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Independent bond percolation on the square lattice satisfies $p_c\le 1-{\mu }^{-1}$, where $\mu$ is its connective constant. If $(1-p)\mu <1$, closed dual graph cycles have a summable large-length tail: their counts are bounded by a polynomial factor times the count of self-avoiding walks. Conditioning a sufficiently large finite box to be open excludes short enclosing graph cycles without changing the law of the remaining edges. With positive conditional probability no enclosing closed dual graph cycle remains, so the origin belongs to an infinite percolation cluster.

For $k=2$, a disconnected graph has a component of size $3\le s\le n/2$. The probability that a fixed $s$-set has no crossing edge is at most $(s/n{)}^{2s}(1-s/n{)}^{2(n-s)}$. The union bound and $\left(\genfrac{}{}{0ex}{}{n}{s}\right)\le (n/s{)}^s[n/(n-s){]}^{n-s}$ bound the failure probability by ${\sum }_{s=3}^{\lfloor n/2\rfloor }(s/n{)}^s=o(1)$. Split this sum at $\sqrt{n}$. The random k-out graph coupling extends the result to every fixed $k\ge 2$.

Suppose random continuous real functions have unique zeros. If pointwise convergence in probability gives opposite strict limiting signs at the two ends of every sufficiently small interval about the target, the intermediate value theorem traps their zeros in those intervals with probability tending to one. The limiting function need not be continuous and the random functions need not be monotone.

The same intensity measure can belong to very different point processes. A Poisson point process with a locally finite intensity measure $\mu$ adds two requirements: each finite-intensity window count has the Poisson distribution of mean $\mu (W)$, and counts in pairwise disjoint observation windows are mutually independent. These requirements determine its probability law; they do not follow from the formula for $\mu$.^[ref]

A randomly shifted lattice point process, a unit-rate Poisson point process, and the shared random rate gives dependent point counts construction all satisfy $\mu (A)=|A|$ on the real line. Nevertheless, their counts in $[0,1)$ have respectively variance zero, one, and two. Counts in the disjoint observation windows $[0,1/2)$ and $[1/2,1)$ have respectively covariance $-1/4$, zero, and $1/4$. These exact calculations show why measuring mean density cannot identify a spacing or statistical dependence model.

A constant random variable takes the same fixed real number $c$ at every outcome in its sample space. Its probability distribution puts mass one at $c$. An almost surely constant random variable may differ on an event of probability zero, while having the same probability distribution.

Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

A cooling schedule for simulated annealing specifies successive effective temperatures and how many proposals to attempt at each. For example, $T_k=T_0{r}^k$ with $0<r<1$, using the geometric sequence, can define stages of equal chosen length. Changing $T$ changes the transition matrix: if the state probability distribution is ${\mu }_n$, then ${\mu }_{n+1}={\mu }_n{P}_{T_n}$, which generally differs from ${\pi }_{T_{n+1}}$. Time spent at one temperature must be compared with its mixing time before equilibrium intuition is justified.

In the five-state barrier example for simulated annealing, fixed $T_{\ast }>0$ gives a finite irreducible Markov chain, so $4$ is eventually visited with probability one. That unbounded-time statement gives no fixed finite deadline. Over any finite number $N$ of trials starting at $1$, the probability of staying at $1$ throughout isEach factor is the rejection probability at $1$. This is an explicit failure event even when every $T_n$ is positive: the best visited objective function value remains $1$, above the global minimum value $0$. More exploratory proposals can help cross a barrier; cooling before those moves occur can preserve a poor local minimum. Successful asymptotic schedules need additional analysis of the allowed moves and barriers.

Historically, Scott Kirkpatrick, Charles Daniel Gelatt Jr. and Mario P. Vecchi connected statistical mechanics with mathematical optimization in 1983. They adapted the Metropolis–Hastings algorithm to cooling searches with symmetric proposals and illustrated computer-design problems.^[ref]

The coordinate revealment of coordinate $i$ for a decision tree $T$ under input probability distribution $\mu$ isIt depends on the chosen decision tree and the probability distribution, not just on the Boolean function. Since no coordinate is queried twice on a graph path, indicator functions giveThe second equality uses linearity of expectation; it does not require independence of random variables or a product measure. The cited paper defines these inspection probabilities and their sum for product measures; the displayed indicator function argument extends the identity to any input probability distribution.^[ref]

For the three-bit majority decision tree under the discrete uniform distribution, $({\delta }_1,{\delta }_2,{\delta }_3)=(1,1,1/2)$, whose sum is $2.5$. Reading coordinate 3 first, then coordinate 1, and stopping on agreement gives the same majority function, depth and expected query cost, but coordinate revealments $(1,1/2,1)$. The asymmetry belongs to the chosen query order. Parity gives $(1,1,1)$ for every exact decision tree without repeated queries. If repeated queries were charged, inspection probabilities would count distinct coordinates; total query count would instead require expected multiplicities.

