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Two random variables have negative covariance when $\mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)]<0$, assuming finite second moments. Opposite deviations from their respective expected values then outweigh matching deviations on average. For example, if $Y=c-X$ for a constant $c$, the bilinear identities of covariance give $\mathrm{Cov}(X,Y)=-\mathrm{Var}(X)$, which is strictly negative when $X$ has positive variance.^[ref]

Sampling without replacement gives a useful example in which the negative sign comes from a shared finite resource. Choose a simple random sample $S$ of exactly $k$ distinct members from a statistical population of size $N$, with $0<k<N$. For each member define the inclusion indicator $I_i=1_{\{i\in S\}}$. Its inclusion probability is $p=k/N$. For different members $i,j$, the joint inclusion probability isSubtracting the product of the expected values givesThe Pearson correlation coefficient follows because each inclusion indicator has variance $p(1-p)$. Once member $i$ is included, only $k-1$ places remain for the other $N-1$ members, reducing member $j$'s conditional probability of inclusion below $k/N$.

For an original exact example, draw two members from four. The six unordered sample pairs are equiprobable. For $I_1,I_2$, the outcomes $(0,0)$ and $(1,1)$ each have probability $1/6$, while $(0,1)$ and $(1,0)$ each have probability $1/3$. Hence $\mathrm{Cov}(I_1,I_2)=1/6-1/4=-1/12$, and their Pearson correlation coefficient is $-1/3$. Matching outcomes remain possible despite the negative covariance.

A fixed total alone does not force every pair among three or more random variables to have negative covariance. If ${\sum }_j{X}_j=c$, thenThis constrains the sum of the off-diagonal entries in a row of the covariance matrix. It does not fix each entry's sign. For a counterexample, let $Z$ be equally likely to be zero or one, and set $X_1=Z$, $X_2=Z$, $X_3=2-2Z$. Their total is always two, yet $\mathrm{Cov}(X_1,X_2)=\mathrm{Var}(Z)=1/4>0$.

In the simple random sample, exchangeability makes the off-diagonal covariances equal, so the fixed-total identity does determine each of them: $(N-1)\mathrm{Cov}(I_i,I_j)=-p(1-p)$. The same distinction between a linear constraint and individual pair signs matters for the joint dependence of leave-one-out consensus residuals.

For a sample or finite statistical population of positive size $n$, the relative frequency of a category represented $n_i$ times is $n_i/n$. It is a proportion and differs from an oscillation frequency.

Sampling bias is a systematic distortion arising when the sampling mechanism makes the observed sample unrepresentative of the target statistical population for the estimand of interest. It is distinct from random sampling variability and from other causes of the bias of an estimator.

Suppose a population has equal-sized groups with values zero and ten, so its mean is five. If independent sampling with replacement chooses the first group with probability $0.9$, the sample mean has expectation one and converges to one as the sample grows. More data reduce its random variation without removing the selection bias. Known selection probabilities can support a design-based correction, but a group never sampled cannot be recovered merely by assigning larger weights to observed units.

A sampling frame is the operational list or other device identifying the units from which a statistical sample can be selected. The intended statistical population, the sampling frame and the responding units can differ. A register of residential addresses, for example, can serve as a sampling frame for a household survey, but may miss newly built homes and include demolished buildings.

If 100 of 1000 target households are absent from the sampling frame, randomly selecting 90 of the 900 listed households gives each listed household inclusion probability $0.1$ and each omitted household inclusion probability zero. Increasing the sample among listed households cannot by itself correct that lack of coverage. Duplicate entries create a different problem: units with more entries can be more likely to be selected. A sound design therefore documents how frame units map to population units and distinguishes coverage errors from sampling variability.

A statistical population is the collection of units, outcomes or distributional cases about which an analysis seeks conclusions. A finite population may consist of $N$ fixed values $y_1,\dots ,y_N$, with population mean ${\mu }_N=N^{-1}{\sum }_i{y}_i$. A superpopulation model instead describes observations through a probability distribution.

These viewpoints support different sources of randomness. In sampling from a fixed finite population, selection is random even if each value is fixed. Under simple random sampling without replacement, every unit has inclusion probability $n/N$, and the sample mean is unbiased for ${\mu }_N$. The accessible sampling frame can differ from the target population, producing a coverage problem that a large sample alone cannot repair.

Statistical sampling selects observations from a target statistical population through an explicit selection procedure or statistical model. It links observed values to the larger collection about which conclusions are sought. The accessible sampling frame and the statistical population need not coincide, which can cause sampling bias even in a large statistical sample.

For a finite statistical population with values $y_1,\dots ,y_N$, suppose every unit has known positive inclusion probability ${\pi }_i$. If $s$ is the selected statistical sample, thenis an unbiased estimator of the total $Y={\sum }_i{y}_i$. Indeed, writing the sum with inclusion indicator functions gives $\mathbb{E}\hat{Y}={\sum }_i{y}_i{\pi }_i/{\pi }_i=Y$. Under simple random sampling without replacement of size $n$, ${\pi }_i=n/N$, so $\hat{Y}/N$ is the sample mean. The selection procedure also determines dependence between selected units and the resulting sampling variability; unbiasedness alone does not specify its size.