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The Aalen–Johansen estimator is a product of empirical state-transition matrices over event times. In competing risks, the estimated cumulative incidence function for cause $j$ jumps by $\hat{S}(t-)d_j(t)/Y(t)$, while overall survival is updated using all event causes. Right censoring changes subsequent risk sets but does not itself create a probability jump.

A Markov chain with at least one absorbing state, such that from each state an absorbing state can be reached. In a finite such chain, absorption occurs with probability 1. Expected times to absorption satisfy a linear system obtained by conditioning on the first transition.

For a finite, time-homogeneous Discrete-time Markov chain with row-stochastic transition matrix $P$, a state $a$ is absorbing exactly when $P_{aa}=1$. Nonnegativity and the row sum then force $P_{aj}=0$ for $j\ne a$: once entered, $a$ is never left. Equivalently, $\{a\}$ is a closed communicating class. A larger closed communicating class may keep moving internally; closure means confinement to a set, whereas absorption at a state means remaining at one point.

An absorbing state is useful for recording a terminal outcome: after a toy routine finishes, keep its outcome in the state space so that the same transition matrix still describes subsequent steps. Calling a finite Markov chain absorbing usually also requires that some absorbing state be accessible from every state. Merely containing an absorbing state is insufficient: another closed communicating class could prevent absorption. See Grinstead and Snell, Introduction to Probability, section 11.2 for this convention.

Consider a constructed routine with unfinished modes $u,v$ and terminal outcomes $A,B$. All the following transition probabilities are chosen for illustration, not estimated from observations. In state order $(u,v,A,B)$, takeThe communicating class $\{u,v\}$ is open, while $\{A\}$ and $\{B\}$ are closed communicating classes consisting of absorbing states. Both terminal outcomes are accessible from both unfinished modes, sometimes by an indirect path.

Let $T=\inf \{n\ge 0:X_n\in \{A,B\}\}$ be the terminal hitting time. Write $h_i={\mathbb{P}}_i(X_T=A)$ for the absorption probability and $t_i={\mathbb{E}}_{i}T$ for the expected hitting time, with the subscript denoting the starting state. First-step analysis givesHere $h_A=1$, $h_B=0$, and $t_A=t_B=0$. Solving these two pairs of linear equations yieldsStarting at $u$ therefore ends at $A$ with probability $0.6$ and takes $3.2$ steps on average to reach either terminal state. The terminal hitting time counts the transition into the outcome, but no later self-transitions. Outcome and duration answer different questions: from $v$, outcome $A$ is less likely, yet termination is faster on average.

The matrix power $P^2$ gives the probability distribution after two steps from $u$ asThus ${\mathbb{P}}_u(T>2)=1/2$, and the probability of having already reached $A$ is $3/8$, still below its eventual absorption probability $3/5$. From either unfinished mode, each next step has terminal probability at least $1/4$. Consequently ${\mathbb{P}}_i(T>n)\le (3/4{)}^n$, proving termination almost surely and finite expected hitting time directly. The Fundamental matrix of a finite absorbing Markov chain packages both calculations and counts visits to unfinished modes.

For two subsystems with independently specified macrostates, suppose every allowed microstate of $A$ can be paired with every allowed microstate of $B$. Their allowed set is a Cartesian product. With equal probability for each pair, Boltzmann's entropy formula givesFor example, take two independent copies of the toy model, each with two labels choosing between its two boxes. Each copy has four configurations and Boltzmann entropy $k_B\ln 4$. Combining their independent labels gives sixteen configurations and Boltzmann entropy $k_B\ln 16=2k_B\ln 4$. This explains why entropy adds even though microscopic counts multiply. ^[ref]

The product assumption matters. A constraint shared between the subsystems can forbid some pairs; for example, fixing only their combined energy generally requires summing over its possible divisions. See entropy maximization under thermal contact for how thermal equilibrium selects the dominant division. Statistical independence cannot be inferred merely from drawing a boundary between two parts.

