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Let $Z$ have the fair Bernoulli distribution and set $X_j=Z$ for every $j$. Each coordinate has the same fair probability distribution as in the independent model, but the coordinates are dependent: $\mathbb{P}(X_1=1,X_2=1)=1/2$, whereas the product of the marginal probabilities is $1/4$. Every future sigma-algebra equals $\sigma (Z)$, so the tail sigma-algebra also equals $\sigma (Z)$.

The tail event of infinitely many heads is now exactly $\{Z=1\}$, with probability $1/2$. For the finite-window events $B_{m,r}$ in infinitely many heads as a tail event,The sample space can even be the same full Cartesian product $\{0,1{\}}^{\mathbb{N}}$ in both models: the dependent probability measure assigns mass $1/2$ to each of the all-zero and all-one sequences. Then $A$ is the same literal tail event in both models. Only the probability measure and its independence properties change.

For $r,s\in [0,1]$ and a random variable $U$ with the uniform distribution on $[0,1]$, set $B_r=1_{\{U\le r\}}$ and $B_s=1_{\{U\le s\}}$. Each has the corresponding Bernoulli distribution, and $\mathbb{P}(B_r\ne B_s)=|r-s|$, because disagreement occurs precisely between the two thresholds. This is a maximal coupling of the two laws.

A Bernoulli random variable records one success as $1$ and one failure as $0$. For $0\le p\le 1$, its probability mass function iswith zero mass elsewhere. The terms success and failure merely name outcomes; success need not be desirable. Any event $A$ produces such a variable through its indicator random variable $B=1_A$, with $p=\mathbb{P}(A)$. A single Bernoulli distribution requires no independence assumption; independence becomes relevant when several trials are combined.

Because $B^2=B$, the expected value and variance follow immediately:For a constructed trial with $p=0.3$, these are $0.3$ and $0.21$. The variance is greatest at $p=1/2$, where it equals $1/4$, since $p(1-p)=1/4-(p-1/2{)}^2$. At $p=0$ or $p=1$, the outcome is certain and the variance vanishes.

The beauty of an indicator random variable is that counting becomes addition: summing such indicators counts successes, and the linearity of expectation gives the expected count even when the indicators are dependent. With independence and a shared $p$, the full count law is the binomial distribution. These models are summarized in Penn State STAT 504, discrete distributions.

A historical anchor is Bernoulli's posthumous Ars conjectandi: its original title page identifies the author and records publication at Basel in 1713. This documents the book's publication, rather than establishing priority for the modern name Bernoulli distribution. See Ars conjectandi, original title page, ETH Library.

On a finite graph with symmetric edge-swap rates, the stirring representation of symmetric exclusion transports the occupations by a random permutation. An independent identically distributed Bernoulli distribution occupation field is invariant under every deterministic permutation, and hence under this random permutation when it is independent of the initial field. Therefore the displayed product measure is invariant, including densities zero and one. Equivalently, an edge swap preserves the number of particles and has the same reverse rate, giving detailed balance.

A Bernoulli process is a sequence of independent and identically distributed random variables with a Bernoulli distribution: $\mathbb{P}(B_n=1)=p$ and $\mathbb{P}(B_n=0)=1-p$. It models repeated Bernoulli trials with a fixed success probability and no dependence between trials. The count $K_n={\sum }_{j=1}^n{B}_j$ has a binomial distribution, expected value $np$, and variance $np(1-p)$; the first success time has a geometric distribution on $\{1,2,\dots \}$. The observations $B_n$ and the accumulated count $K_n$ are different stochastic processes: the observations are independent, while consecutive counts share all earlier observations. ^[ref]

A random experiment with two outcomes, success and failure, whose success indicator has a Bernoulli distribution. Repeated trials sharing success probability $\theta$ give a binomial distribution for their success count if they are independent conditional on $\theta$; integrating over an uncertain common $\theta$ generally makes the trials dependent marginally.

Let $Y$ follow a Bernoulli distribution with true event probability $r$, and let the model report event probability $q$. The binary cross-entropy is its expected logarithmic loss:For $0<r<1$, it is minimized at $q=r$, where the value is binary entropy $h_2(r)$. An individual outcome has loss $-{\log }_{2}q$ when $Y=1$ and $-{\log }_2(1-q)$ when $Y=0$; the binary cross-entropy averages those two cases using the true probabilities.

For $r=1/4$, truthful reporting gives $h_2(1/4)\approx 0.811278$ bits. Reporting $q=1/2$ gives one bit. Reporting $q=0.01$ givesBoth $q=1/4$ and $q=0.01$ select zero if one predicts only the more likely binary outcome. Their predicted probabilities are very different, however, and cross-entropy reveals the cost of underestimating the event that occurs a quarter of the time. Endpoint losses are computed using the realized outcome rather than multiplying zero by an undefined logarithm.

