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The fundamental theorem of calculus says that accumulation and local change undo one another, under appropriate regularity conditions. If $f$ is a continuous function on $[a,b]$ andthen $A$ is differentiable for $a<x<b$ and $A^{\prime }(x)=f(x)$. Conversely, if $G$ is an antiderivative of $f$ on this interval, continuous at its endpoints, thenThe first statement turns an integral into a function with a known derivative; the second turns an entire accumulation into two endpoint values.

For $f(t)=1+2t$, accumulation from zero gives $A(x)=x+x^2$. At $x=2$, its derivative is $A^{\prime }(2)=5$, exactly the current value $f(2)$. The accumulated value there is $A(2)=6$, and by $x=3$ it has reached $A(3)=12$. The accumulated quantity and its rate are different functions, linked by differentiation.

This connection makes the theorem useful far beyond geometric area. It recovers displacement from velocity, gives a probability from a probability density function, and converts many differential equations into integral equations. The striking economy is that information spread across a whole interval can sometimes be recovered from an antiderivative at its two boundaries.

A short explanation of the first part is thatis the local average of $f$. Continuity makes this average approach $f(x)$. For the second part, $G-A$ has zero derivative and is constant by the mean value theorem. These are the ideas behind the proof; the endpoint and regularity conditions make them valid.

An early geometric form of this connection appears in Isaac Barrow's Lectiones geometricae (1670), Lecture X, Proposition 11. The original volume is catalogued by CESIMA, and Michael Nauenberg's historical research reproduces and examines the geometric argument.^[ref][ref]

For a Galton-Watson process with one ancestor and offspring probability generating function $F$, the branching extinction probability $q$ is the smallest fixed point of $F$ in $[0,1]$. A direct way to find it is to compute extinction by generation $n$:Zero is an absorbing state, so extinction by generation $n$ is exactly the event $Z_n=0$. ^[ref]

For $F(s)=1/4+s/4+s^2/2$, the equation isBoth $1/2$ and $1$ are fixed points, but iteration from zero selects $q=1/2$. The first values are $q_1=1/4$, $q_2=11/32$, $q_3=809/2048\approx 0.395020$, and $q_6\approx 0.462186$. Iteration started at $1$ would stay at $1$ forever and would give the wrong extinction answer for this offspring distribution.

The vertical steps in the figure apply the offspring probability generating function; horizontal steps return to the diagonal to start the next iteration. This gives a useful numerical calculation even when solving $F(s)=s$ explicitly is difficult. For the displayed law, every finite iteration is a lower bound for the eventual branching extinction probability; at generation six, about $0.037814$ additional extinction probability remains beyond the current empty families.

Why the smallest fixed point? The ancestor's children's descendant families are independent; given $k$ children, all are empty $n$ generations later with probability $q_n^k$. Averaging gives $q_{n+1}={\sum }_k{p}_k{q}_n^k=F(q_n)$. The increasing extinction events give the limit $q$, and continuity of the probability generating function yields $F(q)=q$. For any other fixed point $r\in [0,1]$, $q_0\le r$ and monotonicity of $F$ imply $q_n\le r$ at every step; hence $q\le r$. ^[ref]

A Galton-Watson process is a branching process that counts individuals by generation. Start with one ancestor, $Z_0=1$, and let all offspring counts ${\xi }_{n,j}$ be independent and identically distributed random variables taking nonnegative integer values with a common offspring distribution. ThenThe empty sum is zero: zero is an absorbing state, so once a generation is empty every later generation is empty. The same offspring distribution applies to every individual in every generation. ^[ref]

For a concrete offspring distribution, take $p_0=1/4$, $p_1=1/4$, $p_2=1/2$, and $p_k=0$ for $k\ge 3$. An individual has no children with probability $1/4$, one with probability $1/4$, and two with probability $1/2$. The offspring expected value is $\mu =5/4$. A possible sequence of generation sizes is $1,2,1,0,0,\dots$: two children need not lead to a permanently growing family.