If host-specific equilibrium mean burden has a gamma distribution, conditional Poisson distributions produce a Poisson-gamma mixture, hence a negative binomial distribution. With mean $m$ and shape $k$, its zero-burden probability is $(1+m/k{)}^{-k}$. Decreasing $k$ increases aggregation: a smaller infected-host prevalence can accompany the same mean burden and more heavily infected individuals.

An almost strongly universal hash family relaxes the exact ordered pair balance of a strongly universal hash family. In the convention used here, choose $h$ with the uniform distribution on a finite set from a family with $q$ possible outputs. Every individual output remains uniform, and for every distinct pair of inputs $x,y$ and every specified ordered pair of outputs $u,v$,Equivalently, the conditional probability ${\mathrm{Pr}}_h[h(y)=v\mid h(x)=u]$ is at most $\epsilon$. This asks how predictable a second output can become after the first is known.

For $q=5$, strongly universal hash families give $\epsilon =1/5$: learning one output leaves each specified second output with conditional probability $1/5$. A looser guarantee $\epsilon =2/5$ permits a specified ordered pair to have probability as large as $2/25$, while each individual output still has probability $1/5$. Summing over the five equal ordered pairs bounds the collision probability by $2/5$, rather than the $1/5$ required of a universal hash family. Thus an almost strongly universal hash family need not be a universal hash family. Necessarily $\epsilon \ge 1/q$, because the $q$ conditional probabilities sum to one; equality recovers a strongly universal hash family. This relaxation is useful in Wegman–Carter authentication, where efficiency can be traded against a quantified forgery bound within a complete message authentication protocol.

At parameter $1/2$, primal and dual bond percolation on the square lattice have the same probability distribution. Suppose an infinite percolation cluster exists. Large boxes meet it with probability tending to one. The square-root trick for positively associated events and quarter-turn symmetry imply that each side has an infinite exterior open arm with probability tending to one. The same holds for dual arms. A union bound then gives positive probability of primal top and bottom arms together with dual left and right arms.

Use a primal box $[-r,r{]}^2$ and a dual box with sides at $\pm (r+1/2)$. Open only primal edges whose endpoints both lie in the primal box, preserving all outward and exterior arm edges. The two primal arms are joined inside. Planarity prevents the two alternating dual arms from joining: a proposed dual connection together with the inner boundary arc separates one primal infinite arm from infinity. Thus the modified configuration has two infinite dual clusters. Its positive probability by finite modification of Bernoulli percolation contradicts uniqueness of the infinite percolation cluster. Consequently the percolation probability at $1/2$ is zero. This is often called Zhang's argument.

Amplitude amplification increases the probability of obtaining an answer accepted by a specified Boolean function. It requires a coherent preparation $A$, its inverse, and a marked-state phase oracle that negates accepted components. Alternating that sign change with a reflection operator about the prepared state directs interference toward the accepted subspace. Brassard, Høyer, Mosca, and Tapp developed the general framework, extending Grover's algorithm beyond a uniform quantum superposition. ^[ref].

If a normalized quantum state has good amplitude $\sin \theta$, the product of the reflection in the bad axis and the reflection in the state line rotates its good-bad plane by $2\theta$. After $k$ iterations the good probability isso $O(1/\sin \theta )$ iterations raise a small good amplitude close to one. Success subsequently oscillates, making the stopping time part of the algorithm.

Compute the exact Grover search on four entries directly. Order the computational basis as $00,01,10,11$ and mark $11$. The Hadamard transform prepares amplitudes $(1/2,1/2,1/2,1/2)$. The marked-state phase oracle changes these to $(1/2,1/2,1/2,-1/2)$, whose arithmetic mean is $\overline{a}=1/4$. The Grover diffusion operator performs inversion about the mean:The three unmarked amplitudes become $1/2-1/2=0$, while the marked amplitude becomes $1/2-(-1/2)=1$. All three stages have squared norm one. Here $\sin \theta =1/2$, so $\theta =\pi /6$ and one iteration gives ${\sin }^2(3\theta )=1$, agreeing with the explicit calculation.

