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Choose two Ising spins with Hamiltonian $H=-J{s}_1{s}_2$, where $J$ independently takes $J_0$ or $2J_0$ with equal prior probabilities and $J_0>0$. This original finite example has no geometrical frustration and no phase transition. Its four spin assignments give $Z_J=4\cosh (\beta J)$. ThereforeThe concave function $\log$ puts $F_{\mathrm{a}}\le F_{\mathrm{q}}$. For this two-value case the claim is just $(Z_1+Z_2)/2\ge \sqrt{Z_1{Z}_2}$, obtained by squaring $\sqrt{Z_1}-\sqrt{Z_2}$. Thus no formal continuation of replicas is needed to see the distinction.

In the annealed disorder ensemble, the stronger bond has probability $Z_{2J_0}/(Z_{J_0}+Z_{2J_0})$, approaching one at low temperature. A quenched disorder collection instead retains one half at each coupling. As $T\to 0$, $F_{\mathrm{q}}\to -3J_0/2$ and $F_{\mathrm{a}}\to -2J_0$: averaging samples and allowing the material variables to equilibrate describe different physics.

Two-state ion-channel gating models an individual gate as a random switch between closed and open states. Write $p(t)$ for its open probability; this is the $X$ in the header equation. An individual gate's state is always closed or open, rather than partly open. A population of identical gates can have a fraction open near $p(t)$, and averaging repeated identically prepared trials estimates the same probability. A finite population's observed fraction fluctuates. See Purves and colleagues, Neuroscience, 2nd edition (2001): Ion Channels Underlying Action Potentials for single-channel steps and their population averages.

At fixed membrane potential $V$, let constant rates $\alpha ,\beta >0$ govern closed-to-open and open-to-closed transitions respectively, in ${\mathrm{s}}^{-1}$. This is a two-state continuous-time Markov chain. Over a short interval $\Delta t$, an initially closed gate opens with probability $\alpha \Delta t+o(\Delta t)$, and an initially open gate closes with probability $\beta \Delta t+o(\Delta t)$. Gain minus loss gives the deterministic ordinary differential equation for the probability:Here $p_{\infty }$ is the fixed-voltage limiting open probability, and $\tau$ is a time constant. The ratio of the rates fixes the limit, while their sum fixes the speed. With initial probability $p(0)=p_0\in [0,1]$,For a short derivation, subtract $p_{\infty }$: the deviation obeys $d(p-p_{\infty })/dt=-(p-p_{\infty })/\tau$, yielding the exponential function. After one time constant, $1/e$ of the initial deviation remains. The solution stays between $p_0$ and $p_{\infty }$, so it remains a valid probability. The rate and time constant formulations are described in Gerstner and colleagues, Neuronal Dynamics, §2.2.2: transition rates and channel kinetics.

For a worked toy condition A, choose $\alpha =\beta =50{\mathrm{s}}^{-1}$ and $p_0=0.1$. Then $p_{\infty }=0.5$, $\tau =0.01\mathrm{s}=10\mathrm{ms}$, andIn a separately prepared condition B, choose $\alpha =200{\mathrm{s}}^{-1}$, $\beta =50{\mathrm{s}}^{-1}$ and the same $p_0=0.1$. Now $p_{\infty }=0.8$, $\tau =4\mathrm{ms}$, and $p(10\mathrm{ms})=0.8-0.7e^{-2.5}\approx 0.742540$. Condition B approaches a higher limit more quickly. These are two stipulated constant-rate conditions with $V$ held fixed in each experiment, not an empirical function relating rates to voltage or a sequential switch between conditions. The initial $p_0=0.1$ need not be the limiting probability at the newly imposed voltage.

In the Hodgkin-Huxley model, the rates depend on membrane potential: ${\alpha }_X(V)$ and ${\beta }_X(V)$. Its $X_{\infty }(V)={\alpha }_X(V)/({\alpha }_X(V)+{\beta }_X(V))$ is the limit for voltage held at that $V$. During an action potential, changing $V$ changes the rates and gates feed back onto $V$ through ionic electric currents; substituting the present voltage into the constant-rate solution does not solve that coupled system. Gating describes availability of a conducting pathway. Even an open ion channel carries zero net ionic electric current at its reversal potential.

