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There is no such value of $p$ in the specified interval. Indeed, $\mathbb{P}(Y=1)=p{e}^{-p}>0$, and the joint probability mass function givesThis contradicts independence for every $0<p<0.8$. Equivalently, $(1,2)$ has joint probability zero but the product of its marginal probabilities is $p{e}^{-p}{p}^2/2>0$.

At the suggested value $r=e^{-r}$, the covariance is zero, but the same conditional probability contradiction still applies. This is an explicit example of how uncorrelated Bernoulli and Poisson variables need not be independent. Zero covariance controls one mixed moment, whereas independence requires the entire joint probability mass function to factor.

Let $(B_j)$ be a finite or countable partition of a set of the sample space into disjoint events, with $\mathbb{P}(B_j)>0$. For any event $A$, the law of total probability isTo prove it, the intersections $A\cap B_j$ are disjoint and have union $A$. Countable additivity and the definition of conditional probability therefore giveIt is enough that the union of the partition events has probability one. Parts of probability zero contribute zero in the intersection formula and can be omitted before writing conditional probabilities; conditioning by division on a zero-probability part would be undefined.

Poisson arrivals see time averages says that an exogenous, homogeneous Poisson process samples prearrival states with the same long-run frequencies as observation over time, provided the observed system cannot anticipate future arrivals and the relevant time averages exist. The system may depend on all past arrivals: an ordinary queue does. The required independence is between future increments of the Poisson process and the history available to the system, including its service information. Assume measurable state observations with ordinary piecewise continuous functions as sample paths; write $X(t-)$ for the state immediately before time $t$.

For a state set $B$, let $A(T)$ count offered arrivals by time $T$ and let $t_k$ be their times. With a positive finite constant arrival rate, the PASTA equality isalmost surely when the time limit exists. The indicator functions count visits on the left and time spent on the right. For a stationary process with ergodicity, the right side is the stationary probability of $B$. It matters that the arriving customer has not yet changed the observed state.

In the M/M/1 utilization and delay example, server utilization is $0.8$. Thus PASTA says that 80 percent of offered arrivals find the server busy and 20 percent find the whole queue empty. If an offered arrival seeing population $n$ is admitted with conditional probability $p(n)$, using fresh randomness independent of that state, the accepted fraction is ${\sum }_n{\pi }_{n}p(n)$, where $\pi$ is the stationary distribution. Multiplying this fraction by the offered rate gives the admitted queue throughput when every admitted customer eventually completes service. Accepted arrivals are selected by state and need not themselves form a Poisson process or see the unconditioned time frequencies.

Arbitrary arrivals need not see time averages. Suppose a customer arrives every two minutes and its service time is exactly one minute, with one server and no other customers. The server is busy half the time, but every arrival finds it idle. A uniformly random initial phase can even make the occupancy a stationary process, without repairing the disagreement. Nevertheless Little's law still gives $L=(1/2)\cdot 1=1/2$, with rate measured in customers per minute. This separates Little's law from PASTA.

Ronald W. Wolff's 1982 paper gave a general PASTA theorem under nonanticipation, rather than requiring the observed system to be independent of its driving Poisson process. Its Theorem 1 compares the two long-run fractions; stationarity is a way to establish convergence, not the central sampling condition. ^[ref]. ^[ref].

Let $T_n=\inf \{t\ge 0:N_t\ge n\}$ be the $n$th arrival time, and set $T_0=0$. In a Homogeneous Poisson process of rate $\lambda >0$, the gaps $W_n=T_n-T_{n-1}$ are independent and identically distributed random variables with an exponential distribution:Here $\mathrm{Exp}(\lambda )$ uses the rate convention. Under the scale convention $\theta =1/\lambda$, the same probability density function is ${\theta }^{-1}{e}^{-w/\theta }$; a simulation function accepting a scale must receive $1/\lambda$. Conversely, summing independent rate-$\lambda$ exponential random variables to form $T_n$ and counting those sums constructs a Homogeneous Poisson process.^[ref]

The first-gap formula follows immediately from $\{W_1>w\}=\{N_w=0\}$ and the Poisson distribution. Independence of all successive gaps is the additional interval-theorem assertion; deterministic-time independent increments alone should not simply be substituted at random arrival times without justification. The memorylessness of the exponential distribution givesAt the toy rate of two per minute, the mean gap is half a minute, the median is $(\log 2)/2\approx 0.346574$ minutes, and the probability of waiting longer than half a minute is $e^{-1}\approx 0.367879$. Here $\log$ is the natural logarithm. Already having waited a minute without an arrival leaves this same conditional probability of waiting another half minute. The exponential distribution describes a model with no aging of the remaining wait; a scheduled arrival with a fixed gap has no such property.

