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A blocked Gibbs sampler draws groups of coordinates jointly from their full conditional distributions. Each update preserves the joint probability distribution; dependence between blocks can still cause a long mixing time.

Where the marginal density $f_X(x)$ is positive, a joint probability density determines the conditional densityIt integrates to one in $y$, and supplies conditional expectations by integrating against it. Its values on a probability-zero set of conditioning points are not fixed by the joint probability distribution.

A coupling of probability distributions $\mu ,\nu$ is a joint probability distribution of random variables $(X,Y)$ whose marginal distributions are $\mu ,\nu$. It lets us choose the statistical dependence between two random constructions while preserving the individual laws. This choice can make a comparison much easier: giving two simulations the same random input may keep their paths close, even though separately generating both paths would hide that relationship.^[ref]

For example, two fair coin tosses can be generated independently, with mismatch probability $1/2$, or by tossing once and setting $Y=X$, with mismatch probability zero. Both constructions have the same two Bernoulli distributions as marginals. Their joint laws differ. A maximal coupling makes the values agree as often as possible; a coalescing coupling arranges permanent agreement of two processes after a random time. Minimizing expected geometric separation instead leads to Wasserstein distance.

For any step $\Delta >0$, set $a=e^{-\theta \Delta }$. The Ornstein-Uhlenbeck transition distribution gives the exact updateusing a fresh independent standard normal random variable for each interval. The update reproduces the joint probability distribution on the selected time grid. It can start from a fixed value or use Ornstein-Uhlenbeck stationary initialization. On an equally spaced grid it is an autoregressive process of order one after subtracting $\mu$. Its stationary variance follows directly from $v=a^{2}v+v_{\infty }(1-a^2)$.

For the stipulated example with $\Delta =0.1$, $a=e^{-0.1}\approx 0.90484$ and the innovation variance is $1-e^{-0.2}\approx 0.18127$. Given $X_t=6$, the next conditional expected value is about $5.61935$ and the conditional standard deviation is about $0.42576$.

The Euler-Maruyama method instead approximates the same stochastic differential equation byFor $0<\theta \Delta <2$ its stationary variance solves $v=(1-\theta \Delta {)}^{2}v+{\sigma }^2\Delta$, givingAt $\Delta =0.1$ in the example, this is $1/0.95\approx 1.05263$, about $5.26\%$ above the continuous model's stationary variance. If $\theta \Delta \ge 2$, this Euler recurrence has no finite stationary variance despite the stable continuous model. Exact transitions avoid this discretization error at sampled times; neither update alone determines the unresolved motion between samples.

A family of random variables is exchangeable if its joint probability distribution is unchanged by finite permutations of its coordinates. Independent and identically distributed random variables are exchangeable, but exchangeable random variables need not be independent random variables.

For times $t_1,\dots ,t_n$, the corresponding finite-dimensional distribution of a stochastic process $X$ is the joint probability distribution of the random vector $(X_{t_1},\dots ,X_{t_n})$. The collection of these distributions records every finite set of coordinates of the process.

Knowing only the one-time marginal distributions is insufficient. Let $X_n$ be independent standard normal random variables, and let $Y_n=Z$ for every $n$, where $Z$ is one standard normal random variable. Every $X_n$ and $Y_n$ has the same standard normal distribution. Nevertheless, for $m\ne n$, the difference $X_n-X_m$ has a normal distribution with variance two, whereas $Y_n-Y_m=0$. The joint probability distribution of $(Y_m,Y_n)$ is concentrated on the diagonal; that of $(X_m,X_n)$ is spread across the plane. A model for tomorrow's value therefore needs the relation between observations as well as their separate distributions.

For a real-valued stochastic process, its finite-dimensional distributions determine its probability law on the product sigma-algebra. With an uncountable index set, a path property such as continuity need not be measurable in that sigma-algebra; specifying a suitable space of sample paths and a regular modification of a stochastic process is a separate step. This distinction is explained in MIT, The basics of stochastic processes, section 1.1.