The useful bridge is from abrupt random molecular events to a smooth population prediction. A channel requiring several independent permissive gates in an ion channel has an open probability equal to their product; this can make channel availability much smaller than each gate's availability. The Hodgkin-Huxley model uses ${\bar{g}}_{\mathrm{Na}}{m}^{3}h$ and ${\bar{g}}_{\mathrm{K}}{n}^4$ for sodium and potassium electrical conductances, respectively.

Model two fair independent coin tosses, recording their order. Here $H$ denotes heads and $T$ tails, andThe power set $\mathcal{P}(\Omega )$ includes every question expressible as a subset of these four outcomes. The mass $1/4=(1/2)(1/2)$ comes from the fairness and independence assumptions. These are exact model probabilities, rather than measured toss frequencies.

Let $A$ be the event that the first toss is heads and $B$ the event that the second toss is heads. ThenThe union means “at least one head,” while the intersection means “both heads.” Their probabilities are $3/4$ and $1/4$. Counting the union through the inclusion-exclusion principle gives $1/2+1/2-1/4=3/4$; alternatively its complement of an event is $\{TT\}$, of probability $1/4$.

On the same sample space, assigning masses $3/8,1/8,1/8,3/8$ to $HH,HT,TH,TT$ would still make each toss fair: $\mathbb{P}(A)=\mathbb{P}(B)=1/2$. But $\mathbb{P}(A\cap B)=3/8\ne 1/4$, so the tosses would be dependent. The event of at least one head would have probability $5/8$. Changing the probability measure changes the answer even when the sample space and the single-toss probabilities stay the same.

Suppose each independent turn makes the current player win with probability $p$, lose with probability $q$, or pass with probability $r$, with $p+q+r=1$ and $r<1$. The players have identical turn rules and alternate after a pass. Before the first pair of turns finishes, the starting player wins with probability $p+rq$. Passing twice returns to the initial state with probability $r^2$. First-step analysis therefore gives $P=p+rq+r^{2}P$, proving the formula. The geometric series of repeated restarts converges because $r<1$; infinite play has probability zero. For $r=1$, nobody ever wins, so that excluded case has no eventual winning probability summing to one.

Inside a rooted $d$-ary subtree of a degree-$(d+1)$ regular tree, pair each cell root with one distinguished child. Keep reinfection inside each pair, suppress infection back into its parent cell, and allow each child cell to be activated only once. Offspring counts of different cells are independent and identically distributed, so cell genealogies form a Galton-Watson process. A cell root has $d-1$ outgoing child-cell edges and its paired vertex has $d$. Starting at the root, the probabilities to send an arrow along one specified outgoing edge from these two vertices before internal extinction are respectivelyThey follow by first-step equations on the three nonempty pair states. The offspring mean is $M=(d-1)a+db$, andContinuity makes $M>1$ at some strictly smaller positive rate. The branching-process extinction criterion then gives positive survival probability for the suppressed process, hence for the original contact process.

A type summarizes a string by how often each symbol occurs. Fix a finite alphabet $\mathcal{A}$ of cardinality $m$ and a positive length $n$. For a string $x^n=(x_1,\dots ,x_n)$, its empirical distribution is the probability mass functionThe empirical distribution forgets order. The strings $0011$ and $1010$ both assign probability $1/2$ to each binary symbol; $0001$ assigns $3/4$ to zero and $1/4$ to one. Thus a long string can have a short summary even though that summary does not reconstruct the original order.

For independent and identically distributed symbols, this loss of order has a useful consequence: all strings with the same empirical distribution have the same probability. The method of types uses this symmetry to turn questions about exponentially many strings into counting questions about relatively few frequency profiles.^[ref]

The count $K_n$ has binomial distribution with mean $n/4$ and variance $3n/16$. The information content per symbol is an affine function of $K_n/n$, so its variance isChebyshev's inequality therefore gives the finite-sample lower boundThe bound is often loose, but it proves that every fixed positive tolerance eventually captures nearly all the probability. Direct summation of the binomial distribution's probability masses gives $P(T_{0.1}^{(640)})\approx 0.999779$. A finite sample does not need to have a nearly full typical set: the 40-symbol example above captures only about $0.639$.

The typical sequence theorem turns a probability concentration into a coding construction. Assign one binary number index to each weakly typical sequence, plus a reserved index for every word outside the set. The decoder recovers every typical word exactly; the reserved index has one predetermined output. The block code usesbits and has reconstruction-error probability at most $1-P(T_{\epsilon }^{(n)})$.

