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For a continuous uniform distribution on an interval of length $L>0$, the probability density is $1/L$. Direct integration givesThus a uniform law on an interval of length $1/2$ has differential entropy $-\ln 2$, while length two gives $\ln 2$. Both laws have total probability one. The shorter interval has larger probability density, making $-\ln f(X)$ negative; this is neither a negative probability nor a negative discrete information entropy.

In bond percolation on a directed graph, independently open each directed edge with probability $p$. The out-cluster of $x$ consists of vertices reachable by open directed paths. In site percolation, independently open vertices and require every vertex of the directed path, including $x$, to be open. The same definitions allow a multigraph. The orientation matters: an infinite underlying connected graph need not contain an infinite out-cluster from a specified root.

A real function $h$ on the vertices of a graph of an unweighted locally finite graph is a discrete harmonic function at $u$ when $\mathrm{deg}(u)h(u)={\sum }_{v\sim u}h(v)$, equivalently $Lh(u)=0$ for the Graph Laplacian. For positive degree this says its value is the average over its graph neighbours; at an isolated vertex of a graph the Laplacian condition is vacuous. On an electrical network the corresponding equation uses conductance weights and the Weighted graph Laplacian. Voltages away from sources and sinks, and hitting probabilities away from their prescribed boundaries, are discrete harmonic functions.

For a generator of a group $g$ of order $M$ and $x=g^y$, the function $f(a,b)=g^a{x}^{-b}$ hides the subgroup $\{(yb,b):b\in {\mathbb{Z}}_M\}$ of ${\mathbb{Z}}_M^2$. Each fiber is an affine coset $\{(yb+k,b):b\in {\mathbb{Z}}_M\}$. Measuring the function register gives its normalized coset state. Applying the positive-exponent quantum Fourier transform to both input registers gives amplitudeThe finite geometric series vanishes off $c_2=-y{c}_1$ and equals $M$ on that line. Thus each allowed output pair has probability $1/M$, independently of the coset offset. If $\gcd (c_1,M)=1$, a modular inverse recovers $y=-c_2{c}_1^{-1}$.

A discrete probability distribution is a probability distribution concentrated on a finite or countable set of possible values $S$: a random variable $X$ with this law satisfies $\mathbb{P}(X\in S)=1$. Its probability mass function assigns a probability to each value:Thus the probability of an event is obtained by adding masses, rather than integrating a probability density function. The possible values need not be equally spaced, or even integers. When the relevant moments are finite, its expected value and variance are

For a constructed example, assign values $0,2,5$ masses $0.5,0.3,0.2$. They sum to one; $\mathbb{P}(X\ge 2)=0.3+0.2=0.5$, $\mathbb{E}X=0(0.5)+2(0.3)+5(0.2)=1.6$, and $\mathbb{E}[X^2]=4(0.3)+25(0.2)=6.2$. Hence the variance is $6.2-1.6^2=3.64$. A mean can lie between possible values: no outcome here equals $1.6$.

The useful organizing question is what an experiment records. One yes/no indicator random variable gives a Bernoulli distribution; a success count in a fixed number of independent attempts gives a binomial distribution; the attempt number of the first success gives a geometric distribution. A Poisson distribution describes an event count under different assumptions. Naming the quantity first helps prevent selecting a familiar formula for the wrong experiment.

A random variable is discrete when some finite or countable set $D\subseteq \mathbb{R}$ has $\mathbb{P}(X\in D)=1$. Its probability mass function is $p_X(x)=\mathbb{P}(X=x)$; these nonnegative masses sum to one. An event determined by the value has probability $\mathbb{P}(X\in A)={\sum }_{x\in A\cap D}{p}_X(x)$. Values outside $D$ may still occur on a zero-probability event without changing this definition.

Countable support need not consist of integers: a random variable constantly equal to $3/2$ is discrete. Integer-indexed tail-sum formulas therefore require a separate integer-valued hypothesis. For the dice sum in random variable, the expected value is ${\sum }_{s=2}^{12}s{p}_S(s)=7$ and the variance is ${\sum }_{s=2}^{12}(s-7{)}^2{p}_S(s)=35/6$; the same answer follows by adding the variances of its independent summands.

At each node a positive stock price moves to $u{S}_t$ or $d{S}_t$, with $0<d<u$, while a bank account grows by a fixed gross factor $R>0$. Assume both successors have positive physical probability, no dividends, a frictionless market with no trading costs, divisible holdings, unrestricted short selling, and borrowing and lending at the same fixed $R$. A self-financing portfolio receives no outside cash after its initial funding; subsequent rebalancing uses its own proceeds.