Grover's original search paper was submitted on 29 May 1996; its algorithm combines selective sign changes with inversion about the mean. The example exposes the beauty of that construction: a change initially invisible to computational-basis measurement becomes complete cancellation of every wrong answer. ^[ref].

The marked-state phase oracle is an assumed operation, whose implementation cost must be counted. Computational-basis measurement yields a candidate label, here $11$; using it to retrieve a record requires the relevant data access. Preparing a quantum superposition does not make all its entries available as classical outputs. The improvement concerns the number of oracle queries needed to obtain an accepted label.

Consider a binary state $X$ and a report $Y$ with the following joint probability mass function:Before the report, $X=0$ is strongly favored: its information entropy is $H(X)=h_2(1/10)\approx 0.468996$ bits, where $h_2$ is binary entropy. A report $Y=1$ makes the two states equally likely, so $H(X\mid Y=1)=1$ bit, strictly larger than before. This report occurs with probability $1/5$. The more common report $Y=0$ determines $X=0$, giving $H(X\mid Y=0)=0$. Therefore the averaged conditional entropy isThe resulting mutual information is $I(X;Y)=h_2(1/10)-1/5\approx 0.268996$ bits. The rare report increases uncertainty about the state; that increase does not make the report useless. Together the reports remove uncertainty on average. These are stipulated probabilities in an exact model, not measured frequencies.

For standard Cauchy distribution probability density function $f$ and $k>0$, this estimates its upper-tail probability without sampling a heavy tail. The function $f(ku)$ decreases on $[0,1]$, while $f(k(1-u))$ increases, so their covariance is nonpositive. The paired estimator is unbiased and has no larger variance than an estimator using the same number $2r$ of independent probability density function evaluations.

An arbitrage is an allowed self-financing portfolio with zero initial wealth, nonnegative terminal portfolio wealth almost surely under the physical probability measure $P$, and strictly positive terminal wealth with positive probability. In a finite model where every state has positive probability, nonnegativity means no loss in any state. Positive expected return alone does not establish arbitrage: losses may still occur.

For the binomial market example, suppose the European call option sells for 13 while its replicating portfolio costs $250/21$. Sell the call, buy its replicating portfolio, and lend the surplus $13-250/21=23/21$ in the bank account. The initial net cost is zero. At maturity the call obligation and portfolio cancel in either state, leaving $(21/20)(23/21)=23/20$. A price below $250/21$ permits the reversed construction, using short selling and borrowing. This is the law of one price made operational; the assumed trading permissions are essential.

The standard arcsine distribution is a continuous probability distribution on $[0,1]$ with probability density functionIt is the beta distribution with both parameters equal to $1/2$. The density grows without bound toward both endpoints, yet neither endpoint has any probability mass. Its name comes from the inverse sine in its cumulative distribution function.^[ref]

A useful way to see the shape is to choose a point uniformly around a circle and project it onto a diameter. Equal angular intervals do not give equal horizontal distances: the projection moves slowly near either end. An ideal harmonic oscillator sampled at a uniformly chosen time in one complete period has the same endpoint concentration after rescaling its position. The distribution translates a simple geometric slowdown into a precise probability calculation.

For an original comparison, take equal-length regions: the middle half $[1/4,3/4]$, and the two outside quarters together. ThenA uniform distribution on $[0,1]$ instead assigns each region probability $1/2$. Both distributions have mean $1/2$; the mean alone hides their different concentration.

Choose $\Theta$ with uniform distribution on $(0,\pi /2)$ and set $X={\sin }^2\Theta$. Since the sine is increasing on this interval,Differentiation gives $f(x)=1/(\pi \sqrt{x(1-x)})$, and $F(0)=0$, $F(1)=1$ prove normalization. This also evaluates the density's improper integral without treating its divergent endpoint values as infinite probability.

For the circle construction, instead choose $\Phi$ with uniform distribution on $[0,2\pi )$ and put $Y=\cos \Phi$. Each $-1<y<1$ has two preimages in one revolution. The change of variables formula adds both contributions:The affine change of variables $X=(1+Y)/2$ gives the standard arcsine distribution. A single monotone-branch calculation without the second preimage would give only half the total probability.

