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An irreducible recurrent discrete-time Markov chain with at least two states has $P_{ii}<1$ everywhere. Removing repeated consecutive states from a positive-probability path proves irreducibility of its discrete-time jump chain. Each sojourn in a state is almost surely finite; infinitely many visits to a recurrent state therefore correspond to infinitely many distinct returning sojourns. The jump chain remains recurrent, although deleting holding times may change periodicity.

A compound lottery first selects branch $k$ with probability $p_k$ and then outcome ${\omega }_j$ with conditional probability $q_{kj}$. Its reduced lottery assigns ${\omega }_j$ the total probability ${\sum }_k{p}_k{q}_{kj}$. The law of total expectation gives the same expected utility in either description:This rule assumes that only the final outcome matters. A preference for a particular randomization procedure requires including that procedure in the outcome description; it cannot be captured by changing the arithmetic of expected utility on a fixed outcome set.

Let independent $D_n$ have $\mathbb{P}(D_n=k)=p{q}^k$, $k\ge 0$, $q=1-p$. The displayed chain has transition probabilities $P_{ij}=p{q}^{i-j+1}$ for $1\le j\le i+1$ and $P_{i0}=q^{i+1}$. Its probability of hitting zero from $i\ge 1$ is one when $p\le 1/2$ and $(q/p{)}^{i+1}$ when $p>1/2$. The first-step boundary equation at $i=1$ is essential for the extra exponent. The chain is positive recurrent for $p<1/2$, with stationary probabilities $(1-p/q)(p/q{)}^j$, null recurrent at $p=1/2$, and transient for $p>1/2$.

For the Bell state $|{\varphi }_{+}⟩$, let both zero-input measurements use the computational basis, and let the one-input bases be rotated by $\theta$ for Alice and $-\theta$ for Bob in the real qubit plane. In the CHSH game, the four conditional winning probabilities are $1,{\cos }^2\theta ,{\cos }^2\theta ,{\sin }^2(2\theta )$. They follow by contraction of the Bell state with vectors from the real orthonormal bases used for the basic measurements. Their average is the displayed expression. On $0\le \theta \le \pi /2$, this beats $3/4$ exactly when $0<\theta <\pi /4$ and reaches $13/16$ at $\theta =\pi /6$, since $t-t^2=1/4-(t-1/2{)}^2$.

Choose a graph vertex $x=(n,x_2,\dots ,x_d)$ on the boundary of a lattice box. Translation and reflection invariance identify the probabilities of $0\leftrightarrow x$ and $x\leftrightarrow 2n{e}_1$. Their intersection implies $0\leftrightarrow 2n{e}_1$, so positive association of random variables gives $C_{2n}\ge \varphi (0\leftrightarrow x{)}^2$. A union bound and lattice symmetries choose such an $x$ with probability at least the box-boundary connection probability divided by the number of boundary graph vertices. This polynomial factor disappears in the exponential rate, identifying boundary and axial inverse correlation lengths.

For bond percolation on the square lattice, write $h_n={\mathbb{P}}_p(0\leftrightarrow (n,0))$ and $g_k={\mathbb{P}}_p(0\leftrightarrow \partial [-k,k{]}^2)$. A maximal-probability boundary graph vertex can be placed at $(k,j)$ by symmetry. Reflection about $x=k$ identifies its connection probability from $0$ with that from $(2k,0)$. The Harris-FKG inequality yields $(g_k/(8k){)}^2\le h_{2k}\le g_{2k}$ for $k\ge 1$.

Let $X=X_1\bigsqcup \cdots \bigsqcup X_r$ and $Y=Y_1\bigsqcup \cdots \bigsqcup Y_s$ be set partitions of the two parts of a bipartite graph. The pair of partitions is $\epsilon$-regular whenThus a uniformly random pair in $X\times Y$ lies in an irregular cell pair with probability at most $\epsilon$.

Let $p_i>0$ be a probability mass function on a finite alphabet with numerical outcomes $a_i$. Prescribe a mean $m$ strictly between the smallest and largest outcomes. Exponential tilting giveswith $\theta$ chosen so that ${\sum }_i{q}_i^{\ast }{a}_i=m$. For any other feasible law $r$, direct substitution yieldsRelative entropy nonnegativity therefore proves optimality and uniqueness. At an extreme prescribed mean, feasible laws may have zero probabilities; a finite tilting parameter need not realize them, although a limit can.

Let $G$ be the event of membership in the first of two groups and $A$ the event of selection. Write $p=P(G)$, $a=P(G\mid A)$ and $q=P(A)>0$, with $0<p,a<1$. Bayes' theorem gives the conditional probabilities $r_1=P(A\mid G)=aq/p$ and $r_0=P(A\mid G^c)=(1-a)q/(1-p)$. Their risk ratio is the displayed expression: the unknown overall selection probability cancels. Equivalently, it is the ratio of the group-membership odds after selection to those before selection.