For nontrivial events, strict attraction under conditional probability is equivalent to a positive covariance of their indicator variables. This condition is symmetric in the two events. Complementing exactly one of the events reverses the strict inequality. This two-event condition should be distinguished from association of an entire family of random variables.

For a binary classifier, the conditional probability of the target condition given a positive classification. If its base rate is $p$, its true-positive probability is $s$, and its false-positive probability is $f$, Bayes' theorem gives $P(E\mid +)=sp/(sp+f(1-p))$. The denominator is positive whenever a positive classification can occur.

If future Bernoulli random variables are conditionally independent given a shared parameter $\Theta$, with success probability $\pi (\Theta )$, their all-success conditional probability is $\pi (\Theta {)}^k$. Taking a posterior expectation proves the identity. For $k\ge 1$, Jensen's inequality and $0\le \pi \le 1$ give $(\mathbb{E}[\pi \mid D]{)}^k\le \mathbb{E}[{\pi }^k\mid D]\le \mathbb{E}[\pi \mid D]$. For two distinct future trials the posterior covariance is $\mathbb{E}[{\pi }^2\mid D]-(\mathbb{E}[\pi \mid D]{)}^2=\mathrm{Var}(\pi \mid D)$. Thus shared parameter uncertainty can induce positive dependence even when trials are conditionally independent.

An urn has $N$ balls, including $K$ of one colour and $N-K$ of the other, with $0<K<N$. After a uniform first draw, an independent decision returns that ball with probability $r$; the second draw is uniform from the remaining contents. Let $B_j$ mean the $j$th draw has the first colour and $R$ mean replacement. The two relevant joint probabilities areApplying Bayes' theorem and cancelling the common factors gives the displayed conditional probability. It is independent of $K$ and satisfiesTwo different observed colours therefore decrease the posterior probability of replacement: removing the first colour makes the second more likely. The replacement decision must be independent of the first draw for this formula; its marginal probability alone does not specify the model.

Probability theory describes uncertainty through a probability space: possible outcomes form a sample space, observable questions are events, and a probability measure assigns them numbers between zero and one. The Kolmogorov axioms make those assignments consistent. They let us combine information about several events, update it through conditional probability, and study random variables without listing every possible outcome separately.

For example, roll two dice whose results are independent random variables, each with a discrete uniform distribution on $\{1,\dots ,6\}$. There are 36 equally likely ordered outcomes. Only $(1,6),(2,5),\dots ,(6,1)$ have sum seven, so that event has probability $6/36=1/6$. The sum itself is not uniformly distributed: sum two has only one outcome. Choosing a probability distribution for the elementary outcomes is part of the model; equal likelihood does not follow merely from counting the named possibilities.

The subject becomes especially useful when complete outcome lists are enormous or infinite. Independence turns a joint calculation into simpler pieces, expected value summarizes a random variable, and the law of large numbers explains how averages can become predictable even when individual observations are uncertain. The central limit theorem describes the scale and shape of many remaining fluctuations. These are distinct claims: a small variance of an average does not make every observation close to its expected value.

There is a satisfying unity here: dice, hidden causes, and Brownian motion all use the same rules for events and conditional probability. A model still needs justification. Its exact deductions describe what its assumptions imply; comparison with observations belongs to statistical inference.

For an edge with endpoints already connected without that edge, including any boundary wiring, the conditional open probability is $p$. Otherwise opening the edge merges two components, so the conditional probability is $p/(p+q(1-p))$. For $q\ge 1$, this conditional probability increases when more other edges are open or more boundary vertices are identified. This is the local mechanism of boundary monotonicity of the random-cluster measure.

A box contains one certainly red ball and $n\ge 1$ independently coloured extra balls, each red with probability $1/2$. Conditional on the first draw being red, a second draw without replacement is red with probability $(n+3)/(2(n+2))$. To see the effect of using a conditional probability, let $R=1+B$, where $B$ has a binomial distribution with parameters $n,1/2$. The conditional probability is $\mathbb{E}[R(R-1)]/(n\mathbb{E}[R])$: histories with more red balls are more likely to produce the first red draw and therefore receive greater weight.

A power law $q(t)=t^{\rho }$ responds to every fixed rescaling $t\mapsto ut$ by the same factor $u^{\rho }$. Regular variation keeps this scaling behaviour in the limit $t\to \infty$, while allowing departures from an exact power law. It is useful when the scaling exponent is stable but a smaller correction continues changing. The correction can even grow without bound.