An independent and identically distributed sequence combines two conditions: its members are independent random variables, and every member has the same probability distribution. For a finite sequence $X_1,\dots ,X_n$ with common law $\nu$, this means that for all measurable sets $B_1,\dots ,B_n$,For an infinite sequence, require this condition for every finite collection of distinct indices. Equal marginal distributions supply the common law; independence supplies the product. Neither condition supplies the other. Grinstead and Snell call the combination an independent trials process.^[ref]

Choose a Bernoulli distribution with success probability $1/2$. Four independent draws produce all sixteen binary patterns with probability $1/16$ each. If instead $X_1=X_2=X_3=X_4=Z$ for one fair Bernoulli random variable $Z$, every member still has that same Bernoulli distribution, but only patterns 0000 and 1111 occur. Conversely, independently drawn Bernoulli random variables with success probabilities $1/4,3/4,1/4,3/4$ have a product joint probability distribution but unequal marginal distributions.

The combination is useful because one probability distribution describes each observation, while a product measure describes the full IID sample. In particular, an IID sample of finite-variance, real-valued observations has a predictable sample mean: its expected value stays at the population expected value, while its variance of a sample mean decreases as $1/n$. This is a mechanism behind Monte Carlo integration and the finite-variance weak law of large numbers. Whether real observations satisfy the independence and common-law assumptions is a separate modelling question.

Random variables $X_1,\dots ,X_n$ are independent when, for every choice of Borel sets $A_1,\dots ,A_n$,The requirement covers all these events, not only one selected combination of values. Equivalently, the generated sigma-algebras are independent sigma-algebras, or the joint probability distribution is the product measure of the marginal distributions. For an infinite family, the same condition must hold for every finite subfamily. For random elements, replace Borel sets by measurable sets in their specified target spaces.

For discrete random variables, checking every entry of the joint probability mass function is enough:Summing this identity over the desired sets proves the event formulation. For general real-valued random variables, checking point probabilities alone can say nothing: a pair with no point masses has $\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)=0$ even when $Y=X$. The all-set definition avoids that trap. See Gautam Iyer, Conditional Probability and Independence, definitions 12, 16 and 18 and proposition 19, with the continuous case treated in his ^[ref].

For a concrete comparison, let $X,Y$ be independent fair bits: each has a Bernoulli distribution with parameter $1/2$. Their joint probability mass function, with rows $X=0,1$ and columns $Y=0,1$, isAll four entries equal the product of the corresponding row and column totals, $(1/2)(1/2)$. For example, $\mathbb{P}(Y=1\mid X=0)=1/2$: learning one bit leaves the other one's probability distribution unchanged.

Now keep a fair $X$ but set $Y=X$. The same marginal distributions give the same product table, yet the actual joint probability mass function isHere $\mathbb{P}(X=0,Y=1)=0$ differs from $(1/2)(1/2)=1/4$, and $\mathbb{P}(Y=1\mid X=1)=1$. Thus matching marginal distributions does not establish independence. Grinstead and Snell formulate the discrete product criterion in Introduction to Probability, section 4.1, definition 4.4, printed page 143.

Independence is stronger than zero covariance. For square-integrable independent random variables, $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$, so they are uncorrelated random variables. The converse fails: take $X$ with a discrete uniform distribution on $\{-1,0,1\}$ and $Y=X^2$. Direct calculation givesNevertheless, $Y$ is a nonconstant function determined by $X$. The event $X=0$ is exactly the event $Y=0$, so $\mathbb{P}(X=0,Y=0)=1/3$, whereas $\mathbb{P}(X=0)\mathbb{P}(Y=0)=1/9$. The positive and negative contributions to the covariance cancel without erasing the statistical dependence.

The usefulness of independence is that a large joint probability distribution can be assembled from small pieces. For $n$ fair independent bits, each of the $2^n$ patterns has probability $2^{-n}$. Products also simplify expected values, and vanishing cross covariances simplify the variance of a sum. Separate measurable functions of independent random variables remain independent: each output event is already an input event, so the defining product rule still applies. These are consequences of a model's independence assumption; whether the assumption describes actual observations is a question for statistical inference.

