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An Alpha-spending function assigns the cumulative Type I error budget to successive information fractions in a group sequential design. Its total at the final analysis is the overall significance level. The rejection boundaries must be calibrated using the joint distribution of the sequential test statistics; independent fixed-level tests do not automatically spend the assigned budget correctly.

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.

In 1924–1925, Walther Bothe and Hans Geiger tested event-level Compton scattering correlations against the statistical conservation proposed by BKS theory.^[ref]

Needle Geiger counters shared a wall. X-rays passed beside it through hydrogen; recoil electrons registered in one counter and scattered photons triggered electron release in the other. Photographic records resolved approximately $10^{-4}\mathrm{s}$.^[ref]

The coincidence excess matched paired-emission expectations. Bothe distinguishes a later directional test from this timing experiment.^[ref]

The two channels have different physical roles. The first responds to the local recoil electron's ionization. The second converts a scattered photon into an electron response, as in the photoelectric effect. A large counter signal is an amplified indication of an interaction, not a direct measurement of the incoming photon's energy. The detection efficiency and accepted directions determine how many genuine partners can be seen. Consequently, failure to record every partner is compatible with paired production.

The relevant distinction is between an average balance and a balance in each event. For an initially stationary free electron, Compton recoil kinematics givesHere $E,E^{\prime }$ are incident and scattered photon energies, $T_e$ is recoil kinetic energy, and ${\mathbf{p}}_e$ is electron momentum. These are paired constraints, not just equations between averages over unrelated detections. Summing correctly paired events preserves the equations; throwing away the pair identities loses information. Two samples can have identical separate energy distributions while differing in their joint probability distribution.

The experiment therefore supplied evidence for the event association required by photon recoil, rather than weighing both outgoing energies in every event. A statistically established excess above accidental coincidences matters; one near-simultaneous pair cannot establish the association. It also cannot by itself settle every question about wave-particle duality.

Preparation, readout and background checks address different failure modes. The direct beam must avoid the second counter, and the counter boundary must prevent a recoil electron from masquerading as its own photon partner. Separate-channel rates constrain accidental coincidences; common disturbances or electrical coupling could instead create a correlated instrumental background. A useful diagnostic is to displace one record in time: under a stationary process assumption, unrelated-pair expectations persist while prompt partners separate. This is the explanatory control modeled below, not an assertion that Bothe performed that particular record-shifting procedure. A background estimate and a statistical comparison are essential to the inference.

The existing coincidence circuit article explains electronic selection, and accidental coincidence derives the accepted-window background. Here pairing can be understood directly from the two recorded time sequences, without imposing a modern detector trigger on the historical readout.

Independence makes the characteristic function of a sum equal the product of the individual characteristic functions.

For mutually independent random variables $X_1,\dots ,X_n$, write $S_n={\sum }_{j=1}^n{X}_j$. ThenThis is the expectation of a product of independent random variables applied to bounded complex exponentials; no moment hypothesis is needed.^[ref]

Identical probability distributions are unnecessary. When the random variables are independent and identically distributed random variables, the product is ${\phi }_X(t{)}^n$. A deterministic rescaling also follows directly from the definition: ${\phi }_{aX+b}(t)=e^{itb}{\phi }_X(at)$ for real $a,b$. Consequently the sample mean has characteristic function ${\phi }_X(t/n{)}^n$, and a sum divided by $\sqrt{n}$ has characteristic function ${\phi }_X(t/\sqrt{n}{)}^n$.

The independence assumption does real work. Let $X$ take $-1$ and $1$ with equal probabilities and set $Y=X$. Both characteristic functions are $\cos t$, but ${\phi }_{X+Y}(t)=\cos (2t)$, not ${\cos }^{2}t$. At $t=\pi /2$ they give $-1$ and zero respectively. Knowing the two marginal probability distributions does not determine the sum without their joint probability distribution.

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.

A cryptographic environment is the interactive external context used to judge a cryptographic protocol in universal composability. It supplies honest parties' inputs, observes their outputs, communicates with the adversary, and ultimately outputs a decision bit. The comparison asks whether this bit's probability distribution changes detectably between the real execution and an ideal execution with a cryptographic simulator.^[ref]

The cryptographic environment models other software using the service, including choices depending on earlier outputs. It can retain private information with which to check later responses. Its view therefore includes the relationship between inputs, outputs and adversarial messages, not just isolated packet statistics.

An original two-bit example shows why that relationship matters. Let $B$ be a random variable with the uniform distribution on a finite set $\{0,1\}$. One system returns $(B,B)$; another returns $(B,C)$ with $C$ an independent uniform bit. Each individual coordinate has the same marginal distribution in both systems. Nevertheless, the test “accept exactly when the two bits agree” has acceptance probabilitiesIts cryptographic distinguishing advantage is $1/2$. Matching each output's marginal does not match their joint probability distribution. An interactive cryptographic environment can perform corresponding consistency checks against its retained inputs.

