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Return to pairwise independence does not imply joint independence, where $X,Y$ are independent fair bits and $Z=X\oplus Y$ is their exclusive or. Given $Z=0$, only $000$ and $110$ remain, each with conditional probability $1/2$. Given $Z=1$, only $011$ and $101$ remain, again each with conditional probability $1/2$. Consequently,The conditional marginal distributions are still fair, so their product has four entries $1/4$ in either case. The actual tables differ: $Y=X$ given $Z=0$, and $Y=1-X$ given $Z=1$. Thus unconditional independence need not survive conditioning. Observing whether two bits agree restricts their possible combinations; no physical interaction between the original trials is required to produce that information constraint.

Let $X,Y$ be independent fair bits and set $Z=X\oplus Y$, their exclusive or. The four possible triples are $000,011,101,110$, each with probability $1/4$. Every coordinate is fair, and every pair has all four outcomes with equal probability. Thus each pair's mutual information is zero, as in pairwise independence does not imply joint independence.

The full random vector has joint entropy ${\log }_{2}4=2$ bits, while the marginal information entropies sum to three bits. ConsequentlyThe product of the marginals predicts probability $1/8$ for every triple, including the four impossible ones. On each possible triple, the ratio of true to product probability is $(1/4)/(1/8)=2$; averaging its base-two logarithm gives the same one bit of relative entropy.

The example gives a concrete use for joint information measures. A lossless source coding scheme that treats the three coordinates separately needs three bits here. If the decoder knows the integer parity rule, the encoder can send just $X,Y$ and let the decoder reconstruct $Z$ exactly. Pairwise independence does not reveal that saving. This is a stipulated source law and an explicit two-bit code, rather than a promise that arbitrary data can be shortened.

The parity construction of pairwise independent bits turns two independent fair bits into three pairwise independent random variables. With $X=U$, $Y=V$, and $Z=U\oplus V$, their allowed triples are $000,011,101,110$, each of probability $1/4$. Every coordinate has the Bernoulli distribution with parameter $1/2$. Every pair has the same joint probability mass function:All twelve pair cells equal the product of the corresponding probabilities in their marginal distributions. This establishes pairwise independent random variables, a requirement involving every pair of value-events, rather than merely one chosen event per coordinate. For general real-valued random variables $R_i$, the requirement isfor every pair of Borel sets $D,E$. For these binary random variables, checking all four cells of each pair suffices: any value-event is a union of the checked cells. In fact, for binary random variables, independence of their one-value events also gives the other cells by taking complements. Siegrist gives the all-set definition and its connection to indicator random variables in Random Services: independence of random variables.

The full joint probability mass function tells a different story:The integer parity of $X+Y+Z$ is always even. Any two bits determine the third: $Z=X\oplus Y$, $X=Y\oplus Z$, and $Y=X\oplus Z$. Each coordinate is independent of either other coordinate individually, but is dependent on the pair together. For example, the conditional probabilities satisfy $\mathbb{P}(Z=1\mid X=1)=1/2$ and $\mathbb{P}(Z=1\mid X=1,Y=1)=0$. See also pairwise independence does not imply joint independence for the general distinction.

The constraint changes the sum's probability distribution without changing its expected value or variance. Let $S=X+Y+Z$. Only $000$ gives $S=0$; each other allowed triple gives $S=2$. Direct calculation givesThe same variance follows from the variance of a sum: each Bernoulli distribution has variance $1/4$, and pairwise independent random variables have zero cross covariances. Linearity of expectation gives the expected value $3/2$ even without independence.

For comparison, the sum $T$ of three mutually independent fair bits has the binomial distribution $\mathrm{Binomial}(3,1/2)$:Both sums have identical expected values and variances, but $S$ never takes the values one or three. Even their third moments differ: $\mathbb{E}[S^3]=6$, whereas $\mathbb{E}[T^3]=27/4$. Pairwise independence preserves these first two sum summaries while allowing a different joint probability distribution and a different sum law. The calculation is an exact finite example.