For discrete random variables $X,Y$, their joint probability mass function assigns a probability to each possible ordered pair. Its entries are nonnegative and sum to one. Marginalization gives the probability mass functions $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$ and $p_Y(y)={\sum }_x{p}_{X,Y}(x,y)$. In a table with rows indexed by $x$ and columns by $y$, these are the row and column totals. The joint probability distribution records how the coordinates occur together, while the marginal distributions describe each separately.

The law of total expectation makes averaging in stages reliable, provided the outer stage uses the actual probabilities of the groups. For an integrable random variable $X$ and a sub-sigma-algebra $\mathcal{G}$,Replacing each outcome by its group's conditional mean preserves the overall expected value. In the definition of conditional expectation, take the event $A=\Omega$ to obtain this identity.^[ref]

For a complete finite example, choose a group $Y=A$ with probability $1/4$ and $Y=B$ with probability $3/4$. Within group $A$, $X$ is $0$ or $4$ with equal conditional probabilities; within group $B$, it is $6$ or $10$ with equal conditional probabilities. Thus the joint probability distribution puts probabilities $1/8,1/8,3/8,3/8$ on $(A,0),(A,4),(B,6),(B,10)$. The conditional means are two and eight. Consequently,Direct averaging of the four values gives the same $0/8+4/8+18/8+30/8=13/2$. Taking the unweighted average of the two group means instead gives five, which describes a different choice of group probabilities. This is why the outer weighting matters.

The formula is useful when conditioning on a first choice makes the remaining experiment simple. It can replace a large outcome enumeration by smaller calculations within groups. In a mixture distribution, it combines component means using mixture weights; in a random sum of independent claims, it separates the number of summands from their individual expected values under the stated independence assumptions. The same averaging step appears in the law of total variance.

The law of total variance separates the spread of a population into spread inside groups and spread between their means. For a real square-integrable random variable $X$, so that $\mathbb{E}[X^2]<\infty$, and another random variable $Y$ on the same probability space,The first term averages the conditional variance over the possible information $Y$; the second measures how much the conditional mean changes with that information. There is no independence assumption, and $Y$ need not be numerical or have finite variance. Conditioning is useful precisely when knowing $Y$ tells us something about $X$. See MIT tutorial on using conditional expectation and variance.

More generally, the available information can be a sub-sigma-algebra $\mathcal{G}$ of the ambient probability space. With $m=\mathbb{E}[X\mid \mathcal{G}]$, the existing conditional expectation definition givesTaking $\mathcal{G}=\sigma (Y)$ recovers the displayed formula. The conditional mean and conditional variance are themselves random variables: before learning the group, we do not yet know which group average or group spread will apply. Their outer expectation and variance turn these varying quantities into numbers.

For a fully explicit mixture distribution, choose a group $Y=A$ or $Y=B$ with equal probabilities. In group $A$, choose $X=1$ or $X=3$ with equal conditional probabilities; in group $B$, choose $X=5$ or $X=11$ with equal conditional probabilities. Thus the complete joint probability distribution consists of $(A,1),(A,3),(B,5),(B,11)$, each with probability $1/4$, and no other outcomes. The conditional means are $2$ and $8$. Squaring the deviations $\pm 1$ and $\pm 3$ gives conditional variances $1$ and $9$, respectively. The law of total expectation gives the overall mean $\mu =(2+8)/2=5$.

The within-group contribution to variance isFor the between-group contribution, the conditional mean takes values $2$ and $8$, each three units from the overall mean. Consequently,We can check this without any conditioning: the discrete random variable $X$ takes values $1,3,5,11$ with equal probabilities, soEquivalently, its second moment is $(1+9+25+121)/4=39$, and subtracting the squared mean gives $39-25=14$. These are population variances, with the stated probabilities as weights, rather than estimates using the corrected denominator of a sample variance.

The figure makes the two sources of spread visible. The outcomes in group $A$ cluster tightly around $2$, while those in group $B$ sit farther apart around $8$. Replacing every outcome by its conditional mean removes the internal spread but leaves the separation between groups. Conversely, subtracting each group's conditional mean aligns the two groups around zero. Averaging only the conditional variances would miss nine units of variance; retaining only the conditional means would miss five. Neither summary alone describes the whole population.