The probability generating function $F(s)=1/4+s/4+s^2/2$ gives the exact generation probability distributions through branching-process generating-function iteration. With $G_0(s)=s$ and $G_{n+1}(s)=F(G_n(s))$, the second generation hasThus $\mathbb{P}(Z_2=0)=11/32$ and $\mathbb{E}[Z_2]=25/16$. The figure shows full generation probability mass functions, computed by polynomial function composition, through generation six; no simulated frequencies are used.

The contrast is striking: the expected generation size in a Galton-Watson process grows as $(5/4{)}^n$, while the branching extinction probability is $1/2$. By generation six, $\mathbb{P}(Z_6=0)\approx 0.462186$ although $\mathbb{E}[Z_6]\approx 3.814697$. Larger surviving families carry the rising expected value; that mean is an average over possible families, not a forecast of one family's path. The Galton-Watson extinction fixed point turns an unbounded number of future generations into a single equation for this probability.

The historical question was the disappearance of surnames passed through male descendants. In Watson and Galton, On the Probability of the Extinction of Families, pages 139–140, Francis Galton formulates a constant-law model counting sons who reach adulthood, and Watson computes later generation probability distributions by composing generating functions. Those are assumptions of their model, rather than a claim that actual human reproduction follows this offspring distribution.

For independent means $\theta \sim N({\mu }_{\theta },t_{\theta })$ and $\varphi \sim N({\mu }_{\varphi },t_{\varphi })$, the first mean's conditional precision is $m/v+1/t_{\theta }$ and conditional precision-weighted mean is ${\sum }_{i\le m}{x}_i/v+{\mu }_{\theta }/t_{\theta }$; replace $m$ by $n-m$ for the second mean. With $v\sim \mathrm{IG}(a_0,b_0)$ its conditional law is $\mathrm{IG}(a_0+n/2,b_0+S_m/2)$. The discrete split has conditional probabilities proportional to $e^{-S_m/(2v)}$. These full conditional distributions give a Gibbs sampler including $m=n$. Averaging the sampled indicators of that configuration estimates its probability; averaging its conditional probabilities gives an estimate by Rao-Blackwellization.

For a finite-dimensional discounted gain random vector $X$, absence of arbitrage means $h\cdot X\ge 0$ almost surely only when $h\cdot X=0$ almost surely. Let $N=\{h:h\cdot X=0\text{ almost surely}\}$ and minimize $F(h)=\mathbb{E}{e}^{-|X{|}^2-h\cdot X}$ on the orthogonal complement $N^{\perp }$. The Gaussian damping makes $F$ finite and differentiable without physical integrability assumptions on $X$. Every nonzero direction in $N^{\perp }$ has negative gains on an event of positive probability; compactness of directions proves coercivity. At a minimizer, differentiation gives $\mathbb{E}[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]=0$. Normalizing the positive weight produces an equivalent martingale measure with integrable gains. This is a constructive one-period proof of the fundamental theorem of asset pricing even on an infinite probability space.

Given a real unit-vector Gram matrix $Y_{ij}=⟨v_i,v_j⟩$, draw a standard Gaussian random vector $Z$ and set $y_i=\mathrm{sign}⟨v_i,Z⟩$. Each coordinate is almost surely a sign; zero projections have probability zero. The same random separating hyperplane is used for every coordinate, so the signs need not be independent. Their pair expectations follow the Gaussian sign-correlation identity.