An eventually positive measurable function $q$ has regular variation at infinity with index $\rho \in \mathbb{R}$ ifThe argument $ut$ is in the domain of a function for sufficiently large $t$. Each $u$ is held fixed as the limit is taken; allowing $u$ to change with $t$ is a different assertion. This is the scalar definition recalled by ^[ref]. The index is unique: taking $u=2$ determines $2^{\rho }$.

For example, on $t>1$ the function $q(t)=t^2\log t$ hasThus $q$ has regular variation of index two. Yet $q(t)/t^2=\log t\to \infty$: regular variation does not require asymptotic equivalence to a constant times a power law. Doubling $t$ gives $q(2t)/q(t)=4(1+\log 2/\log t)$. At $t=100,10^4,10^6$ this is approximately $4.60206,4.30103,4.20069$, respectively. The ratio approaches four slowly, even though the exponent is exactly two in the limiting scaling law.

The slowly varying function isolates this correction. Set $\ell (t)=q(t)/t^{\rho }$. ThenConversely, multiplying a slowly varying function by $t^{\rho }$ gives regular variation of index $\rho$. Hence the decomposition $q(t)=t^{\rho }\ell (t)$ is exact on its eventual domain of a function; it is the scaling relation, rather than the decomposition itself, that is asymptotic.

A Pareto distribution gives a decreasing example with a direct interpretation. For $x\ge 1$, take the tail probability $\mathbb{P}(X>x)=x^{-2}$. Its regular variation index is $-2$. Doubling a threshold divides its tail probability by four, andFor a tail probability with regular variation of index $-\alpha$, the same conditional probability tends to $2^{-\alpha }$. An exponential distribution behaves differently: its corresponding conditional probability is $e^{-\lambda t}\to 0$. This comparison explains why scaling limits are useful in extreme value theory; it does not establish that an observed dataset follows a particular tail model. The role of regular variation in heavy-tailed distributions is discussed in Lindskog, Resnick and Roy, introduction, p. 1.

An exponential function $q(t)=e^t$ is not regularly varying with any finite index: at $u=2$, $q(2t)/q(t)=e^t\to \infty$, rather than $2^{\rho }$. Boundedness alone is also insufficient, as the oscillating counterexample in slowly varying function shows. Checking a few rescaling ratios at finite arguments can illustrate a proposed law, but cannot prove its limiting definition.

Jovan Karamata developed this theory in his 1930 paper and returned to its fundamental results in 1933. The opening of his 1933 paper identifies the earlier work; pp. 57–58 connect the rescaling limit with a power law and a slowly varying factor. See J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bulletin de la Société Mathématique de France 61 (1933), 55–62. This gives a precise language for “almost a power” without discarding corrections that remain large.

Let $G$ be the event of membership in the first of two groups and $A$ the event of selection. Write $p=P(G)$, $a=P(G\mid A)$ and $q=P(A)>0$, with $0<p,a<1$. Bayes' theorem gives the conditional probabilities $r_1=P(A\mid G)=aq/p$ and $r_0=P(A\mid G^c)=(1-a)q/(1-p)$. Their risk ratio is the displayed expression: the unknown overall selection probability cancels. Equivalently, it is the ratio of the group-membership odds after selection to those before selection.

When both selection probabilities lie strictly between zero and one, the selection odds ratio is $[r_1/(1-r_1)]/[r_0/(1-r_0)]$ and generally depends on $q$. Thus compositions alone do not identify that odds ratio or provide sample sizes for a standard error. When $a<p$, the first group's observed selection rate is lower; this comparison alone does not establish a causal effect of group membership.

A random number generation technique that retains the unused equally likely outcomes after a rejected trial in rejection sampling. Instead of discarding those outcomes, it uses their conditional probability to obtain a new uniform distribution on a finite set for the input state of the next trial.

For a normalized quantum state $\psi$ and a time-independent Hamiltonian operator $H$, consider ideal projective measurements of the orthogonal projection $P=|\psi ⟩⟨\psi |$ at intervals $\Delta t=T/N$. Write $U=e^{-iH\Delta t/\hslash }$. A successful check leaves the post-measurement state $\psi$ up to global phase. Thus each successive success, conditional on earlier successes, has conditional probability $|⟨\psi ,U\psi ⟩{|}^2$, and multiplication gives the displayed formula. Equivalently,This is the probability of the entire sequence of successful checks. It underlies the quantum Zeno effect.