This binomial market excludes arbitrage exactly when $d<R<u$. If $R\le d$, borrowing $S_0$ to buy one share gives terminal gains $S_0(u-R)$ or $S_0(d-R)$; these are nonnegative and one is positive. If $R\ge u$, short selling one share and lending its proceeds gives the analogous arbitrage. Strict inequalities allow a positive risk-neutral probability for each state. The two traded payoffs also span both successor states, giving market completeness relative to the information generated by this tree.

A time-homogeneous Discrete-time Markov chain is a sequence of random variables $Y_0,Y_1,\dots$ indexed by nonnegative integers, with the displayed conditional probabilities. Here ${\mathcal{F}}_n$ is its natural filtration and $P$ is a row-stochastic transition matrix. The Markov property removes dependence on earlier observations once $Y_n$ is specified; the time-homogeneous Markov property makes $P$ independent of $n$.

Consider states $S$ (sunny) and $R$ (rainy), with the constructed transition matrixRows give today's state and columns tomorrow's, in order $S,R$. These numbers define an illustrative model, with one step interpreted as one day; they are not estimates of real weather. Let $q_n=\mathbb{P}(Y_n=S)$. The law of total probability gives the recurrence relationSubtracting $2/3$ makes the deviation multiply by $0.4$ at each step. ThusStarting sunny gives $q_0=1$, $q_1=0.8$, $q_2=0.72$ and $q_3=0.688$; starting rainy gives $q_0=0$ and $q_2=0.56$. The two predictions approach the same stationary distribution while retaining a shrinking trace of the initial probability distribution.

An expected hitting time answers a different question. Starting sunny, let $T_R=\inf \{n\ge 0:Y_n=R\}$. Every day before the first rain remains sunny, soThis geometric distribution has finite expected value. A first-step calculation gives $h_S=1+0.8h_S+0.2h_R$, with $h_R=0$, hence $h_S=5$ days. The same result follows from the geometric series ${\mathbb{E}}_S{T}_R={\sum }_{n\ge 0}0.8^n=5$. This is a mean waiting time, not a prediction that rain first occurs exactly five days later.

A probability row vector $\pi$ is a stationary distribution when $\pi P=\pi$. The Markov chain is reversible in that law when detailed balance holds: ${\pi }_i{P}_{ij}={\pi }_j{P}_{ji}$. Multiplying those equations along a finite path gives equal forward and reversed stationary path probabilities.

A discrete uniform distribution gives every element of a nonempty finite set $S$ the same probability. Write $n=|S|$ for its cardinality. A random variable $X$ with this law has support of a probability distribution $S$ and probability mass functionThe masses sum to $n(1/n)=1$. Conversely, if all $n$ masses equal $p$, their sum is $np=1$, so necessarily $p=1/n$. Thus equal chances determine the whole law.^[ref]

Counting now computes an event's probability: for any set $A$ of possible values,In particular, when $A\subseteq S$, the answer is $|A|/|S|$. The intersection matters: values outside the support of a probability distribution contribute zero. This turns a probability question into a combinatorics question, provided that equal chances are justified by the model.

For an ideal fair six-sided die, $S=\{1,2,3,4,5,6\}$ and each face has probability $1/6$. The event of an even integer is $\{2,4,6\}$, so its probability is $3/6=1/2$; the event $X\ge 5$ has probability $2/6=1/3$. Direct summation gives the expected value $\mathbb{E}X=21/6=7/2$ and the variance $\mathrm{Var}(X)=91/6-(7/2{)}^2=35/12$. The mean $7/2$ is a balance point between possible outcomes. These are exact consequences of stipulated fairness, rather than measured frequencies from physical dice.

Uniformity describes probabilities, not distances between values. For a second constructed model, stipulate equal-chance selection from $S=\{1,2,4,8\}$. Every value has probability $1/4$, despite the unequal gaps $1,2,4$. Here $\mathbb{P}(X\ge 4)=2/4=1/2$, while the expected value and variance areNeither the midpoint of the extreme values, $9/2$, nor a formula using only the cardinality gives these moments. For a general numerical support of a probability distribution, use its actual values:

Choosing an item from a finite list with equal chances is a basic building block of simple random sampling. A bijection that relabels the items preserves equal probabilities, although changing numerical labels can change the expected value and variance. A function that merges labels can destroy uniformity: for the fair-die model, $Y=\min (X,3)$ has masses $1/6,1/6,4/6$ at $1,2,3$, since four faces map to three. Always count the original equally likely outcomes.