In independent bond percolation with parameter $p$, every edge of a graph is open with probability $p$ and closed with probability $1-p$. The choices are independent random variables with a Bernoulli distribution. All vertices remain present; only the usable connections change. The percolation clusters are the connected components of a graph formed by the open edges, including isolated vertices.

A specified simple graph path with $m$ edges is an open path with probability $p^m$, by independence. Connecting its endpoints is a different event: other graph paths may also connect them. The four-edge example in percolation theory makes that distinction explicit. Interpreting open edges as usable channels makes the model useful for connectivity questions; it does not by itself assign a travel time, flow rate, or capacity to a channel.

The binary Brier score applies squared-error loss to a reported success probability $p$ and the observed indicator $y\in \{0,1\}$. It is a loss, so smaller is better. Its values lie in $[0,1]$: predicting the outcome with certainty gives zero, while assigning it zero probability gives one. Unlike the logarithmic scoring rule loss, the binary Brier score remains bounded even for a confidently wrong forecast.

For a Bernoulli distribution with true success probability $q$, averaging the two squared losses givesThe report controls only $(p-q{)}^2$; the remaining term is the variance of the Bernoulli random variable. Consequently the Brier score is a strictly proper scoring rule. At $q=0.7$, the expected loss is $(p-0.7{)}^2+0.21$, with unique minimum $0.21$ at $p=0.7$. Reporting $p=0.2$ instead gives expected loss $0.46$, despite its better loss on the particular outcome $Y=0$ above.

For several possible categories, the Brier score sums the squared differences between reported probabilities and observed category indicators. With two categories that sum is $2(p-y{)}^2$, so its numerical values are twice the binary convention used here, with the same optimizing report. Brier's original ^[ref] averages these category-sum losses over forecasts. Strictly Proper Scoring Rules, Prediction, and Estimation by Tilmann Gneiting and others (Section 3.1, Example 1, and Table 1) gives the category-sum and binary forms as negative rewards; negating them gives the losses described here.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

The characteristic function of a real random variable $X$ isIt determines the probability distribution of $X$.

Here $t\in \mathbb{R}$ and $i^2=-1$. By Euler's formula, its real part and imaginary part are $\mathbb{E}[\cos (tX)]$ and $\mathbb{E}[\sin (tX)]$. The complex exponential has absolute value one, so this expected value exists for every real random variable, even without finite absolute moments. Moreover,The bound uses the triangle inequality for a vector-valued integral; the last identity is complex conjugate symmetry.^[ref] Consequently the real part is an even function and the imaginary part is an odd function.

For a discrete probability distribution with values $x_j$ and probabilities $p_j$, the characteristic function is ${\sum }_j{p}_j{e}^{it{x}_j}$: a weighted average of points on the unit circle in the complex plane. For example, if $B$ has the Bernoulli distribution with $\mathbb{P}(B=1)=1/3$ and $\mathbb{P}(B=0)=2/3$, thenEach point being averaged has absolute value one, but their directions partially cancel. At $t=\pi$, the same average is $1/3$; at $t=2\pi$, it returns to one. The characteristic function is thus neither a probability density function nor a probability mass function: its argument is a frequency parameter, and its values can be complex numbers. This Bernoulli distribution has no probability density function with respect to Lebesgue measure, yet its characteristic function is defined at every frequency.

For a probability density function $f$, the same definition becomes ${\phi }_X(t)={\int }_{\mathbb{R}}{e}^{itx}f(x)dx$, the Fourier transform with the displayed sign convention. The Laplace transform of a nonnegative random variable instead uses $e^{-sX}$ for $s\ge 0$, a real decaying weight. For the Bernoulli distribution above, it is $2/3+e^{-s}/3$; neither transform is the probability density function. The uniqueness theorem for characteristic functions explains why the oscillating averages still retain the entire probability law, while the characteristic function of a sum of independent variables turns combining laws into multiplication. These two properties make the representation useful even when directly computing a sum's probability distribution is cumbersome.

If $m\ge 1$ is an integer and $\mathbb{E}[|X{|}^m]<\infty$, the characteristic function has continuous derivatives through order $m$, withThus finite absolute moments justify recovering moments from derivatives.^[ref]

For a short justification of differentiation under the integral sign, the difference quotient for the $(k-1)$st derivative has integrand $(iX{)}^{k-1}{e}^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X{|}^k$, since $|e^{iu}-1|\le |u|$ for real $u$. Lower absolute moments are finite because $|X{|}^k\le 1+|X{|}^m$. Apply the dominated convergence theorem successively, then use the same integrable bound to obtain continuity of each derivative.