Two modifications clarify what each contribution measures. Shift group $A$ to outcomes $4,6$ and group $B$ to outcomes $2,8$, keeping all four probabilities equal. Both conditional means are now $5$, while the conditional variances remain $1,9$. The between-group contribution vanishes, and total variance is just $5$. Group membership still tells us how uncertain an outcome is, even though the best point prediction stays the same. Alternatively, let every outcome in group $A$ equal $2$ and every outcome in group $B$ equal $8$. Now both conditional variances vanish, and all nine units of total variance come from group separation. Knowing the group predicts the outcome perfectly. These limiting cases distinguish differences in average outcome from differences in internal spread; the original mixture distribution contains both.

The same calculation answers a prediction question. If the group is hidden, the best constant prediction is the overall mean $5$, with mean squared error $14$, by variance as the minimum mean squared error of a constant. If the group is revealed, predict $2$ for $A$ and $8$ for $B$. The errors become $-1,1,-3,3$, and their mean squared error is $5$. Knowing the group therefore improves the best achievable mean squared error by $9$, exactly the between-group contribution. This is an interpretation relative to the information available: the remaining uncertainty could shrink further if additional information were revealed.

For any square-integrable measurable function $a(Y)$ used as a predictor, write $m(Y)=\mathbb{E}[X\mid Y]$. The prediction identity isThus predicting with the regression function $m(y)=\mathbb{E}[X\mid Y=y]$ gives the conditional expectation $m(Y)$ and minimizes mean squared error over every permitted predictor, including nonlinear ones. This minimizes average squared error, not the error for each individual outcome. See Berkeley Statistics 201A notes, sections 36–37.

Why do the two pieces add cleanly? Put $R=X-m(Y)$. Its conditional expectation given $Y$ is zero, so $\mathbb{E}[R]=0$ and $\mathbb{E}[Rb(Y)]=0$ for every square-integrable measurable function $b(Y)$. In particular, $R$ and $m(Y)$ are uncorrelated random variables. Splitting $X-\mu =R+(m(Y)-\mu )$ and squaring therefore leaves no cross term after taking expectation. Geometrically, the conditional expectation is an orthogonal projection in the L2 space, and the law of total variance is its Pythagorean theorem for centered random variables.

In our example, the four products $R(m(Y)-5)$ are $3,-3,-9,9$, which average to zero. Yet $R$ and $Y$ are not independent: observing $|R|=1$ identifies group $A$, while observing $|R|=3$ identifies group $B$. This is a concrete instance of heteroscedasticity, with different conditional variances despite zero conditional means of the errors. A group can predict uncertainty even when it cannot predict the error's sign.

The law of total variance is useful whenever aggregated data conceal distinct groups. Changing mixture weights changes both contributions, so the weighting must match the population being described. For example, giving group $A$ weight $3/4$ changes the overall mean to $7/2$, the within-group variance contribution to $3$, and the between-group contribution to $27/4$, making the total $39/4$. Group differences account for spread and prediction gains; assigning them a causal explanation requires a separate causal inference argument.

Linkage disequilibrium means that knowing an allele at one genetic locus changes the probability of an allele at another genetic locus in population haplotypes. It compares their joint probability distribution with the product of their marginal distributions. This is useful because a genetic marker can carry information about another genetic locus: tag SNPs exploit this redundancy, without establishing a causal genetic association.

Consider two genetic loci with alleles $A,a$ and $B,b$. Sample one haplotype, rather than an unphased diploid genotype. Write its four haplotype frequencies as $f_{AB},f_{Ab},f_{aB},f_{ab}$, summing to one, and set the marginal allele frequencies to $p=f_{AB}+f_{Ab}$ and $q=f_{AB}+f_{aB}$. The linkage disequilibrium coefficient isThe second expression follows by substituting $f_{ab}=1-f_{AB}-f_{Ab}-f_{aB}$. The entire two-locus joint probability distribution is thenEvery row and column sum is unchanged when $D$ varies. Linkage equilibrium means $D=0$: all four haplotype frequencies equal products of allele frequencies, so the two allele indicators are independent random variables. For $p>0$, the conditional probability $P(B\mid A)=q+D/p$ shows directly how positive $D$ enriches $B$ among copies carrying $A$, while negative $D$ depletes it. Positive $D$ also enriches $ab$; negative $D$ enriches $Ab$ and $aB$. Swapping the labels $A$ and $a$ reverses the sign of $D$, so neither sign is intrinsically favorable.