An indicator random variable for membership in a set $A\subseteq S$ has a Bernoulli distribution with parameter $|A|/n$; counting that event over $r$ independent repetitions gives a binomial distribution with parameters $r$ and $|A|/n$. Equal chances among individual values need not produce equal chances among aggregated counts.^[ref]

Suppose the support of a probability distribution consists of all consecutive integers from $a$ to $b$, including both endpoints, with $a\le b$. Its cardinality is $n=b-a+1$, and a discrete uniform distribution on it satisfiesFor a short derivation, set $K=X-a$, whose possible values are $0,\dots ,n-1$. The finite sums giveand henceAdding $a$ shifts the expected value by $a$ and leaves the variance unchanged. For completeness, the first sum follows by pairing $k$ with $n-1-k$. For the second, sum $(k+1{)}^3-k^3=3k^2+3k+1$ from $k=0$ to $n-1$: the left side is $n^3$, so inserting the first sum gives the stated result.

The edge case $a=b$ has $n=1$: $X=a$ with probability one, expected value $a$, and variance zero. On an arithmetic progression $\{a,a+h,\dots ,a+(n-1)h\}$ with $h>0$, the same argument instead gives expected value $a+h(n-1)/2$ and variance $h^2(n^2-1)/12$. Equal spacing is the extra hypothesis behind these formulas; the discrete uniform distribution on $\{1,2,4,8\}$ requires the general sums above.

A finite open coordinate set witnesses an increasing event when prescribing those coordinates open forces the event, regardless of the remaining configuration. The disjoint occurrence of increasing events $A\square B$ means that $A$ and $B$ have disjoint finite witnesses. For finite-coordinate events this is the usual disjoint-occurrence definition. The Van den Berg-Kesten inequality bounds its probability by $\mathbb{P}(A)\mathbb{P}(B)$ under an independent Bernoulli distribution product measure.

Independence factors the events that both variables exceed a threshold or both lie below it. This gives the two cumulative distribution functions above, including for distributions with atoms. The minimum formula is most naturally expressed through survival probabilities, while the maximum formula uses the CDFs directly.

Two electrons of opposite spin occupying the same spatial orbital. The mean occupation product measures its probability at a specified site. In the two-electron dimer, the sum over sites is also the probability of any doubly occupied site, since two sites cannot both be doubly occupied.

The ground-state probability of both electrons occupying one site is $P_{\mathrm{double}}=\frac{1}{2}[1-U/\sqrt{U^2+16t^2}]$. It equals the derivative of the relative ground energy with respect to $U$. It tends to zero for $|t|/U\to 0$ and to $1/2$ for $|t|/U\to \infty$.

For constant diffusion coefficient $D$ and drift velocity $v\ne 0$, the balance $|v|t\sim \sqrt{Dt}$ gives the characteristic length $D/|v|$ and time $D/v^2$. The fundamental Gaussian distribution of the advection-diffusion equation has a fixed-position exponential factor $\exp [-v^{2}t/(4D)]$. Total probability is still conserved: this describes probability moving away from a fixed observation region, rather than disappearance of the particle. At zero drift, this finite exponential time scale is absent.

For independent and identically distributed real random variables with cumulative distribution function $F$, the Dvoretzky-Kiefer-Wolfowitz inequality gives, for every $n\ge 1$ and $\epsilon >0$,The supremum is the Kolmogorov distance: a single event controls all thresholds simultaneously. The bound holds for discrete, continuous, and mixed probability distributions.^[ref]

For one fixed threshold, Hoeffding's inequality gives the same expression as an upper bound on $\mathbb{P}(|F_n(x)-F(x)|>\epsilon )$. That fixed-threshold result alone does not justify moving the supremum inside the probability. The DKW inequality is the stronger statement needed for a simultaneous empirical distribution confidence band.

Along the dyadic partitions of an interval of $[0,1]$, the sums ${\sum }_i|{\Delta }_{i}M{|}^p$ of a continuous local martingale tend to zero in probability for $p>2$. If $0<p<2$ and these sums are eventually bounded almost surely, then $M$ is constant on $[0,1]$: compare them with the sums defining quadratic variation, using the maximum increment, which tends to zero by uniform continuity.

Deleting an edge of a locally finite recurrent graph leaves recurrent Markov chain connected components. On an infinite component choose a root away from the deleted edge's endpoints, so its vertex degree does not change. The Rayleigh monotonicity principle increases resistance to wired outer boundaries, and effective resistance to infinity therefore remains infinite. Finite components are recurrent Markov chain too. This is a recurrence-type comparison; at an endpoint, changes in the vertex degree factor prevent a blanket monotonicity claim about first-return probabilities.