For the Bernoulli distribution with parameter $1/3$, direct differentiation gives ${\phi }_B^{\prime }(0)=i/3$ and ${\phi }_B^{\prime \prime }(0)=-1/3$, agreeing with $\mathbb{E}[B]=\mathbb{E}[B^2]=1/3$. Its variance is therefore $1/3-(1/3{)}^2=2/9$. Existence and continuity of a characteristic function alone do not authorize these moment calculations: the standard Cauchy distribution, whose expected value is undefined, still has a characteristic function. The stated finite absolute moment hypothesis is a sufficient condition for the derivative formula, not an assumption built into the definition of a characteristic function.

Let $P$ be the Bernoulli distribution with success probability $0.2$ and $Q$ the Bernoulli distribution with success probability $0.8$. For $0\le s\le 1$,The two summands have product $0.16$. Their arithmetic mean is at least their geometric mean, with equality at $s=1/2$. ThusThe same argument gives $C=-\ln (2\sqrt{p(1-p)})$ for complementary parameters $0<p<1$. At $p=1/2$ the laws agree and $C=0$.

With equal prior probabilities, count the successes $K$ among $n$ observations. The likelihood ratio is $Q^{\otimes n}/P^{\otimes n}=4^{2K-n}$, so choose $Q$ when $K>n/2$ and $P$ when $K<n/2$. If $n$ is even, either decision or a fair tie randomization gives the same average error at $K=n/2$. Under $P$, $K$ has the binomial distribution with parameters $n,0.2$. Symmetry therefore gives the exact optimal average errorThe last term applies only for even $n$; it represents half the tie probability. Direct finite sums give $p_1^{\ast }=0.2$, $p_5^{\ast }=0.05792$ and $p_{20}^{\ast }\simeq 0.00157912$. Twenty observations reduce the optimal average error to about 0.158%. The finite-sample Chernoff error bound at $n=20$ is $\frac{1}{2}(0.8{)}^{20}\simeq 0.00576461$, a guarantee rather than the exact error.

Reducing the exact error from $p_{20}^{\ast }$ to $p_{50}^{\ast }\simeq 1.29363\times 10^{-6}$ illustrates the value of repeated evidence. The example assumes a single fixed coin law throughout the experiment. If repeated sensor readings share an unmodelled common error, increasing their number need not supply this much independent information.

Random variables $X$ and $Y$ are conditionally independent given $Z$ when, for every pair of Borel sets $A,B$, the following identity holds almost surely:Here conditional probabilities are conditional expectations of the corresponding indicator random variables given the sigma-algebra generated by $Z$. This defines the condition even when $\mathbb{P}(Z=z)=0$. In terms of regular conditional distributions, the conditional joint law is the product of the conditional marginal distributions for almost every conditioning value.

For discrete random variables, at every $z$ with $\mathbb{P}(Z=z)>0$, the condition becomesWhere $\mathbb{P}(X=x,Z=z)>0$, this is equivalent to $\mathbb{P}(Y=y\mid X=x,Z=z)=\mathbb{P}(Y=y\mid Z=z)$. Once $Z$ is known, observing $X$ provides no further information about $Y$.^[ref]

For a worked example, suppose a latent variable $T$ chooses between two environments with $\mathbb{P}(T=0)=\mathbb{P}(T=1)=1/2$. Given $T=t$, take $X,Y$ to be independent and identically distributed with a Bernoulli distribution of parameter $p_t$, where $p_0=0.1$ and $p_1=0.9$. Think of two separate trials run under the same quiet or active environment. Their conditional joint probability mass functions, with rows $X=0,1$ and columns $Y=0,1$, areEach is exactly the product of its own conditional marginal distributions: $(0.9,0.1)$ for $T=0$, and $(0.1,0.9)$ for $T=1$. Thus $X,Y$ are conditionally independent given $T$.

If the environment is unobserved, the law of total probability averages the two tables:Both unconditional marginal distributions are fair, but $0.41\ne 0.5\times 0.5$. The law of total expectation gives

Observing $X=1$ makes the active environment more plausible, so it raises $\mathbb{P}(Y=1\mid X=1)$ to $0.41/0.5=0.82$. Knowing the actual environment removes this source of statistical dependence. The covariance induced by a shared latent variable lemma explains the calculation without listing all the outcomes.

This is why conditional independence is useful in statistical modelling: a model can multiply simple conditional laws and then average over a shared latent variable. The resulting observations can remain statistically dependent. Conditional independence graphs organize such information relations; their interpretation as causes requires additional assumptions from causal inference. Conditioning describes the information available to an observer. It does not, by itself, describe what would happen if an experimenter changed a variable.