In the computational definition, the cryptographic environment has the specified polynomial time bound indexed by the cryptographic security parameter. For each real adversary, one cryptographic simulator must handle every allowed cryptographic environment; choosing a new simulator after learning which particular context will test it changes the requirement.^[ref]

This breadth explains the usefulness of universal composability: a service can be analysed within a larger application context. The conclusion still concerns the interactions and cryptographic leakage admitted by the stated execution model.

A cryptographic transcript is the specified record of messages exchanged during a cryptographic protocol execution. Its content depends on the purpose: a public communication record differs from a participant's complete private view, which can also include its input, secret state and randomness. Order, message boundaries and the identities or roles associated with messages can matter.

For a concrete standardized use, Transport Layer Security 1.3 authenticates a transcript hash computed from selected handshake messages in their prescribed order. It includes handshake type and length headers, but excludes record-layer headers. Its retry path specifies a special replacement for the first client message.^[ref]

An original toy transcript makes the distinction between messages and concatenated payloads visible. Suppose a sender transmits ab and then c. The payload sequence is $(\mathtt{ab},\mathtt{c})$. Another execution transmits a and then bc. Both flatten to ASCII bytes 61 62 63, even though they have different first messages. A length-prefixed encoding preserves the distinction as 02 61 62 01 63 versus 01 61 02 62 63. A hash cannot recover boundaries that were omitted before hashing.

Direction can be equally important. The records “initiator sends yes” and “responder sends yes” contain identical payloads but assert different behavior. A format used to authenticate those assertions must bind the relevant roles, either explicitly in the encoded record or through an unambiguous protocol context. This toy rule is not a replacement for the TLS transcript format.

In a security proof, the observer's view and its correlations must be specified. A cryptographic simulator that samples messages with correct individual frequencies may still produce the wrong joint probability distribution with the cryptographic environment's retained inputs. A cryptographic transcript also discloses only the observations it records: a transcript omitting timing cannot by itself justify a security claim against timing observations.

A transcript hash is a compact fingerprint of the encoded record under the relevant cryptographic hash function assumptions. An unkeyed hash alone does not identify the speaker: a digital signature or message authentication code must bind it to an authenticated context when that is the intended guarantee.

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.

A standard real Gaussian random matrix has independent and identically distributed standard normal entries. Each column is an orthogonally invariant random vector: its probability density function is proportional to $\exp (-\Vert z{\Vert }_2^2/2)$. Consequently multiplying on either side by a fixed orthogonal matrix preserves the joint distribution. Its singular values are random; the expressions $\sqrt{m}\pm \sqrt{n}$ for an $m\times n$ matrix describe useful spectral scales, not deterministic endpoints that hold for every realization.

A generative model describes how observations can be produced. It can specify a probability distribution $p_{\theta }(x)$ explicitly, or provide a procedure that generates samples whose distribution is determined by parameters $\theta$. Producing observations and assigning a tractable numerical probability to every observation are separate capabilities.

An original two-bit example defines a joint probability distribution through conditional probabilities. LetGenerate $X_1$ first, then generate $X_2$ using the applicable conditional probability. The four possible outputs have probabilitieswhich sum to one. The procedure generates both complete observations and a dependence between their components. Always outputting the most probable observation, 11, would produce a different distribution: it would discard all the other outcomes rather than sample the stipulated model.

A generative adversarial network gives a different implementation. A generator transforms an input random variable into a candidate observation, while a discriminator supplies a training signal by distinguishing generated samples from training examples. The original paper develops this as an adversarial learning procedure.^[ref]

The model's samples must be assessed against the intended distribution or task. A visually plausible observation alone does not establish that the model represents the full range of observations well; distributional coverage and the probability of different outcomes matter alongside individual sample appearance.

A group sequential design evaluates accumulating data at several planned interim analyses instead of committing to one final test. At each analysis, a prespecified rule can stop for efficacy, stop for futility, or continue collecting observations. The design specifies both the timing of analyses and the sequential stopping boundaries.

Repeated inspection changes the probability of rejecting a Null hypothesis. Even two independent tests, each at significance level $0.05$, reject at least once with probability $1-0.95^2=0.0975$ under their common Null hypothesis. Tests on accumulating observations are dependent, so their rejection boundaries require their actual joint distribution rather than that independence calculation. An Alpha-spending function distributes the permitted type I error over information fractions; it does not make unchanged fixed-sample critical values valid at every analysis.

For a simple valid construction with $K$ planned analyses, choose separate rejection events whose null probabilities are each at most $\alpha /K$. The union bound makes the probability of any rejection at most $\alpha$, whatever their dependence. More efficient boundaries exploit the dependence between successive test statistics. A Futility boundary removes rejection paths, and a Conditional error principle can justify a later design adaptation while preserving the original error guarantee. The possibility of early stopping also affects point estimation and confidence intervals; fixed-sample formulas do not automatically retain their usual interpretation after selection.

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.

The joint probability distribution of random variables $X_1,\dots ,X_n$ is the probability distribution of the random vector $(X_1,\dots ,X_n)$. It records both the marginal distributions and their dependence.

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.