The Efron–Stein inequality bounds the variance of a complicated output by the effect of replacing one input at a time. Let $X_1,\dots ,X_n$ be mutually independent random variables, not necessarily identically distributed, and $Z=f(X_1,\dots ,X_n)$ for a measurable function $f$, with $\mathbb{E}{Z}^2<\infty$. Take $X_1^{\prime },\dots ,X_n^{\prime }$ with $X_i^{\prime }$ having the same probability distribution as $X_i$, all $2n$ inputs mutually independent. DefineUse the same function $f$, keeping every other input fixed. ThenThe expected value averages over both the original inputs and their replacements. An infinite right side gives a trivial bound; under the stated square-integrable condition every term is finite.^[ref]

For a concrete four-state example, take independent Bernoulli random variables with $\mathbb{P}(X_1=1)=0.3$ and $\mathbb{P}(X_2=1)=0.6$, and set $Z=\max \{X_1,X_2\}$. This output records whether either input is one. The states $(0,0),(1,0),(0,1),(1,1)$ have respective probabilities $0.28,0.12,0.42,0.18$, soReplacing $X_1$ changes $Z$ only when $X_2=0$ and $X_1^{\prime }\ne X_1$. A fresh Bernoulli draw can repeat the old value: the replacement is not a forced flip. By independence, the expected value of the squared change isSimilarly,Thus Efron–Stein inequality gives $0.2016\le (0.168+0.336)/2=0.252$. The larger second contribution reflects both its replacement probability and how often the other input leaves the output sensitive to it.

For the same inputs, consider instead $W=X_1+2X_2$. The variance of a sum of independent random variables givesHere $\mathbb{E}[(W-W_1^{\prime }{)}^2]=0.42$ and $\mathbb{E}[(W-W_2^{\prime }{)}^2]=1.92$, so the Efron–Stein inequality is exact: $(0.42+1.92)/2=1.17$. More generally, for $W=c+{\sum }_i{g}_i(X_i)$ with square-integrable summands, each contribution $\frac{1}{2}\mathbb{E}[(W-W_i^{\prime }{)}^2]$ equals $\mathrm{Var}(g_i(X_i))$ and these variances add exactly.

Independence is essential. Let $X_1=X_2=B$ for a single Bernoulli random variable with parameter $1/2$, and take $W=X_1+X_2=2B$. Its variance is one. If each coordinate is nevertheless replaced by an independent draw from its marginal Bernoulli distribution, each squared-change expected value is $1/2$, giving the false proposed upper bound $1/2$. Replacement now creates off-diagonal states $(0,1)$ and $(1,0)$ that the original input never visits. The original coordinates fail the hypothesis of Efron–Stein inequality.

The attraction is local calculation: a function may involve many inputs, yet changing one can have a simple effect. Each half squared-change expected value supplies a coordinate's contribution to an upper budget for the total variance. Applying Chebyshev's inequality then bounds a tail probability using that budget. These contributions need not partition the actual variance, as the maximum example shows.

The name commemorates Bradley Efron and Charles M. Stein. Their 1981 paper The Jackknife Estimate of Variance appeared in The Annals of Statistics, volume 9, issue 3, pp. 586–596.^[ref]

In the straight-trajectory dipole model, retain the Coulomb-coupling convention of $V(t)=Z{e}^2(bx+vty)/(b^2+v^2{t}^2{)}^{3/2}$. In SI units the coupling denoted $e^2$ here is $e^2/(4\pi {ϵ}_0)$. First-order time-dependent perturbation theory expresses a transition through the Fourier transform of this pulse at the transition angular frequency.

For positive $\omega$ put $\beta =\omega b/v$. The two Cartesian pulse transforms arewhere $K_{\nu }$ is a Modified Bessel function of the second kind. One may derive these from the Fourier transform of $(1+s^2{)}^{-1/2}$, differentiating in its scale and integrating its spatial derivative by parts. For $\beta \ll 1$, $\beta K_1(\beta )\to 1$ while $\beta K_0(\beta )\to 0$, so the transverse impulse gives the probability in the title. If the $x$ matrix element vanishes, the smaller longitudinal term can be the leading nonzero contribution.

The conditions include $b$ large compared with the atomic extent for the electric-dipole approximation, a small total transition probability for first-order theory, and a collision time $b/v$ short compared with $1/|\omega |$. Large $\beta$ instead gives exponentially suppressed excitation in this model through the large-argument behavior of $K_{\nu }$.