A logistic regression can specify each binary response conditionally on recorded past outcomes and baseline covariates. A chain rule for probabilities gives the path likelihood as the product of the conditional Bernoulli distribution masses. This permits dependence within a person's history without treating their outcomes as unconditionally independent. The model must specify the initial history and which past features enter its conditional mean.

Let $X,Y$ be independent random variables, each with a Bernoulli distribution of parameter $1/2$, and put $S=X+Y$. Unconditionally, the four ordered pairs $(0,0),(0,1),(1,0),(1,1)$ each have probability $1/4$. On the event $S=1$, only $(0,1)$ and $(1,0)$ remain, each with conditional probability $1/2$. ThusKnowing that exactly one trial succeeded makes observing one outcome informative about the other. This violates conditional independence given $S$, even though the original random variables are independent. Given $S=0$ or $S=2$, both variables are constant random variables and are conditionally independent there; that does not repair the failure at $S=1$, an event of positive probability.

The conditional covariance here is $\mathbb{E}[XY\mid S=1]-\mathbb{E}[X\mid S=1]\mathbb{E}[Y\mid S=1]=0-1/4=-1/4$. Both conditional variances are $1/4$, so the conditional correlation coefficient is $-1$: within the selected group, $Y=1-X$. These conclusions follow from restricting a finite sample space, without positing any physical interaction between the trials.

This example and the shared-environment example establish both logical directions: independent random variables need not be conditionally independent, and conditionally independent variables need not be independent. When $S$ is represented as the joint result of two inputs in a causal directed acyclic graph, conditioning on it has the collider structure associated with collider bias. A probability calculation by itself establishes a dependence relation, not a physical causal mechanism or an intervention.

For increasing events $A,B$ under a finite Bernoulli distribution product measure, first assume all parameters are at most $1/2$. Write $A_0\subseteq A_1$ and $B_0\subseteq B_1$ for last-coordinate sections. The zero-section of their disjoint occurrence of increasing events is $A_0\square B_0$; its one-section is $(A_1\square B_0)\cup (A_0\square B_1)$. The intersection in this union contains $A_0\square B_0$. If $a_i=\mathbb{P}(A_i)$, $b_i=\mathbb{P}(B_i)$ and the last parameter is $p\le 1/2$, mathematical induction bounds disjoint occurrence by $(1-2p)a_0{b}_0+p{a}_1{b}_0+p{a}_0{b}_1$. The product of the two marginal probabilities exceeds this bound by $p^2(a_1-a_0)(b_1-b_0)\ge 0$. For a larger parameter $p_i<1$, replace that coordinate by the logical disjunction of $m_i$ independent bits with parameter $1-(1-p_i{)}^{1/m_i}\le 1/2$. Disjoint original witnesses lift to disjoint bit witnesses. Apply the small-parameter inequality and then pass to degenerate parameters by continuity.

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.

Independent marking is the hypothesis that guarantees the Poisson thinning theorem. Equal marginal retention probabilities do not guarantee it. Consider one Bernoulli random variable $B$ with success probability $p\in (0,1)$, independent of a Poisson process, and retain every arrival when $B=1$ and none when $B=0$. Each arrival is retained with probability $p$, but the marks share the same coin. In an interval whose original count has Poisson distribution of mean $m>0$, the retained count is $R=BN$ and the rejected count is $D=(1-B)N$, soThe extra variance shows that $R$ is not Poisson distributed; the negative covariance proves dependence. Its no-arrival probability is $1-p+p{e}^{-m}$, rather than $e^{-pm}$. For the toy values $p=1/2,m=2$, both rules have retained mean $1$, but the common-coin rule has variance $2$ and no-arrival probability $(1+e^{-2})/2\approx 0.567668$, compared with variance $1$ and no-arrival probability $e^{-1}\approx 0.367879$ for independent marking. These formulas follow by conditioning on $B$ and using the law of total variance.

An arrival-dependent or adversarial retention rule can change the law just as strongly. Keeping every second arrival gives $R_t=\lfloor N_t/2\rfloor$. Its successive retained gaps are sums of two independent exponential random variables, hence have the Erlang distribution of shape two and rate $\lambda$, not an exponential distribution. The resulting renewal process and its count variance are analyzed in Every second Poisson arrival has underdispersed counts, and the two dependent streams in alternating split of a Poisson process. As an even more restrictive rule, keeping only the first arrival ever gives $R_t=1_{\{N_t\ge 1\}}$, a Bernoulli distribution at each $t>0$, so it cannot have a positive-mean Poisson distribution. A thinning rule must specify how decisions depend on arrivals and on one another; the independent-marking conclusion cannot be applied automatically to arbitrary deletion.

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

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

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