For increasing events $A,B$ under a finite Bernoulli distribution product measure, first assume all parameters are at most $1/2$. Write $A_0\subseteq A_1$ and $B_0\subseteq B_1$ for last-coordinate sections. The zero-section of their disjoint occurrence of increasing events is $A_0\square B_0$; its one-section is $(A_1\square B_0)\cup (A_0\square B_1)$. The intersection in this union contains $A_0\square B_0$. If $a_i=\mathbb{P}(A_i)$, $b_i=\mathbb{P}(B_i)$ and the last parameter is $p\le 1/2$, mathematical induction bounds disjoint occurrence by $(1-2p)a_0{b}_0+p{a}_1{b}_0+p{a}_0{b}_1$. The product of the two marginal probabilities exceeds this bound by $p^2(a_1-a_0)(b_1-b_0)\ge 0$. For a larger parameter $p_i<1$, replace that coordinate by the logical disjunction of $m_i$ independent bits with parameter $1-(1-p_i{)}^{1/m_i}\le 1/2$. Disjoint original witnesses lift to disjoint bit witnesses. Apply the small-parameter inequality and then pass to degenerate parameters by continuity.

On $|{\Phi }^{+}⟩$, measure ${\sigma }_z\cos \theta +{\sigma }_x\sin \theta$, assigning bit $a$ to eigenvalue $(-1{)}^a$. Choose Alice's angles $2(i-1)\delta$ and Bob's angles $(2i-1)\delta$, with $\delta =\pi /(2N)$. The Bell state correlation is $\cos (\theta -\varphi )$. Each adjacent disagreement term and the last agreement term therefore has probability ${\sin }^2(\pi /(4N))$. For $N\ge 2$, their total violates the chained modular Bell inequality bound one, and it tends to zero as $N\to \infty$. At $N=1$ the value is exactly one.

// Target: probability-and-statistics.bigb

Let distinct nonorthogonal pure states have overlap magnitude $0<c<1$, and let one quantum channel take $n$ identical inputs to $m$ outputs. Let ${ϵ}_u,{ϵ}_v$ be the respective probabilities that a test for the ideal $m$-copy pure state fails. Trace distance contraction, its triangle inequality and the pure-target upper bound on trace distance give $\sqrt{1-c^{2m}}\le \sqrt{{ϵ}_u}+\sqrt{1-c^{2n}}+\sqrt{{ϵ}_v}$. Squaring with the Cauchy-Schwarz inequality gives the displayed strategy-independent positive bound. It establishes both a worst-case bound and an equal-prior average bound. Averaging rotated pairs establishes a positive Haar-average bound for unknown qubit inputs, without assuming independent clone errors.

Let $A$ be a fair bit, let $N$ be an independent Bernoulli random variable with $\mathbb{P}(N=1)=p$, and set $B=A\oplus N$, where $\oplus$ is exclusive or. Thus $N=1$ flips the first bit. This constructs a binary symmetric channel with a fair input, for $0\le p\le 1$. The joint probability mass function isThe column sums show that $B$ is also a fair bit, whatever $p$ is. Both marginal distributions therefore have information entropy one bit. Given $A$, only the flip decision remains unknown, giving conditional entropy $H(B\mid A)=h_2(p)$, whereis the binary entropy, with $h_2(0)=h_2(1)=0$. The chain rule for information entropy and the definition of mutual information now give

Take $p=1/10$. The four joint probabilities, in row order, are $(0.45,0.05,0.05,0.45)$. Directly from the definition of joint entropy,Conditioning on either value of $A$ gives probabilities $0.9$ and $0.1$ for the two values of $B$, henceAdding $H(A)+H(B)=2$ would overcount by exactly $I(A;B)$. These are average uncertainties under the stipulated probability distribution; the four possible pairs are still represented by four distinct outcomes.

There is another way to see the decomposition: $(A,B)\mapsto (A,A\oplus B)=(A,N)$ is a bijection. It merely relabels the four outcomes, so it preserves joint entropy. The independence of $A$ and $N$ then gives $H(A,B)=H(A,N)=H(A)+H(N)=1+h_2(p)$. Describing the initial bit and the flip decision separates shared uncertainty from new uncertainty.