For this example and $\epsilon =0.1$, $n=40$ needs 34 bits but permits error up to about $0.3611$. At $n=640$, 570 bits suffice and the error bound is about $0.0002215$. These are existence calculations for an explicitly enumerated collection, rather than claims about an efficient practical encoder. The reduction comes from protecting the whole high-probability collection, rather than just its single most likely member.

A bound tending to zero with one factor's Gowers U3 norm, uniformly over all finite abelian groups, is false. On $G={\mathbb{F}}_2^m$, choose a random independent sign at each point. A cube with eight distinct vertices has sign product of expectation zero. The probability of a collision among its vertices is at most $\left(\genfrac{}{}{0ex}{}{8}{2}\right)/|G|$, since each equality of distinct vertices imposes a nonzero linear equation in the increments. Thus some sign function $f$ has $\Vert f{\Vert }_{U^3}^8\le 28/|G|$. But the four-term arithmetic progression is $(x,x+d,x,x+d)$, so choosing factors $(f,1,f,1)$ gives correlation identically one. As $m\to \infty$, the norm tends to zero. Bounded torsion, as in cyclic groups, avoids this obstruction.

Two square-integrable random variables $X,Y$ are uncorrelated random variables when $\mathrm{Cov}(X,Y)=0$. Independent random variables with finite second moments are uncorrelated random variables, but zero covariance does not generally imply independence.

For an explicit counterexample, let $X$ take $-1$, $0$, $1$ with equal probabilities, and let $Y=X^2$. Then $\mathbb{E}X=\mathbb{E}{X}^3=0$, so $\mathrm{Cov}(X,Y)=\mathbb{E}{X}^3-\mathbb{E}X\mathbb{E}{X}^2=0$. Nevertheless, $Y$ is completely determined by $X$: $\mathbb{P}(Y=0\mid X=0)=1$, while $\mathbb{P}(Y=0)=1/3$. Thus these are uncorrelated random variables with statistical dependence. In this example, opposite contributions from the odd function $x^3$ cancel; the even function $x^2$ still retains information about $|X|$. The last panel of the covariance diagram shows the relationship.

There are useful special cases. Uncorrelated jointly normal variables are independent, and the Bernoulli independence from zero covariance result supplies another setting where the converse holds. Their distribution assumptions do the extra work; zero covariance alone does not.

Uniform convergence in probability means that the supremum of the absolute discrepancy on the entire index set tends to zero in probability. It controls evaluation at random indices in that set, unlike pointwise convergence in probability. On a fixed compact index set it is a special case of uniform convergence on compacts in probability.

For $a,b>1$, the maximum of $r$ on $[0,1]$ occurs at $(a-1)/(a+b-2)$. If one parameter is one, the maximum is attained at the corresponding endpoint; if both are one, $r$ is constant. Let $M=\sup r$. Accept an independent uniform distribution proposal $X$ when another uniform draw is at most $r(X)/M$. The accepted probability density function is the beta distribution, and acceptance probability is $B(a,b)/M$.

For $n\ge 2$, choose $n-2$ independent random variables, each with the uniform distribution on a finite set $\{1,\dots ,n\}$, and decode their values as a Prüfer sequence. Every sequence has probability $n^{-(n-2)}$, and the bijection gives each labelled tree the same probability. This produces a uniform spanning tree of the complete graph without repeatedly proposing sets of edges and rejecting those that fail to be trees. It supplies random inputs for testing implementations of breadth-first search and a tractable model of random connections. ^[ref]

This distribution is uniform over fixed-label trees, rather than over unlabelled trees. With four labels, the star graph shape has probability $4/16=1/4$ and the path graph shape has probability $12/16=3/4$. Also, for a fixed label $v$, the Prüfer degree formula givesHere the binomial distribution counts the independent code positions equal to $v$. The expected value agrees with the average vertex degree from the degree-sum formula. Individual labels have the same distribution by symmetry, though their vertex degrees are constrained together by their fixed sum. When $n=1$, simply return the single isolated vertex.

For independent and identically distributed random variables $U_1,\dots ,U_n$ with the continuous uniform distribution on $(0,1)$, the $r$th order statistic has Beta distributionIts probability density function, mean, and variance areThe probability density function is zero outside $[0,1]$; values assigned to its endpoints do not change any probability. Peter Kempthorne's MIT lecture notes derive the uniform order statistic law and these moments. ^[ref].