For example, fix the allele frequencies at $p=0.60,q=0.40$. At linkage equilibrium, the haplotype frequencies in the order $AB,Ab,aB,ab$ are $(0.24,0.36,0.16,0.24)$. Positive linkage disequilibrium $D=0.12$ instead gives $(0.36,0.24,0.04,0.36)$, with $P(B\mid A)=0.60$; negative linkage disequilibrium $D=-0.12$ gives $(0.12,0.48,0.28,0.12)$, with $P(B\mid A)=0.20$. Counting each allele separately would miss the difference.

Nonnegative haplotype frequencies constrain $D$. Define the nonnegative bounds $L=\min (pq,(1-p)(1-q))$ and $U=\min (p(1-q),(1-p)q)$. The four entries above giveThus $|D|\le 1/4$ globally, but a particular pair of allele frequencies usually permits a smaller, asymmetric interval. The standardized linkage disequilibrium convention used here retains the sign:This convention requires $0<p,q<1$; if either genetic locus has only one allele in the population, $D=0$ but normalized measures are undefined. Some presentations report only $|D^{\prime }|$. At $D^{\prime }=1$ or $-1$, at least one of the four haplotypes has zero frequency; this alone does not prove that genetic recombination has never occurred.

For the indicator random variables $X=1_A$ and $Y=1_B$, their covariance is $D$, and their variances are $p(1-p)$ and $q(1-q)$. The allelic correlation measure is their squared correlation coefficient,The subscript distinguishes it from the recombination fraction $r$ in linkage-disequilibrium decay under recombination. Squaring loses the association's sign. In the illustrated positive and negative examples, $r_{\mathrm{LD}}^2=0.25$ in both, whereas $D^{\prime }=0.75$ and $-0.50$, respectively. For another original example, take $p=0.10,q=0.50,f_{AB}=0.10$: then $D=0.05,D^{\prime }=1$, yet $r_{\mathrm{LD}}^2=1/9$. Every $A$ carries $B$, but most copies carrying $B$ lack $A$. The frequency constraints and the distinction between these normalizations are analyzed in Mathematical properties of the r² measure of linkage disequilibrium by Jenna M. VanLiere and others (sections 2.1 and 2.5, pp. 131–132 and 135–136).

Genetic linkage describes transmission through meiosis; linkage disequilibrium describes a population's haplotype frequencies. Nearby genetic loci on a chromosome can be in linkage equilibrium. Conversely, mixing populations can create linkage disequilibrium even between genetic loci on different chromosomes. For example, mix equal numbers of independently sampled haplotypes from two populations, each internally at linkage equilibrium, with $p=q=0.80$ in one and $p=q=0.20$ in the other. The pooled allele frequencies are $p=q=0.50$, but $f_{AB}=(0.64+0.04)/2=0.34$, giving $D=0.09$. This is a consequence of population stratification, not a measurement of the recombination fraction.

Measured haplotype frequencies also depend on the sample. Finite sample size, related individuals, genetic-study ascertainment, genotyping error and uncertain haplotype estimation can change an estimate. A missing rare haplotype can give empirical $|D^{\prime }|=1$ without demonstrating its absence from the population. Report the sample definition and allele frequencies alongside the association measure; a nonzero sample estimate alone does not establish nonzero population linkage disequilibrium.