The same geometry explains the ideal harmonic oscillator $y(t)=A\cos (\omega t+\varphi )$, with amplitude $A>0$ and angular frequency $\omega >0$. Choose $T$ uniformly in any complete period of duration $2\pi /\omega$. Its oscillation phase modulo $2\pi$ is uniform, and $Y=y(T)$ has densityNear a turning point, $|dy/dt|=\omega \sqrt{A^2-y^2}$ is small, so a fixed position interval takes more time to cross. For the chosen amplitude $A=2\mathrm{cm}$, exactly one third of the period is spent in $|y|\le 1\mathrm{cm}$, and two thirds outside that central interval. These are calculations for an ideal sinusoid, rather than measured oscillator data.

The sampling assumption matters. Taking one observation every full period returns the same position each time. It does not produce this probability distribution. Uniformly random observation times and a fixed regularly spaced sampling scheme are different models; aliasing can conceal the motion altogether.

Paul Lévy used a related geometric construction in his work on stochastic processes: project a uniformly chosen point on a circle onto its diameter to divide a segment. The projection description appears explicitly in §5 of his paper, following its arcsine density calculation.^[ref]

Inverting $F(x)=p$ gives the quantile functionConsequently, inverse transform sampling uses $X={\sin }^2(\pi U/2)$ with $U$ uniform on $(0,1)$. The quartiles are $(2-\sqrt{2})/4$, $1/2$ and $(2+\sqrt{2})/4$.^[ref]

The first and third quartiles are about $0.1464$ and $0.8536$, so the central half of the probability spans about $0.7071$ of the unit interval. Under a uniform distribution, its width would be $0.5$.

The symmetric probability distribution gives $\mathbb{E}[X]=1/2$. The angle representation gives a short calculation of the second moment:Use ${\sin }^4\theta =(3-4\cos 2\theta +\cos 4\theta )/8$ to evaluate the integral. A uniform distribution on $[0,1]$ has variance $1/12$; the larger variance $1/8$ quantifies the greater spread toward the ends.

More generally, for a nonnegative integer $n$, the beta function yieldsFor $n=0$ this is one; for $n=1,2$ it reproduces the values above. The central binomial coefficient thus appears in an exact moment calculation for a continuous probability distribution.

A planar Brownian motion range has zero two-dimensional Lebesgue measure almost surely. Assuming finite expected area, Brownian scaling gives half that expected area for each half-time range. Inclusion-exclusion then makes their expected overlap zero. Translate their ranges by the midpoint position; the independent Brownian arms at a deterministic time make the expected overlap the integral of the squared point-hitting probabilities for a half-time range. Those probabilities vanish almost everywhere, so the half-time expected area, and hence the full-time expected area, vanish. This argument does not assert zero hitting probability for every individual point solely from an almost-everywhere integral conclusion.

Let $Z$ have the fair Bernoulli distribution and set $X_j=Z$ for every $j$. Each coordinate has the same fair probability distribution as in the independent model, but the coordinates are dependent: $\mathbb{P}(X_1=1,X_2=1)=1/2$, whereas the product of the marginal probabilities is $1/4$. Every future sigma-algebra equals $\sigma (Z)$, so the tail sigma-algebra also equals $\sigma (Z)$.

The tail event of infinitely many heads is now exactly $\{Z=1\}$, with probability $1/2$. For the finite-window events $B_{m,r}$ in infinitely many heads as a tail event,The sample space can even be the same full Cartesian product $\{0,1{\}}^{\mathbb{N}}$ in both models: the dependent probability measure assigns mass $1/2$ to each of the all-zero and all-one sequences. Then $A$ is the same literal tail event in both models. Only the probability measure and its independence properties change.

