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Let $U$ have the uniform distribution on $[0,1)$ and place a point at every $k+U$, $k\in \mathbb{Z}$. This is a simple point process, locally finite, and a stationary point process: a translation only changes the phase modulo one, preserving its uniform distribution. For every Borel set $A$, the Tonelli theorem givesIts consecutive points are exactly one unit apart, despite having the same intensity measure as a unit-rate Poisson point process. This is the one-dimensional translated-grid construction.^[ref]

Every half-open interval of length one contains exactly one point. For $A=[0,1/2)$ and $B=[1/2,1)$, $\Pi (A)+\Pi (B)=1$, each count has the Bernoulli distribution with parameter $1/2$, andMore generally, an interval of length $L=m+r$, with $m$ a nonnegative integer and $0\le r<1$, contains $m$ or $m+1$ points with probabilities $1-r$ and $r$. Its count variance is $r(1-r)$, while a unit-rate Poisson point process has count variance $L$.

For independent $U,V$ uniform on $[0,1]$, real roots occur on $U\le V^2$, a region of probability $1/3$. On that region both roots are nonpositive. Both lie in $[-1,0]$ exactly when $\max (0,2V-1)\le U\le V^2$. Its area is $1/12$, so the conditional probability of both absolute values being at most one is $1/4$. Joint probability density integration turns algebraic discriminant and root constraints into simple planar areas.

A real-valued random variable is a measurable function $X$ from a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ to the real numbers with their Borel sigma-algebra. More general targets give random elements. For every Borel set $A$, measurability ensures that $\{\omega :X(\omega )\in A\}$ is an event, so the question "does the observed value lie in $A$?" has a defined probability.

The function $X$ is fixed; its input outcome is uncertain. This separates what we measure from how the uncertainty is modeled. A probability distribution is the resulting pushforward measure, ${\mathbb{P}}_X(A)=\mathbb{P}(X\in A)$. A cumulative distribution function, probability mass function, or probability density function describes that measure in an appropriate setting. The random variable and its probability distribution are different objects: the former also retains its relationship to other measurements on the same sample space.

For two independent uniform dice, let $U,V$ record their results and define $S=U+V$ and $M=\max (U,V)$. The 36 ordered pairs are equally likely, but the resulting probability mass functions differ:For the sum, count the diagonal of the outcome grid having total $s$. For the maximum, subtract the $(m-1{)}^2$ pairs with both results below $m$ from the $m^2$ pairs with both results at most $m$. These examples show how a deterministic function transforms a simple input probability distribution into a nonuniform output one.

The corresponding expected values are $\mathbb{E}S=7$ and $\mathbb{E}M=161/36$. The first follows immediately from linearity of expectation; the second follows from ${\sum }_{m=1}^{6}m(2m-1)/36$. Neither number must be an attainable result. An expected value describes an average under the probability measure, not a prediction that the next observation will equal it.

The distinction between a random variable and its probability distribution matters for dependence. $U$ and $V$ have the same discrete uniform distribution and are independent random variables. $U$ and a second copy of $U$ also have that same marginal probability distribution, but they always agree. Their joint probability distribution is concentrated on the diagonal; the marginals alone cannot say which relationship holds. This is why a model for several measurements must specify their joint law or an appropriate independence assumption.

Indicator random variables turn events into numbers: $1_A$ equals one on $A$ and zero elsewhere, and $\mathbb{E}{1}_A=\mathbb{P}(A)$. This connects event counting with expected value and variance, and lets tools for random variables answer questions about probabilities. More complicated functions of observations can describe totals, waiting times, or estimation errors using the same framework.

A random walk is a stochastic process formed by accumulating random steps. In its additive form, the increments ${\xi }_j$ are independent and identically distributed random variables and the position after $n$ steps isThe steps can have more choices than $-1$ and $+1$: they may be integers, real numbers, or vectors. A random walk on a graph instead describes steps between vertices along edges; its available moves can depend on the current vertex.

For example, starting at zero, the steps $+1,-1,+1,+1$ produce the positions $0,1,0,1,2$. This is one sample path; other step sequences can have the same endpoint. A simple random walk on the integer line makes this distinction particularly tangible: repeated Bernoulli trials determine both a whole sample path and an endpoint with a binomial distribution. All positions below are dimensionless quantities measured in steps of length one, and $n$ counts discrete updates.

The model is useful because it separates an average tendency from accumulated fluctuations. A fair simple symmetric random walk has zero expected value at every time, yet its standard deviation grows: cancellation across many possible sample paths does not keep each walker near zero. The two figures below compare sample paths with exact endpoint probabilities. For the basic additive construction and its independent increments, see MIT, Stochastic Processes I, section 2.

By I. J. Good.

Published in the Journal of the Royal Statistical Society, Series B, volume 14, number 1, pp. 107–114.

Publisher website: rss.onlinelibrary.wiley.com/doi/10.1111/j.2517-6161.1952.tb00104.x

Cited passage: the published summary describes encouraging accurate probability estimates and a connection with information theory.