In the normal linear model, assume $\epsilon \sim N_n(0,{\sigma }^{2}I)$ is independent of $X$. Independent rows of $X$ with multivariate normal distribution $N_p(0,I_p)$ give a simple random design for Debiased Lasso. Let $p\to \infty$, $\log p=o(n)$ and $s\log p=o(\sqrt{n})$, with fixed positive noise scale. Choose the Lasso penalty and all Nodewise Lasso penalties as sufficiently large constant multiples of $\sqrt{\log p/n}$, including the noise scale in the former. Uniform sample cross-correlations are bounded by this scale, so zero is the nodewise solution with probability tending to one. Column squared norms divided by $n$ converge uniformly to one. To check the Compatibility condition for the Lasso directly, put $\hat{\Sigma }=X^{T}X/n$ and $\eta ={\max }_{j,k}|{\hat{\Sigma }}_{jk}-1_{\{j=k\}}|$. On the Lasso cone condition, $\Vert \delta {\Vert }_1\le 4\Vert {\delta }_S{\Vert }_1$, and the Cauchy-Schwarz inequality givesFor $s\ge 1$, entrywise concentration of a Gaussian Gram matrix gives $\eta \le D_0\sqrt{\log p/n}$ with probability tending to one for sufficiently large fixed $D_0$. The stated sparsity condition implies $s\eta \to 0$ on this event, so one can take ${\varphi }^2\ge 1/2$ eventually. The conditional Gaussian tail bound for the noise score and the Basic inequality for the Lasso then give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le Cs\sqrt{\log p/n}$ with probability tending to one. Consequently the debiased-Lasso remainder bound is at most $As\log p/\sqrt{n}$ with probability tending to one, for a fixed constant $A$. These assumptions allow both $p$ and $s$ to grow, for example $p=n^2$, $s=\lfloor n^{1/4}\rfloor$.

The Gaussian integral gives the exact area under a Gaussian function. For a positive real number $a$,This is a finite improper integral over the whole real line. For example, $I(1)=\sqrt{\pi }\approx 1.772454$ and $I(4)=\sqrt{\pi }/2\approx 0.886227$. Increasing $a$ from one to four narrows the curve and halves its area, while its height at zero stays one. The change of variables $u=\sqrt{a}x$ explains the factor $a^{-1/2}$.

The useful consequence is an exact normalization. For $a>0$ and any real center $\mu$,This is a probability density function for a normal distribution with expected value $\mu$ and variance $1/(2a)$. A narrower probability density function must be taller to keep total probability one. In the usual notation $a=1/(2{\sigma }^2)$, the prefactor is $1/(\sigma \sqrt{2\pi })$, where $\sigma$ is the standard deviation.

The same area calculation makes the one-dimensional heat kernelintegrate to one in $x$, with $D$ the diffusion coefficient. A spreading Gaussian function can therefore preserve total mass while its height falls. This links the Gaussian integral with both the normal distribution and the diffusion equation.

Finite intervals lead to the error function, rather than this single constant:The complementary tail is described by the complementary error function. These definitions and their real-axis limits are given in the NIST reference. ^[ref]

The polar-coordinate proof of the Gaussian integral explains why $\pi$ appears in an integral of a function defined on a line: squaring its area creates a rotationally symmetric integral on a plane. Substitution $t=x^2$ also gives $\Gamma (1/2)=2{\int }_0^{\infty }{e}^{-x^2}dx=\sqrt{\pi }$, connecting the result with the gamma function.

A short concentration inequality argument explains the bound. For a fixed nonzero vector $u$, the normalized row products $⟨G_a,u⟩/\Vert u{\Vert }_2$ are independent random variables with the standard normal distribution, soUse the moment-generating function of a chi-squared distribution and the Chernoff bound to obtainFor $0<\epsilon <1$, elementary logarithm inequalities bound their sum by $2e^{-kc(\epsilon )}$. Apply the union bound to the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ nonzero pair differences to get failure probability at most $n(n-1)e^{-kc(\epsilon )}\le \delta$. Independence of different pairs is unnecessary; they use the same map. This combines fixed-vector concentration with a finite-set union bound. ^[ref]

Draw a $k\times D$ random matrix $G$ whose entries are independent random variables with the standard normal distribution, and setThe factor $1/\sqrt{k}$ makes the expected value of the squared L2 norm equal to the original squared L2 norm. This Gaussian Johnson–Lindenstrauss transform is a linear map, generally not an orthogonal projection: its rows are not required to be orthogonal vectors. ^[ref]

For the fixed finite set $S$, let $0<\delta <1$ be an allowed failure probability, and defineWith $\log$ denoting the natural logarithm, a sufficient integer target dimension isThen one draw of this specified random matrix satisfies all the squared Euclidean distance bounds simultaneously with probability at least $1-\delta$. Since $c(\epsilon )\ge {\epsilon }^2/12$, the simpler sufficient bound $k\ge \lceil 12{\epsilon }^{-2}\log (n(n-1)/\delta )\rceil$ also works. Both are sufficient bounds for this argument; they do not claim the smallest possible target dimension.