Two players independently choose venues $A$ or $B$. Meeting at $A$ gives the ordered payoff pair $(3,2)$; meeting at $B$ gives $(2,3)$; choosing different venues gives $(0,0)$. As throughout this article, the row player's payoff comes first:

Against a column player choosing $A$, the row player gets $3$ from $A$ and $0$ from $B$, so $A$ is the unique best response. Against $B$, the row player gets $0$ from $A$ and $2$ from $B$, so $B$ is the unique best response. The column player likewise prefers to match the row choice. Thus $(A,A)$ and $(B,B)$ are pure strategy Nash equilibria, while each mismatched strategy profile has a profitable unilateral deviation.

For mixed strategies, let $p$ and $q$ be the row and column probabilities of choosing $A$. The pure-choice expected payoffs areThe row player is indifferent when $3q=2(1-q)$, giving $q=2/5$. The column player is indifferent when $2p=3(1-p)$, giving $p=3/5$. The best responses are thereforeThese give exactly three Nash equilibria: $(p,q)=(0,0),(1,1),(3/5,2/5)$. If either player uses both choices, their indifference condition places the other player's probability strictly between zero and one, so both conditions must hold; otherwise both choices are pure and must match. Notice whose equation determines whose probability: the row player's indifference fixes the column player's $q$, and the column player's indifference fixes the row player's $p$.

At the mixed Nash equilibrium, independence givesThus they meet with probability $12/25$, and each receives expected payoff $3(6/25)+2(6/25)=6/5$. Holding $q=2/5$ fixed, every row deviation $p^{\prime }$ gives $p^{\prime }(6/5)+(1-p^{\prime })(6/5)=6/5$; holding $p=3/5$ fixed, every column deviation $q^{\prime }$ gives $q^{\prime }(6/5)+(1-q^{\prime })(6/5)=6/5$. This directly verifies the Nash equilibrium inequalities.

Both coordinated pure strategy outcomes strictly improve both players' expected payoffs over $(6/5,6/5)$: $(3,2)$ and $(2,3)$ each do so. The mixed Nash equilibrium therefore fails Pareto efficiency, even though no unilateral deviation helps. Switching together can help when switching alone cannot. Neither coordinated outcome improves both players over the other, and both are Pareto efficient, even when a lottery over joint action profiles is allowed: any positive weight on another outcome reduces the payoff of the player receiving $3$.

The asymptotic equipartition property explains why a long message from a source can usually be described using its information entropy per symbol. Individual messages can have very different probabilities, but the messages carrying almost all the total probability have similar logarithmic costs per symbol. The Bernoulli typical-set example makes the distinction between an individual message and a large collection explicit.

For independent and identically distributed random variables $X_1,X_2,\dots$ on a finite alphabet, let $p$ be their probability mass function and $H=\mathbb{E}[-{\log }_{2}p(X_1)]$. ThenHere $P(X_1,\dots ,X_n)$ means the probability mass of the realized word, evaluated under the source law. The product rule for independent random variables turns its information content into a sum:The strong law of large numbers gives the limit. Zero-probability symbols do not occur; on the positive-probability support of a finite alphabet, these information costs are finite. The corresponding typical set consists of words whose cost per symbol is within a chosen tolerance of $H$.

For a strictly stationary process with a finite alphabet, the constant-limit extension requires ergodicity. The Shannon-McMillan-Breiman theorem then replaces the one-symbol information entropy with the information rate. Stationarity without ergodicity in an information source shows why stationarity alone is insufficient.^[ref]

Prepare independent candidate blocks by one-bit heralding of a remote state preparation block, but send only the first successful block's index. Success eventually occurs with probability one; a prefix code for a rare geometric success uses at most $n+e/(e-1)$ expected bits per block. The expected singlet state consumption is $n{2}^n$. Alternatively, limiting the number of trials to $n{2}^n$ gives failure probability at most $e^{-n}$ and a fixed message length $n+{\log }_{2}n+O(1)$. Both versions approach one classical bit per remotely prepared qubit in their respective expected-length or high-success senses.

// Target: quantum-theory.bigb

If a martingale has increments with $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then each one-sided deviation of magnitude $t>0$ from $M_0$ has probability at most $\exp (-t^2/(2{\sum }_i{c}_i^2))$. If all $c_i$ vanish, the martingale is constant. The inequality gives concentration from bounded increments without requiring independent increments.