When both selection probabilities lie strictly between zero and one, the selection odds ratio is $[r_1/(1-r_1)]/[r_0/(1-r_0)]$ and generally depends on $q$. Thus compositions alone do not identify that odds ratio or provide sample sizes for a standard error. When $a<p$, the first group's observed selection rate is lower; this comparison alone does not establish a causal effect of group membership.

For a normalized quantum state $\psi$ and a time-independent Hamiltonian operator $H$, consider ideal projective measurements of the orthogonal projection $P=|\psi ⟩⟨\psi |$ at intervals $\Delta t=T/N$. Write $U=e^{-iH\Delta t/\hslash }$. A successful check leaves the post-measurement state $\psi$ up to global phase. Thus each successive success, conditional on earlier successes, has conditional probability $|⟨\psi ,U\psi ⟩{|}^2$, and multiplication gives the displayed formula. Equivalently,This is the probability of the entire sequence of successful checks. It underlies the quantum Zeno effect.

A binary repetition code represents one bit by repeating it $n$ times, using codewords $0^n$ and $1^n$. Their Hamming distance is $n$, and the code rate is $1/n$. For odd $n$, the majority function recovers the input whenever fewer than half the positions have flipped. Even $n$ allows a tied vote, so a decoder must specify how it handles that case.

For $n=3$, the received words $000,001,010,100$ decode to zero; $111,110,101,011$ decode to one. The decoder therefore corrects every single flip. This guarantee concerns the number of flipped positions, independent of any probabilistic model.

If each position passes independently through a binary symmetric channel with crossover probability $q$, a decoded error requires at least two flips. The binomial distribution givesAt $q=0.1$, the uncoded error probability is $0.1$, while the three-bit code's decoded error probability is $0.028$. This improvement costs three transmitted bits per message bit. For $0<q<1/2$, $3q^2-2q^3<q$; for $q>1/2$, one should first exploit the known tendency to invert symbols rather than applying the same majority rule blindly.

Independent flips are an essential assumption for the displayed formula. If a single common disturbance flips all three positions together with probability $q$, repetition provides no improvement. More efficient error-correcting codes distribute redundancy across larger message blocks; repetition is useful as a transparent example of the geometry and the reliability–rate tradeoff.

Let $c_0$ be the initial monetary balance in the bank account, negative for borrowing. Matching successor financial payoffs gives the linear systemSubtracting determines the stock holding $\Delta$ in the header; substitution determines $c_0$. The initial cost is $V_0=\Delta S_0+c_0$. For the European call option with $S_u=120$, $S_d=80$, strike price 100 and payoffs 20 and zero,The debt becomes 40 at maturity: half a share minus 40 yields $60-40=20$ in the up state and $40-40=0$ in the down state. Holding these positions until maturity is self-financing. Repeating this two-equation construction at earlier nodes gives backward option pricing and the unique price in the complete market.

The discrete replication method was developed in Cox, Ross and Rubinstein's 1979 paper, whose Sections 2–3 explain borrowing, statewise matching and why physical probability drops out of relative option valuation. Their symbols for physical and pricing probabilities differ from the $p,q$ convention here. ^[ref]

Replication in experimental design repeats a treatment on multiple independent experimental units, so that variation between those units can inform a treatment contrast and its standard error. Repeated measurements on one experimental unit can improve measurement precision without providing additional independent treatment replication. The experimental unit is identified by where the treatment is applied, rather than by how many responses are recorded.^[ref]

For a constructed example, grow one plant in each of eight separate pots. Apply treatment A to four pots and treatment B to four pots, without sharing drainage water between pots. Measure the length of three leaves per pot. The pot is the experimental unit; the leaf is the observational unit. There are four treatment replicates per group, eight experimental units overall, and $8\cdot 3=24$ measurements. The three leaves in one pot are nested measurements, because they share the same plant and treatment application. The setup assumes that pots do not affect one another; simply putting them in separate containers does not establish statistical independence.

Randomization chooses which experimental units receive which treatments; replication supplies multiple experimental units receiving the same treatment. These solve different problems. In the balanced allocation above, choose the four A pots uniformly among the $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ possible subsets. A particular pot has probability $1/2$ of receiving A. Given that one pot receives A, a second receives A with conditional probability $3/7$, rather than $1/2$. Thus the assignments are not independent random variables, even though the experimental units can supply independent responses conditional on those assignments in a suitable statistical model.

An actual ecology experiment illustrates why the distinction matters. In Stuart H. Hurlbert's account, twelve experimental ponds initially received zero, fifty or 450 fish, with five, four and three ponds respectively. Preparation included assessing variation between ponds before introducing fish; allocation considered both location and existing differences between ponds. The zero-fish ponds supplied the scientific control. These were repeated treatment applications to ponds, not twelve samples from one pond.^[ref]

The original article followed the southern California ponds for ten months. Its readouts included zooplankton populations and total phytoplankton; it reported decreases in some zooplankton populations and increased total phytoplankton.^[ref] Repeated sampling within a pond helps describe its response through time, while ponds supply treatment replication. Unequal group sizes and initial pond variation matter for the statistical model; the treatment counts alone do not establish equal variances or independent responses.