This is a probability statement about a map drawn independently of the fixed finite set. It implies existence because the success probability is positive, but permits failed draws. The set cannot be chosen after inspecting the map with this same guarantee. In particular, no single draw preserves every possible finite set in the ambient Euclidean space when $k<D$.

Take the realized statistical sample $(-1,0,1)$, the Gaussian density kernel $\varphi (u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$ and smoothing bandwidth $h=1/2$. ThenAt $x=0$, the three contributions are $0.03599398$, $0.26596152$, and $0.03599398$, because $\varphi (0)=0.39894228$ and $\varphi (2)=e^{-2}/\sqrt{2\pi }=0.05399097$. Their sum isThis is a probability density function value, not the probability of observing exactly zero. Each of the three component curves has area $1/3$, despite having a different height at zero.

Let $\Phi$ be the cumulative distribution function of the standard normal distribution. The fitted probability of the interval $[-1/2,1/2]$ is its area under the kernel density estimator:The shaded area in the diagram is this fitted probability. Its agreement to a few digits with $1/3$ is incidental: the empirical measure gives this interval mass $1/3$ exactly, whereas kernel density estimation redistributes that mass into neighboring intervals.

Stop the Exponential martingale for Brownian motion $e^{\theta B_s-{\theta }^{2}s/2}$ when $B$ first reaches $\delta$ or at $t$. On the hitting event its value is at least $e^{\theta \delta -{\theta }^{2}t/2}$, so that event has probability at most $e^{-\theta \delta +{\theta }^{2}t/2}$. Minimize over $\theta >0$ and apply the same bound to $-B$. The union bound gives the displayed inequality for $t>0$; at zero the probability is zero.

Gaussian process classification links latent variables from a Gaussian process to class probabilities. A binary probit regression likelihood is ${\prod }_i\Phi ((2y_i-1)f_i)$ under conditional independence. The posterior distribution of the latent values is not a normal distribution.

For independent $N(0,1)$ increments, a random walk has $S_j\sim N(0,j)$. If $B$ is a bounded measurable set, its occupation time probability obeys $\mathbb{P}(S_j\in B)\le |B|/\sqrt{2\pi j}$. Consequently $\mathbb{E}{\sum }_j{1}_B(S_j)/j<\infty$, and Tonelli's theorem makes this series finite almost surely. Kronecker lemma then gives the displayed zero occupation density. This also shows why its empirical occupation probabilities cannot converge to a probability measure simultaneously on all bounded intervals and the whole real line: continuity from below would contradict the preserved total mass one.

For centered, unit-variance jointly Gaussian random variables with correlation coefficient $\rho$, the displayed identity holds. Realize the pair as projections onto unit vectors of angle $\theta =\arccos \rho$. Rotational symmetry of the two-dimensional Gaussian random vector direction makes sign disagreement have probability $\theta /\pi$. The expectation is therefore $1-2\theta /\pi =(2/\pi )\arcsin \rho$. Perfectly correlated and anticorrelated endpoints follow directly; the sign at zero does not affect the result.

A genetic genetic carrier has one disease-associated recessive allele and one normal allele at a genetic locus, and is unaffected under a fully penetrant recessive model. Two genetic carriers have affected-offspring probability $1/4$ under Mendelian segregation; a genetic carrier and an individual with no disease allele have affected-offspring probability zero under that same mutation-free model.

Why can two individually fair allele transmissions fail to be independent? Genetic loci on the same chromosome can travel together through meiosis. This tendency is genetic linkage; in an ordinary two-locus model, detectable genetic linkage means a recombination fraction below $1/2$. A genetic marker can therefore help track a neighboring genetic locus through a pedigree. See Clark, Douglas and Choi, Biology 2e: chromosomal theory and genetic linkage.