A concrete dataset is the 2005 A haplotype map of the human genome by The International HapMap Consortium. The International HapMap Consortium used genotyping platforms including Illumina BeadArray to obtain SNP genotype calls from 269 DNA samples across four collections: 30 parent–offspring trios from Ibadan, 30 trios from Utah, 45 unrelated individuals from Beijing and 44 from Tokyo. Parent–offspring trios and the PHASE algorithm supplied haplotype estimation, rather than direct observation of every chromosome. In Figure 7, 36 SNPs resolve seven haplotypes in a chromosome 2 region in the Utah panel; seven tag SNPs capture the 36 SNPs at $r_{\mathrm{LD}}^2\ge 0.8$. The readout demonstrates reusable allelic correlation, not causal effects.

Controls required over 80% complete calls, at most one discordance among five duplicate genotypes, at most one Mendelian segregation error per trio panel, and a Hardy-Weinberg equilibrium test with $P>0.001$; blind reassessment checked performance. Minor allele frequencies of at least 0.05 were targeted, the collections were not population-representative, and inferred haplotypes retain phase uncertainty. These limits restrict extrapolation of linkage disequilibrium.^[ref]

Genetic recombination erodes linkage disequilibrium by reshuffling alleles between haplotypes, while preserving marginal allele frequencies in the idealized model. Let $f_{AB,t}$ denote the haplotype frequency in the gamete pool of generation $t$, let $p,q$ denote its marginal allele frequencies, and let $r\in [0,1/2]$ be the constant recombination fraction between two autosomal genetic loci. Assume random mating within one population, nonoverlapping generations, fair Mendelian segregation, equal reciprocal gamete production, the same $r$ in both sexes, and a deterministic population large enough to neglect genetic drift. Exclude natural selection, mutation and gene flow. ThenHere $p$ and $q$ remain constant: recombining changes which alleles occur together, not their total numbers. The other haplotype frequencies change by $+r{D}_t$ for $Ab,aB$ and $-r{D}_t$ for $ab$. Consequently the joint probability distribution stays within its allowed bounds and approaches the product of its marginal distributions when $r>0$.

For the original positive example $p=0.60,q=0.40,D_0=0.12$, a recombination fraction $r=0.10$ gives $D_1=0.108$ and $D_{10}\approx 0.04184$. The negative example $D_0=-0.12$ has exactly the opposite sequence: genetic recombination reduces the magnitude without reversing the sign. With $r=0$, the initial linkage disequilibrium persists in this model; even independent assortment at $r=1/2$ only halves $D$ each generation. A population association does not vanish after one generation merely because the genetic loci are unlinked.

The geometric sequence gives a generation scale: for $r>0$, the continuous interpolation reaches $|D_0|/2$ at $t_{1/2}=\log (1/2)/\log (1-r)$. Actual generations are integers; the first at or below half is $\lceil t_{1/2}\rceil$. For small $r$, $t_{1/2}\approx (\log 2)/r$. Because the denominator of allelic correlation stays fixed, $r_{\mathrm{LD},t}^2=(1-r{)}^{2t}{r}_{\mathrm{LD},0}^2$; its decay is faster than the magnitude of $D_t$.

A short derivation uses random mating to assign frequency $2f_{AB}{f}_{ab}$ to the double heterozygote $AB/ab$ and $2f_{Ab}{f}_{aB}$ to $Ab/aB$. Relative to $r=0$, each loses or gains an $AB$ gamete fraction $r/2$, respectively. Their net contribution is $r(f_{Ab}{f}_{aB}-f_{AB}{f}_{ab})=-rD$. The other genotypes produce no net change in $f_{AB}$, proving the recurrence and, by iteration, its solution.

This temporal model does not identify elapsed generations from physical separation alone. The recombination fraction depends on genetic recombination, including recombination hotspots, while population stratification, natural selection, gene flow or genetic drift can create or maintain linkage disequilibrium. Repeated sampling across time would also require accounting for sample size and haplotype estimation. A cross-sectional plot of association against distance between genetic loci is therefore a different observation from the idealized trajectory plotted here.

A marginal density is a probability density function for one component of a joint probability distribution. Integrating the joint probability density over the other coordinates gives it; for two components, $f_X(x)=\int f_{X,Y}(x,y)dy$.

Marginalization sums or integrates a joint probability distribution over unwanted coordinates to obtain a marginal distribution.