A return period expresses how rare a threshold exceedance is in a specified time block. Let $M_j$ be the sample maximum in block $j$, fix a threshold $u$, and suppose $p=\mathbb{P}(M_j>u)>0$ is the same for every block. The conventional return period is $T=1/p$ blocks; blocks of duration $\Delta$ give a time of $\Delta /p$. Thus an annual exceedance probability $p=0.01$ means a 100-year return period. This names a frequency scale, not a schedule or a guarantee of one exceedance in each century. The annual convention and the misleading “once every 100 years” interpretation are discussed in USGS: the 100-year flood.

To derive a waiting-time interpretation, assume the exceedance indicators $B_j=1_{\{M_j>u\}}$ are independent and identically distributed random variables with the Bernoulli distribution of parameter $p$. Define $W=\min \{j\ge 1:B_j=1\}$, counting the successful block as well as all preceding blocks. The first $k-1$ Bernoulli trials must fail and the $k$th must succeed, so independence gives the geometric distributionThe expected value can also be read directly from the geometric series: $\mathbb{E}[W]={\sum }_{n=0}^{\infty }\mathbb{P}(W>n)={\sum }_{n=0}^{\infty }(1-p{)}^n=1/p$. This counts blocks from the starting boundary, not the exact time of an exceedance within its block. Counting only failed blocks instead would give expected value $T-1$. The waiting law and its memorylessness of the geometric distribution are developed in Kyle Siegrist, Random Services: the geometric distribution.

For a horizon of $n$ complete years, at least one exceedance is the complement of no exceedances. Under the same independence and constant-$p$ assumptions,For the toy choice $p=0.01$, this gives $9.6\%$ in 10 years, $26.0\%$ in 30 years, $39.5\%$ in 50 years, and $63.4\%$ in 100 years. Even after 100 years there is a $36.6\%$ chance of no exceedance. These numbers are exact model calculations rounded to one decimal place, not measured frequencies. The median waiting count is 69 years, since $1-0.99^{68}<1/2\le 1-0.99^{69}$; the expected value is 100 years. A mean and a median answer different questions.

The exceedance count $N_n={\sum }_{j=1}^n{B}_j$ has the binomial distribution under independence, with expected value $np$. Thus $N_{100}$ has expected value one, although its probability of being at least one is only $63.4\%$: some realizations contribute several exceedances. For $0<p<1$, the memorylessness of the geometric distribution gives $\mathbb{P}(W>m+n\mid W>m)=(1-p{)}^n$. A long quiet spell does not make the next independent year's exceedance overdue, and an exceedance this year does not force a quiet interval. Two specified consecutive years both exceed with probability $p^2=0.0001$ in this toy model.

A reproducible tabletop experiment makes the assumptions tangible. Put 100 identical slips numbered 1 to 100 in a bag; after thoroughly mixing, draw one and record whether it is 100. Replace the slip and mix again before every draw. Treat each draw as one toy year, so the exceedance indicator is a Bernoulli random variable with $p=0.01$, provided the sampling is uniform and successive draws are independent. Repeat complete 30-draw runs and record the fraction containing at least one 100; separately repeat runs stopped at the first 100 and record their waiting counts. The first fraction targets $1-0.99^{30}$; the waiting-count histogram targets $0.99^{k-1}(0.01)$. Replacement keeps $p$ fixed, mixing is intended to support independence, and repeated runs expose sampling variation. This is a proposed demonstration of the toy model; no experimental results are asserted here.

Stationarity supports a time-invariant annual probability distribution, but does not imply independence. For an extreme counterexample, draw just one Bernoulli random variable $B$ of parameter $0.01$ and set $B_j=B$ for every year. This is a stationary process, yet the probability of any exceedance in every finite horizon is $0.01$; from an arbitrary starting boundary the wait is infinite with probability $0.99$. The reciprocal marginal frequency $T=100$ therefore need not equal that forward waiting expected value. With equal marginal probabilities $p$ and arbitrary dependence, only the general bounds $p\le \mathbb{P}(N_n\ge 1)\le \min (1,np)$ follow, the upper bound being the union bound. Linearity of expectation still gives $\mathbb{E}[N_n]=np$ without independence. Mere zero correlation does not restore the product formula; even pairwise independence does not imply joint independence. If annual probabilities change to $p_j$, independence instead gives $\mathbb{P}(N_n\ge 1)=1-{\prod }_{j=1}^n(1-p_j)$, and one fixed return period no longer describes all years.