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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.

In finite-population sampling, a simple random sample of fixed size $n$ without replacement is a subset $S$ of $N$ labelled units drawn so that every size-$n$ subset has probabilityThe sample is the selected subset; simple random sampling is the procedure that generates it. This definition does not require the observed values to be distinct, only the selected population labels. The without-replacement convention is important because an independent repeated draw from a population is a different design.^[ref]

One construction randomly orders all $N$ labels and takes the first $n$. Each particular subset can occupy those positions in $n!(N-n)!$ permutations, out of $N!$, proving the probability above. Equivalently, draw successively and uniformly from the remaining labels, then discard their order.

For $N=5,n=2$, there are $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ possible samples. A particular unit belongs to four of them, so its inclusion probability is $2/5$; a particular pair is one complete sample, so its joint inclusion probability is $1/10$. These are properties of the procedure before observing which subset was obtained.

A simple random sample need not resemble the whole population in every realization. For fixed values $(0,0,0,0,100)$, six of the ten size-two samples contain only zeros. Thus the sample mean is zero with probability $0.6$, although its expected value is the finite population mean $20$. The remaining four samples each have mean $50$, giving $0.6\times 0+0.4\times 50=20$. Unbiasedness concerns repeated selection, rather than a guarantee about any one sample. Coverage of the population labels and faithful implementation of the selection rule remain essential.