}\) \(= \frac{12}{15 \times 3}(47.57) = 12.69\).
Stratified. \(\sum_j \bar y_{.j}^2 = 12.96 + 84.64 + 295.84 = 393.44\), so
\[ S_{wst}^2 = \frac{1}{n(k-1)}\Big[\sum\sum y_{ij}^2 - k\sum_j \bar y_{.j}^2\Big] = \frac{2166 - 5(393.44)}{12} = \frac{198.8}{12} = 16.57, \]and \(\text{Var}_{st} = \frac{k-1}{nk}S_{wst}^2 = \frac{4}{15}(16.57) = 4.42\).
Systematic. \(\sum_i \bar y_{i.}^2 = (13^2 + 38^2 + 28^2 + 31^2 + 40^2)/9\) \(= 4958/9 = 550.89\), so
\[ S_{wsy}^2 = \frac{1}{k(n-1)}\Big[\sum\sum y_{ij}^2 - n\sum_i \bar y_{i.}^2\Big] = \frac{2166 - 3(550.89)}{10} = 51.33, \] \[ \text{Var}_{sys} = \frac{nk-1}{nk}S^2 - \frac{n-1}{n}S_{wsy}^2 = \frac{14}{15}(47.57) - \frac23(51.33) = 10.18 . \]Directly. The five possible means deviate from 10 by \(-5.67, 2.67, -0.67, 0.33, 3.33\); their squares add to 50.89, and \(\text{Var}_{sys} = 50.89/5 = 10.18\) (exactly \(458/45\)), as Theorem 2 says.
Comparison. \(\text{Var}_{st} = 4.42\) < \(\text{Var}_{sys} = 10.18\) < \(\text{Var}_{SRS} = 12.69\). Stratified sampling is the most efficient (2.87 times SRSWOR), then systematic (1.25 times SRSWOR). The correlations explain it: within systematic samples \(\rho = -0.156\), below the boundary \(-1/(nk-1) = -0.071\), so systematic beats SRSWOR; but \(\rho_{wst} = 0.652 > 0\), so stratified beats systematic.
Rounding note. The textbook rounds the sample means before squaring and gets \(\sum \bar y_{i.}^2 = 550.95\), \(S_{wsy}^2 = 51.315\) and \(\text{Var}_{sys} = 10.19\); at full precision they are 550.89, 51.33 and 10.18.
| Stratum | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| I | 0 | 1 | 1 | 2 | 5 | 4 | 7 | 7 | 8 | 6 |
| II | 6 | 8 | 9 | 10 | 13 | 12 | 15 | 16 | 16 | 17 |
| III | 18 | 19 | 20 | 20 | 24 | 23 | 25 | 28 | 29 | 27 |
| IV | 26 | 30 | 31 | 31 | 33 | 32 | 35 | 37 | 38 | 38 |
The population is divided into groups called clusters; an SRS of clusters is selected and every unit within the chosen cluster is observed (single-stage cluster sampling).
Clusters are usually natural groupings like villages, schools, blocks, or households.
| Stratified | Cluster | |
|---|---|---|
| Internal variation | Strata homogeneous within (low \(S_h^2\)) | Clusters heterogeneous within |
| Between groups | Strata differ (high between) | Clusters similar to each other |
| Sample from | Every stratum | Some clusters only |
| Goal | Reduce variance | Reduce cost |
A cluster sample is generally less precise than SRS of the same total sample size (when clusters are formed by proximity) but is much cheaper to administer.
To study primary schools in a district: list the schools, randomly select 20 schools, and survey all teachers and students in those 20 schools.
National Sample Survey rural blocks: villages are clusters; a random sample of villages is selected and all households in the chosen villages are surveyed.
Sampling done in stages. At the first stage, primary sampling units (PSUs, e.g., districts) are selected. Within each selected PSU, secondary sampling units (SSUs, e.g., villages) are selected. The process can continue further — three-stage, four-stage etc.
Indian Census-NSSO style: select states (1st stage), districts within states (2nd stage), villages within districts (3rd stage), households within villages (4th stage).
Education survey: select schools (PSU), then classes within schools (SSU), then students within classes.
The investigator is given quotas from each subgroup (gender, age, etc.) to fill, but is free to choose which individuals fill those quotas. It is a non-probability method.
Market researcher told: "interview 30 working women aged 25–34". Picks any 30 willing women — no random selection.
Exit poll instructions: "interview 50 men and 50 women voters from each constituency". No random scheme — interviewer picks who is convenient.
In PPS sampling, units are selected with probabilities proportional to a known size measure \(X_i\) (e.g. villages chosen with probability proportional to population, or shops to floor area). Large units — which usually contribute most to the total — get a higher chance of selection, improving efficiency when \(Y_i\) is roughly proportional to \(X_i\).
Selection probability of unit \(i\): \( \pi_i = X_i / \sum_{k} X_k. \)
which is unbiased for the population total \(Y\); dividing the summand by the size restores an unbiased scale.
Selection methods: (i) Cumulative-total method — cumulate the sizes, draw a random number in \((0, \sum X_k]\), and pick the unit whose cumulative interval contains it; (ii) Lahiri's method — a rejection scheme that avoids cumulation.
Four villages with populations 200, 300, 100, 400 (total 1000) have selection probabilities \(0.20, 0.30, 0.10, 0.40\). Cumulative totals are 200, 500, 600, 1000; a random number 640 falls in the interval \((600, 1000]\), so the fourth village is selected.