For $n=5$, the minimum, sample median, and sample maximum correspond to $r=1,3,5$. Their probability density functions are $5(1-t{)}^4$, $30t^2(1-t{)}^2$, and $5t^4$, respectively. The sample median therefore has Beta distribution $\mathrm{Beta}(3,3)$, mean $1/2$, and variance $1/28\approx 0.0357143$. Its cumulative distribution function isFor instance, $\mathbb{P}(U_{(3)}\le 1/4)=53/512\approx 0.103516$. The mean of the minimum is $1/6$, and the mean of the sample maximum is $5/6$. The fixed observed sample median $0.4$ in the order statistic example is one possible outcome; $1/2$ is the average over repeated statistical samples from this uniform distribution.

Each curve has area one. The extreme order statistics favor the ends, while the sample median favors the center. The endpoint density height $5$ is legitimate: the probability density function gives area per unit length, rather than point probability. In particular, a continuous uniform distribution has zero probability of producing an endpoint.

For a short proof, the binomial distribution counting identity under order statistic gives $\mathbb{P}(U_{(r)}\le t)={\sum }_{k=r}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k(1-t{)}^{n-k}$. Taking its derivative cancels consecutive terms, leaving the displayed probability density function. Recognize its normalizing factorial coefficient as that of the Beta distribution, then substitute $\alpha =r$, $\beta =n+1-r$ into the Beta distribution mean and variance formulas. Although the original observations are independent, their order statistics generally are dependent; see the positive covariance of two uniform order statistics.

A uniform prior on a finite-volume parameter region assigns constant probability density there and zero outside. On $[0,1]$ it is $\mathrm{Beta}(1,1)$, giving a beta distribution posterior after binomial distribution data. Uniformity depends on the coordinate: a nonlinear reparametrization introduces a Jacobian and generally does not preserve a uniform density.

If independent edges of a locally finite graph receive fair orientations, exploring outward from a root tests each fresh reached-to-unreached boundary edge with success probability $1/2$. Deferred decisions give exactly the exploration law of the root cluster in bond percolation of density $1/2$. Thus the two reachable graph vertex sets have the same distribution. An infinite out-cluster contains an infinite simple directed ray by the König infinity lemma. Geometrically biased orientations do not generally give the same uniform edge-success law.

A uniform random permutation of $n$ distinct objects gives each of the $n!$ permutations the same probability. Restricting the order to any fixed subset gives a uniform ordering on that subset. Relative ranks encode such a permutation bijectively by independent choices of an insertion position.

On a connected locally finite recurrent graph, root Wilson's algorithm at a fixed graph vertex $r$ and process a countable enumeration of the graph vertices. Every walk hits the existing tree almost surely, since it contains $r$. Each attached graph path is finite, and the union is a connected acyclic spanning subgraph. Its law is the infinite uniform spanning tree.

It is also the common free and wired finite-volume limit of a sequence along every graph exhaustion. For a finite edge set, run a finite initial segment of the chosen enumeration containing all its endpoints. Its membership is then permanently decided, because later attachments have new internal graph vertices. Only finitely many finite random-walk trajectories were used, and their graph vertices and graph neighbours fit inside all sufficiently large exhaustion sets. Coupling then gives convergence of those finite-dimensional edge laws for both boundary conventions. Finite Wilson's algorithm is root/order independent, so the limiting law is too. There is no uniform counting measure on all infinite spanning trees; the definition is this limiting probability law.

For Kraus operators satisfying ${\sum }_i{A}_i^{†}{A}_i=I$, the map $V|\varphi ⟩={\sum }_i{A}_i|\varphi ⟩\otimes |i⟩$ is an isometry. Implement it by a unitary operator on the system and a quantum ancilla, initially in an extra unused state $|0⟩$. Projecting the output quantum ancilla onto $|i⟩$ gives joint unnormalized state $A_i\rho A_i^{†}\otimes |i⟩⟨i|$. The outcome probability is $\mathrm{Tr}(A_i\rho A_i^{†})$, and the conditional system state is $A_i\rho A_i^{†}$ divided by this probability. The unused level can be merged into the first orthogonal projection without changing any outcome on the prepared state, giving a complete projective measurement with exactly the original outcome labels.

// Target: quantum-theory.bigb

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.