Take a diploid individual heterozygous at two genetic loci, with phase-known genotype $AB/ab$: one homologous chromosome carries alleles $A,B$, the other $a,b$. The parental haplotypes are $AB,ab$; the recombinant haplotypes are $Ab,aB$. Assume fair Mendelian segregation, equal probabilities within each reciprocal pair of gametes, and no allele-dependent differences in gamete recovery. If the recombination fraction is $r=0.2$, the gamete probabilities are $0.4,0.4,0.1,0.1$ in that order. Each genetic locus still transmits its two alleles equally, but $P(AB)=0.4$ exceeds $P(A)P(B)=0.25$: fairness at each genetic locus does not imply independence between genetic loci.

At $r=0$, only the parental haplotypes occur, each with probability $0.5$: this is complete genetic linkage. At $r=0.5$, all four haplotypes have probability $0.25$, giving independent assortment. A recombination fraction near $0.5$ alone cannot establish whether the genetic loci lie on different chromosomes or far apart on one chromosome.

Genetic linkage concerns joint transmission through meiosis. Linkage disequilibrium concerns haplotype frequencies in a population, while genetic association compares allele or genotype frequencies with a phenotype. These questions require different observations; pedigrees directly reveal co-transmission.

A genetic map turns observations of inherited haplotypes into an ordered account of genetic loci along a chromosome. Its distances describe genetic recombination, which makes the map useful for locating a genetic locus relative to known genetic markers. A genetic map measures inheritance separation; separation measured in base pairs describes positions along DNA. Recombination hotspots help explain why equal numbers of base pairs need not give equal genetic map distances. See T. A. Brown, Genomes, 2nd edition (2002), chapter 5: Mapping Genomes.

Write $d$ for the expected value of the number of chromosomal crossovers encountered along one transmitted chromatid between the chosen genetic loci, measured in morgans; $100d$ is the distance in centimorgans. This counts exchanges along that transmitted chromatid, rather than all exchanges in the four-chromatid meiotic bundle. For a short interval with very few multiple exchanges, the recombination fraction satisfies $r\approx d$: about $1\%$ recombinant gametes then corresponds to about one centimorgan.

For a longer interval, hidden even numbers of chromosomal crossovers make that approximation inaccurate. Under the Haldane mapping function, $d=0.2$ morgans gives $r\approx 0.164840$; conversely $r=0.2$ gives $d\approx 0.255413$ morgans, or $25.5413$ centimorgans. Applying the inverse formula to an already specified genetic map distance would confuse a count with a probability.

Genetic map distances add along ordered adjacent intervals because the total exchange count is the sum of its interval counts. For independent random variables describing counts in disjoint intervals of the Haldane mapping function model, the endpoint recombination fraction instead obeysExactly one interval must change parental origin. Thus $r_1=r_2=0.2$ gives $r_{12}=0.32$, although the two genetic map distances add to approximately $0.510826$ morgans. The simple composition formula requires the stated independence; crossover interference can invalidate it.

Genetics studies how genes are inherited, how their alleles vary, and how that variation contributes to phenotypes. Its central distinction is between an inherited description, the genotype, and what an organism actually expresses, the phenotype. Following alleles through a pedigree answers a family-level question; following allele frequencies through generations leads to population genetics and evolution.

The beauty of Mendelian segregation is that discrete inherited copies can generate predictable statistical patterns without making every offspring identical. A single genetic locus already connects meiosis, probability and observable variation. The Hardy-Weinberg principle then supplies a population baseline: it separates reshuffling inherited copies from processes such as natural selection that change their frequencies. These simple models are useful precisely because their assumptions can be stated and checked.

Ascertainment in a genetic study specifies how individuals or families enter a genetic study. If $A$ is the event of inclusion, then $P(X\mid A)=P(X)P(A\mid X)/P(A)$. Selection on disease phenotype can change an unconditional pedigree likelihood. When comparing genetic marker probabilities conditional on a specified disease pattern, selection depending only on that pattern cancels from the linkage likelihood ratio.