At $p=0$, the bits are identical: $B=A$, so $H(B\mid A)=0$ and $I(A;B)=1$. At $p=1$, they are complementary: $B=1-A$, with exactly the same conditional entropy and mutual information. In either case, observing one determines the other. At $p=1/2$, all four joint probabilities are $1/4$; the bits are independent, $H(A,B)=2$, and $I(A;B)=0$. Perfect disagreement is therefore just as informative as perfect agreement.

For $p>1/2$, guessing $A=1-B$ succeeds with probability $p$, better than guessing $A=B$. Changing $B$ to $1-B$ replaces $p$ by $1-p$ and preserves mutual information, since the relabeling is invertible. What matters is how well the observation distinguishes the input, rather than how often its symbol matches the input. The same symmetry makes the binary entropy curve peak at independence, while the mutual information curve reaches its minimum there.

Consider a stipulated repeated-attempt model, with independent Bernoulli trials of success probability $p\in (0,1]$. An unsuccessful attempt costs three units of work, and a successful attempt costs one. Let $B_i$ be its success indicator function, $X_i=3-2B_i$ its cost, and $N=\inf \{i\ge 1:B_i=1\}$ the first success. Observing $X_i$ reveals $B_i$, so $N$ is a stopping time for the costs' natural filtration. The geometric distribution gives $\mathbb{P}(N\ge i)=(1-p{)}^{i-1}$ and $\mathbb{E}[N]=1/p$, including the deterministic case $p=1$. ThusAt $p=1/4$, the average is four attempts and ten units of work. An independent check uses the pathwise identity $S_N=3(N-1)+1$: there are $N-1$ failures and one final success. No success probabilities or costs in this example are measured data.

Here $N$ is not independent of the observed costs. In particular, conditioning on $N=n$ forces every preceding cost to be three and the final cost to be one, so $\mathbb{E}[S_N\mid N=n]=3n-2$, generally different from $n\mathbb{E}[X_1]=(5/2)n$ at $p=1/4$. Nevertheless the unconditional identity holds. This distinction explains why conditioning on the total count as though the original cost law remained unchanged is a faulty proof.

Choose each $r$-element subset of the natural numbers as a hyperedge independently with probability $p$, where $0<p<1$. This defines a countable random uniform hypergraph. For any finite pattern in the hypergraph extension property, an outside vertex realizes it with probability $q=p^{|\mathcal{A}|}(1-p{)}^{|(\genfrac{}{}{0ex}{}{S}{r-1})\setminus \mathcal{A}|}>0$. Distinct outside vertices use disjoint edge decisions, so the probability that the first $N$ candidates all fail is $(1-q{)}^N\to 0$. Taking a countable intersection over all finite patterns proves the extension property almost surely. The bounds $0<p<1$ are necessary for all patterns to have positive probability.

A coupling of probability distributions $\mu ,\nu$ is a joint probability distribution of random variables $(X,Y)$ whose marginal distributions are $\mu ,\nu$. It lets us choose the statistical dependence between two random constructions while preserving the individual laws. This choice can make a comparison much easier: giving two simulations the same random input may keep their paths close, even though separately generating both paths would hide that relationship.^[ref]

For example, two fair coin tosses can be generated independently, with mismatch probability $1/2$, or by tossing once and setting $Y=X$, with mismatch probability zero. Both constructions have the same two Bernoulli distributions as marginals. Their joint laws differ. A maximal coupling makes the values agree as often as possible; a coalescing coupling arranges permanent agreement of two processes after a random time. Minimizing expected geometric separation instead leads to Wasserstein distance.

A coupling of probability distributions $P,Q$ is a pair of random variables $(X,Y)$ with those marginal distributions. On a finite set, and more generally on a standard Borel space, the least achievable mismatch isA maximal coupling attains the minimum; maximal here refers to agreement. An arbitrary coupling can have a much larger mismatch.^[ref]

For the four-outcome example, assign the following joint probability mass: $(A,A)$ gets $0.1$, $(B,B)$ gets $0.1$, $(C,C)$ gets $0.2$, and $(D,D)$ gets $0.3$. The remaining $0.3$ must have $X=A$; place $0.2$ on $(A,B)$ and $0.1$ on $(A,C)$. Row sums give $P$, column sums give $Q$, and the diagonal contains $0.7$. Agreement can reach 70%; mismatch cannot fall below 30%.

To see the lower bound, for any event $E$,The two indicator functions agree whenever $X=Y$. To attain the bound on a finite set, keep mass $\min \{P(x),Q(x)\}$ at $(x,x)$ and couple the remaining excesses. The two residual probability distributions live on disjoint subsets, so their part necessarily mismatches. When the probability distributions are identical, the residual part disappears entirely.

