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Consider a finite population consisting of $N$ elements $y_1, y_2, \cdots, y_N$. Throughout the paper we will assume that $N = nk$. A systematic sample of $n$ elements is drawn by choosing one element at random from the first $k$ elements $y_1, \cdots, y_k$, and then selecting every $k$th element thereafter. Let $y_{ij} = y_{i + (j - 1)k}(i = 1, \cdots, k; j = 1, \cdots, n)$; obviously systematic sampling is equivalent to selecting one of the $k$ "clusters" $$C_i = \{y_{ij}; j = 1, \cdots, n\}$$ at random. From this it follows that the sample mean $\bar y_i = 1/n \sum^n_{j = 1} y_{ij}$ is an unbiased estimate for the population mean $\bar y = 1/N \sum^k_{i = 1} \sum^n_{j = 1} y_{ij}$ and that $\operatorname{Var} \bar y_i = 1/k \sum^k_{i = 1} (\bar y_i - \bar y)^2$. We will denote this variance by $V^{(1)}_{sy}$ indicating by the superscript that only one cluster is selected at random. $V^{(1)}_{sy}$ can be written as \begin{equation*}\tag{1}V^{(1)}_{sy} = S^2 - \frac{1}{k} \sum^k_{i = 1} S^2_i, \text{where} S^2 = \frac{1}{N} \sum^k_{i = 1} \sum^n_{j = 1} (y_{ij} - \bar y)^2,\end{equation*} \\ \begin{equation*} S^2_i = \frac{1}{n} \sum^n_{j = 1} (y_{ij} - \bar y_i)^2.\end{equation*} It is natural to compare systematic sampling with stratified random sampling, where one element is chosen independently in each of the $n$ strata $\{y_1, \cdots, y_k\}, \{y_{k + 1}, \cdots, y_{2k}\}, \cdots$, and with simple random sampling using sample size $n$. The corresponding variances of the sample mean will be denoted by $V^{(1)}_{st} V^{(n)}_{ran}$ respectively. We consider now the following generalization of systematic sampling which appears to have been suggested by J. Tukey (see [3], p. 96, [4], [5]). Instead of choosing at first only one element at random we select a simple random sample of size $s$ (without replacement) from the first $k$ elements and then every $k$th element following those selected. In this way we obtain a sample of $ns$ elements and, if $i_1, i_2, \cdots, i_s$ are the serial numbers of the elements first chosen, the sample mean $1/s(\bar y_{i_1} + \cdots + \bar y_{i_s})$ can be used as an estimate for the population mean. This sampling procedure is clearly equivalent to drawing a simple random sample of size $s$ from the $k$ clusters $C_i(i = 1, \cdots, k)$. It therefore follows (see, for example, [2], Chapter 2.3 to 2.4) that the sample mean is an unbiased estimate for the population mean and that its variance, which we denote by $V^{(s)}_{sy}$, is given by begin{equation*} \tag{2}V^{(s)}_{sy} = \frac{k - s}{ks} \frac{1}{k - 1} \sum^k_{i = 1} (\bar y_i - \bar y)^2 = \frac{1}{s} \frac{k - s}{k - 1} V^{(1)}_{sy}.\end{equation*} Again, it is natural to compare this sampling procedure with stratified random sampling, where a simple random sample of size $s$ is drawn independently in each of the $n$ strata $\{y_1, \cdots, y_k\}, \{y_{k + 1}, \cdots, y_{2k}\}, \cdots$ or with simple random sampling employing sample size $ns$. We denote the corresponding variances of the sample mean (which in both cases is an unbiased estimate for the population mean) by $V^{(s)}_{st} ,V^{(ns)}_{ran}$ respectively. From well-known variance formulae (see, for example, [2], Chapters 2.4 and 5.3) it follows that \begin{equation*}\tag{3}V^{(s)}_{st} = \frac{1}{s} \frac{k - s} {k - 1} V^{(1)}_{st},\\ V^{(ns)}_{ran} = \frac{N - ns}{s(N - n)} V^{(n)}_{ran} = \frac{1}{s} \frac{k - s}{k - 1} V^{(n)}_{ran}. \end{equation*} Thus the relative magnitudes of the three variances $V^{(s)}_{sy}, V^{(s)}_{st}, V^{(ns)}_{ran}$ are the same as for $V^{(1)}_{sy}, V^{(1)}_{st}, V^{(n)}_{ran}$, of which comparisons were made for several types of populations by W. G. Madow and L. H. Madow [6] and W. G. Cochran [1]. Some of the results will be reviewed in Section 3. The object of this note is to compare systematic sampling with $s$ random starts, as described above, with systematic sampling employing only one random start but using a sample of the same size $ns$. To make this comparison we obviously have to assume that $k$ is an integral multiple of $s$, say $k = ls$. The latter procedure then consists in choosing one element at random from the first $l$ elements $\{y_1, \cdots, y_l\}$ and selecting every $l$th consecutive element. We denote the variances of the sample mean of the two procedures by $V^{(s)}_k, V^{(1)}_l$ respectively, indicating by the subscript the size of the initial "counting interval." (In our notation $V^{(s)}_{sy} \equiv V^{(s)}_k$.) We shall show in Section 4 that $V^{(1)}_l = V^{(s)}_k$ in the case of a population "in random order," but $V^{(1)}_l < V^{(s)_k$ for a population with a linear trend or with a positive correlation between the elements which is a decreasing convex function of their distance apart. Some numerical results on the relative precision of the two procedures will be given in Section 5 for the case of a large population with an exponential correlogram.

The purpose of this note is to present a lemma which will settle a question of completeness left open in Section 6 of the above mentioned paper [5]. We give two applications of the lemma, (i) by proving that, in addition to Herbach's results, also the standard $F$-test for $\sigma^2_{ab} = 0$ is a uniformly most powerful similar test, (ii) by pointing out that the standard form introduced in [5] together with our lemma provide convenient tools to prove that in a balanced model II design (with the usual normality assumptions) the standard estimates of variance components are minimum variance unbiased. This result is well known ([2], [3]) and it has in fact been pointed out by Graybill and Wortham [3] that a completeness argument may be used to demonstrate the minimum variance property of the usual estimators for the variance components. The present lemma shows that the estimators do indeed have the necessary completeness property. We will follow Herbach's notation throughout.