The Rayleigh distribution is the probability distribution of the Euclidean norm of a two-dimensional random vector with independent zero-mean normal random variables of equal variance as components. If $X,Y\sim N(0,{\sigma }^2)$ are independent, thenHere $\sigma$ is the standard deviation of each component, not of the radius $R$. At $\sigma =1$ this gives the unit-scale probability density function $r{e}^{-r^2/2}$.^[ref]

This geometry turns two signed errors into one nonnegative distance. For example, suppose horizontal and vertical positioning errors are independent $N(0,4{\mathrm{cm}}^2)$ random variables. Their means both vanish, but the distance from the intended point is usually positive. The Rayleigh distribution with $\sigma =2\mathrm{cm}$ gives the probability of landing inside any centered disk. These are chosen model assumptions, rather than measurements of a particular instrument.

An apparently paradoxical feature is that the most likely location per unit area is the origin, while the radial mode is $\sigma$. A narrow annulus farther out has more area. Near zero that growing circumference outweighs the falling planar density; farther out the exponential function wins. Distinguishing densities measured against different elements of area or length resolves the paradox.

For $r\ge 0$, integration gives the survival function and cumulative distribution functionSolving $F_R(r_p)=p$ gives the quantile function^[ref]

For the chosen positioning model $\sigma =2\mathrm{cm}$,The disk of radius one component standard deviation therefore contains about 39.35% of outcomes; the familiar one-dimensional normal interval percentage does not apply to a two-dimensional disk. The 95% and 99% radial thresholds areA circle defined by $R=r_p$ itself still has probability zero; its enclosed disk carries the stated probability.

The closed-form quantile function also gives inverse transform sampling: if $U$ has uniform distribution on $(0,1)$, then $\sigma \sqrt{-2\log U}$ has Rayleigh distribution. Using $1-U$ instead gives the same law. Endpoints are excluded so that the logarithm stays finite.

A reactive strategy cooperates with probability $p$ after the opponent cooperated and probability $q$ after the opponent defected. It ignores its own previous action. Independent behavioural randomization gives a four-state joint-action Markov chain. The reduced marginal recurrences depend on the prescribed initial action distribution.

For normalized linearly independent pure states, their Gram matrix $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ is positive definite. The displayed reciprocal vectors satisfy $⟨{\chi }_i|{\psi }_j⟩={\delta }_{ij}$. The operator $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$ has nonzero eigenvalues equal to those of $G^{-1}$: use the polar factorization of the matrix whose columns are ${\psi }_i$. Choosing $0<ϵ\le {\lambda }_{\min }(G)$ gives the POVM effects $E_i$ and $E_{?}=I-ϵR$. Each conclusive outcome has probability $ϵ$ on its own state and zero on every other input. The inconclusive effect preserves positivity, so the construction is a physical measurement in quantum mechanics.

The recombination fraction is the probability that a transmitted gamete carries a recombinant haplotype between two specified genetic loci, relative to the parent's known haplotype phase. Write it as $r$ here; the symbol $\theta$ is also common in genetic linkage analysis. In the ordinary symmetric two-locus model, $0\le r\le 1/2$, with $r=0$ for complete genetic linkage and $r=1/2$ for independent assortment.

For phase-known genotype $AB/ab$, fair Mendelian segregation and equal reciprocal gamete probabilities giveFor example, 1,000 independent informative transmissions with observed counts $400,400,100,100$ in this order have empirical recombination fraction $(100+100)/1000=0.2$. Those illustrative counts match the expected values in the toy model, but actual counts fluctuate. The genetic linkage diagram shows both the parental and recombinant classes, so their separate probabilities are not confused with the total $r$.

The recombination fraction measures the probability of an odd number of origin switches along the transmitted chromatid. It is neither the expected value of the exchange count nor the probability of at least one exchange: even positive counts can be invisible at the two endpoints. Converting an observed recombination fraction to a genetic map distance therefore needs an explicit model such as the Haldane mapping function. The distinction between transmitted-chromatid counts and four-chromatid exchange counts is discussed in Weeks, Tang and Kwon (2009), Casares' map function: No need for a 'corrected' Haldane's map function, Genetica 135, 305–307.

For a bounded continuous adapted process $H$ and a Brownian motion $B$, ${\int }_t^{t+h}{H}_{s}d{B}_s/(B_{t+h}-B_t)$ converges in probability to $H_t$ as $h\downarrow 0$. The Itô isometry makes the error numerator small relative to $\sqrt{h}$; the normal distribution of the denominator controls its chance of being too close to zero.

For the induced square lattice subgraph in $[0,m]\times [0,n]$, $h_p(m,n)$ is the probability of an open graph path from its left boundary to its right boundary under independent bond percolation of density $p$. Width and height count lattice spacings. Dual bond percolation exchanges failure with a top-to-bottom dual crossing of dimensions $(m-1)\times (n+1)$, with the usual exterior boundary vertices.

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.

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 $p>1/2$.

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.

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.