Judging probability through resemblance to a typical case or generating process. Resemblance need not incorporate prevalence or evidence reliability, so it can yield plausible-seeming but poorly calibrated probability judgments.

In the Driven two-level Rabi model, the transition probability is $P(t)={\Omega }^2{\sin }^2(\sqrt{{\Omega }^2+{\delta }^2}t/2)/({\Omega }^2+{\delta }^2)$. Maximizing over time at each detuning gives a Lorentzian line shape, with first maximizing time $\pi /\sqrt{{\Omega }^2+{\delta }^2}$ and full width at half maximum $2|\Omega |$ in angular frequency. The amplitude parameter $\Omega$ must use the specified Rabi angular frequency convention.

This envelope differs from a spectrum obtained with a fixed pulse duration: optimizing the time separately removes the sine-squared factor, while fixing time retains it. A finite linewidth here describes detuning-limited transfer under coherent driving, without requiring dissipative decay.

A return level answers the inverse question to a return period: which threshold has the specified rarity? Let $F(z)=F_{\mathrm{annual}}(z)=\mathbb{P}(M_j\le z)$ be the common cumulative distribution function of annual sample maxima in a model with stationarity. For a continuous, strictly increasing $F$ and $T>1$, the $T$-year return level isIt has the same units as $M_j$, whereas $T$ is a count of annual blocks. Independence between years is unnecessary for this one-year quantile definition; it is needed for the geometric waiting and multi-year formulas under return period. For a discrete probability distribution, use the generalized quantile function $F^{-1}(q)=\inf \{z:F(z)\ge q\}$. Then $\mathbb{P}(M_j>z_T)\le 1/T$, with equality not guaranteed at jumps; an exact requested exceedance probability may be unattainable.

For a concrete standard model, choose the maxima convention of the Gumbel distribution, with location $\mu$ and scale $b>0$:The second formula follows by taking logarithms twice and solving $F(z_T)=1-1/T$. The maxima cumulative distribution function and its inverse are given in NIST: extreme value type I distribution, maximum case. With the deliberately chosen toy values $\mu =10$ and $b=2$ in arbitrary units, $z_{10}=14.501$, $z_{100}=19.200$, and $z_{1000}=23.815$, rounded to three decimals. In particular, $19.200$ is approximately the annual quantile at cumulative probability $0.99$; it is not a prediction of the largest observation in the next 100 years. Under independence, the probability that all those annual sample maxima stay below the exact $z_{100}$ is $0.99^{100}\approx 0.366$.

The time block matters. If a toy year contains $m$ independent daily blocks, each with exceedance probability $q$ at one fixed threshold, its annual exceedance probability is $p=1-(1-q{)}^m$, giving annual return period $1/[1-(1-q{)}^m]$. The reciprocal $1/q$ counts daily blocks. Dependence among daily exceedances requires their joint law instead; a threshold crossed on several successive days still produces only one exceedance indicator for that year's sample maximum. The peaks-over-threshold method and declustering of extremes address a different choice of extreme-event sampling.

A return level plot graphs $T$ horizontally and $z_T$ vertically, often using a logarithmic scale for $T$. To read it, choose a return period, move up to the curve, and read the corresponding return level. Conversely, a threshold's horizontal coordinate gives the reciprocal of its annual exceedance probability. The usual plot of fitted quantiles against return periods, and optional pointwise confidence intervals, are documented in Mathieu Ribatet, POT package: return level plot.

The near-straight shape has a mathematical explanation: since $-\log (1-1/T)\sim 1/T$ as $T\to \infty$, the Gumbel distribution gives $z_T=\mu +b\log T+o(1)$. Multiplying a large return period by ten therefore increases the return level by approximately $b\log 10$, about $4.605$ units for this toy curve. A curve indexed by rarity puts several thresholds on one common scale while retaining their measurement units.

For a fitted curve, a pointwise confidence interval concerns uncertainty in the estimated quantile at a specified $T$; it is distinct from a prediction interval for a future annual sample maximum, and pointwise intervals need not form a simultaneous confidence band. A far-right curve value can also be an extrapolation beyond the observation record: extending the line adds no observations to support the chosen tail model. The diagram above uses fixed toy parameters, so it shows neither estimation uncertainty nor an observational record. Its interpretation throughout assumes the stated annual block and stationarity.

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.

For any sequence of events, its limsup of events satisfiesSet $C_N={\bigcup }_{n\ge N}{A}_n$. These are decreasing events with intersection ${\mathrm{limsup}}_n{A}_n$. Continuity of probability from above givesThis is also the bounded-indicator form of the reversed Fatou lemma. The inequality can be strict: independent fair-coin head events each have probability $1/2$, while their limsup of events has probability $1$ by the Second Borel-Cantelli lemma.

For independent site percolation or bond percolation on a rooted locally finite graph, $\theta (p)$ is right continuous on $[0,1)$. Finite-radius connection events have polynomial probabilities in $p$ and decrease to infinite connection. Thus $\theta$ is their infimum by continuity from above of a measure. This infimum of continuous functions is upper semicontinuous; parameter monotonicity from the monotone coupling of Bernoulli percolation then gives right continuity.