This is a choice of joint construction, not a claim that independently drawn observations are usually equal. Two independent random variables with the same continuous probability distribution can disagree with probability one even though their total variation distance is zero. The construction of a maximal coupling deliberately introduces statistical dependence to reveal how similar the marginal laws are.

The covariance measures the average product of two centered deviations. For real square-integrable random variables $X$ and $Y$,Deviations with the same sign contribute positively; deviations with opposite signs contribute negatively. Their magnitudes and probabilities determine the net result. Taking $Y=X$ recovers the variance. The definitions and algebraic properties are presented in Orloff and Bloom, MIT 18.05, Covariance and Correlation, sections 2–2.1.

For a worked example, let $X,Z$ be independent random variables, each equally likely to be $-1$, $0$ or $1$, and put $Y=2X+Z$. All nine pairs $(X,Z)$ have probability $1/9$. The expected values are zero, while linearity of covariance and the vanishing covariance of independent random variables giveThe shared $X$ term makes large $X$ accompany large $Y$ on average. Replacing $2X$ by $-2X$ reverses the covariance to $-4/3$ without changing either variance.

The covariance is useful whenever deviations are added or subtracted. Expanding the square givesFor the worked model, these are $20/3$ and $4/3$. Positive covariance increases variation in the sum while reducing variation in the difference: the same shared fluctuation reinforces one combination and partly cancels in the other. The identity applies beyond independent random variables.

The size of a covariance depends on the measurement scale. For constants $a,b,c,d$, linearity of covariance gives $\mathrm{Cov}(aX+b,cY+d)=ac\mathrm{Cov}(X,Y)$. Adding offsets has no effect; rescaling either measurement rescales the covariance. The Pearson correlation coefficient divides by both standard deviations to describe the standardized linear relationship.

Let $A,B$ be indicator random variables with conditional independence given a latent variable $T$, and suppose $\mathbb{E}[A\mid T]=\mathbb{E}[B\mid T]=g(T)$. ThenFor a short proof, conditional independence gives $\mathbb{E}[AB\mid T]=g(T{)}^2$. The law of total expectation gives $\mathbb{E}[AB]=\mathbb{E}[g(T{)}^2]$ and $\mathbb{E}[A]=\mathbb{E}[B]=\mathbb{E}[g(T)]$. Subtract their product to obtain the claimed variance. Since $0\le g(T)\le 1$, all the required expected values exist.

In the two-environment example of conditional independence, $g(T)$ is $0.1$ or $0.9$ with equal probabilities. Hence $\mathbb{E}[g(T)]=0.5$ and $\mathrm{Var}(g(T))=((0.1-0.5{)}^2+(0.9-0.5{)}^2)/2=0.16$. The two trials have zero conditional covariance in either environment, yet their unconditional covariance is positive because their success probabilities rise and fall together.

More generally, square-integrable conditionally independent random variables satisfyby the law of total covariance. Equal conditional means reduce this to the displayed variance; unequal conditional means need not give a nonnegative covariance. The equal-mean result explains correlated component evolutionary states in a coeval binary population without correlation between initial masses: variation in a shared environment can supply the entire covariance.

The coverage probability is the repeated-sampling probability that a random confidence interval contains the fixed true statistical parameter. A finite-sample level-$1-\alpha$ procedure has coverage probability at least $1-\alpha$ at every admissible statistical parameter; exact nominal coverage means equality. An asymptotically calibrated procedure approaches the nominal value as sample size grows, under its stated assumptions. Higher-order Edgeworth expansions can quantify the discrepancy from nominal coverage probability when their regularity conditions hold.

Observed coverage in finitely many repetitions is itself random. For $m$ independent repetitions of the exact normal mean confidence interval with known variance, the count $K$ of covering confidence intervals has binomial distribution $\mathrm{Binomial}(m,0.95)$. At $m=40$, its expectation is $38$, butConsequently, the 39 successes in the figure are compatible with exact 95% coverage probability. The guarantee follows from the pivotal quantity's probability distribution; the finite observed fraction is an illustration. Also distinguish a confidence interval for a fixed mean from a prediction interval for a future observation: the latter must accommodate the new observation's variation as well as estimation uncertainty.

Assume independent individual event times with positive time-constant hazard multipliers and a common baseline hazard whose cumulative hazard tends to infinity. Transforming each event time by that cumulative hazard then gives independent exponential event times with corresponding rates, so all event orders exist. Their complete-order probabilities select each next label proportionally to its remaining multiplier. Marginalizing over the position of a deleted label leaves the order law of the other exponential times, hence the same sequential formula without that label. This proves the deletion identity used when unobserved ranks are summed out. It does not supply a likelihood function for censoring times; ignoring censoring in observed survival analysis still needs independent censoring.