Replication in experimental design repeats a treatment on multiple independent experimental units, so that variation between those units can inform a treatment contrast and its standard error. Repeated measurements on one experimental unit can improve measurement precision without providing additional independent treatment replication. The experimental unit is identified by where the treatment is applied, rather than by how many responses are recorded.^[ref]

For a constructed example, grow one plant in each of eight separate pots. Apply treatment A to four pots and treatment B to four pots, without sharing drainage water between pots. Measure the length of three leaves per pot. The pot is the experimental unit; the leaf is the observational unit. There are four treatment replicates per group, eight experimental units overall, and $8\cdot 3=24$ measurements. The three leaves in one pot are nested measurements, because they share the same plant and treatment application. The setup assumes that pots do not affect one another; simply putting them in separate containers does not establish statistical independence.

Randomization chooses which experimental units receive which treatments; replication supplies multiple experimental units receiving the same treatment. These solve different problems. In the balanced allocation above, choose the four A pots uniformly among the $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ possible subsets. A particular pot has probability $1/2$ of receiving A. Given that one pot receives A, a second receives A with conditional probability $3/7$, rather than $1/2$. Thus the assignments are not independent random variables, even though the experimental units can supply independent responses conditional on those assignments in a suitable statistical model.

An actual ecology experiment illustrates why the distinction matters. In Stuart H. Hurlbert's account, twelve experimental ponds initially received zero, fifty or 450 fish, with five, four and three ponds respectively. Preparation included assessing variation between ponds before introducing fish; allocation considered both location and existing differences between ponds. The zero-fish ponds supplied the scientific control. These were repeated treatment applications to ponds, not twelve samples from one pond.^[ref]

The original article followed the southern California ponds for ten months. Its readouts included zooplankton populations and total phytoplankton; it reported decreases in some zooplankton populations and increased total phytoplankton.^[ref] Repeated sampling within a pond helps describe its response through time, while ponds supply treatment replication. Unequal group sizes and initial pond variation matter for the statistical model; the treatment counts alone do not establish equal variances or independent responses.

Sampling without replacement selects distinct labelled units: after a unit is selected it is unavailable for the remaining draws. Distinct labels can carry equal numerical values, so repeated values do not necessarily imply a repeated unit. The rule alone does not specify selection probabilities; a procedure favoring particular units is still sampling without replacement but need not be simple random sampling.

In uniform sequential sampling without replacement, the next label is chosen uniformly among those remaining. For a population of $N$ labels, any ordered sequence of $n$ distinct labels has probability $1/[N(N-1)\cdots (N-n+1)]$. Each size-$n$ subset has $n!$ possible orders, so its probability is ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. This gives the fixed-size simple random sampling design. Equivalently, take the first $n$ labels of a uniformly random permutation. Restricting a uniformly random permutation to a specified label set preserves uniformity; biased modulo mappings should not replace uniform selection.

The draw positions have identical marginal probability distributions, but exhibit statistical dependence. If the first label is $i$, the conditional probability that the second label is $i$ is zero, whereas its marginal probability is $1/N$. Thus these draws are not independent and identically distributed when $N>1$ and $n>1$. For a marked subset of labels, counting size-$n$ subsets with binomial coefficients gives the hypergeometric distribution. The resulting negative covariance explains the finite population correction for both counts and sample means.

A scoring rule turns a probability distribution forecast and an observed outcome into a number. For example, a forecast of rain with probability $0.7$ should express uncertainty more usefully than a bare prediction of rain. A scoring rule lets us compare such forecasts after observations arrive, and its choice determines whether exaggerating confidence pays. A proper scoring rule rewards an honest account of uncertainty in expected value.

There are two sign conventions. A reward $S(P,x)$ is maximized; a loss function $L(P,x)$ is minimized. The conversion $L=-S$ reverses every comparison. Here the linear probability score is a reward, while the binary Brier score and the negative logarithmic scoring rule are losses. Calling all three a "score" without specifying the direction would hide the central difference.

Throughout the binary examples, the random variable $Y$ has a Bernoulli distribution with actual success probability $q=0.7$, and the report $p\in [0,1]$ assigns probability $p$ to $Y=1$ and $1-p$ to $Y=0$. The value $q=0.7$ is stipulated by this statistical model; it is not inferred from data or a claim that a forecasting system is calibrated. In practice the true predictive distribution is unknown, and the relevant $q$ is the conditional probability of success given the information available when the forecast is issued. Propriety evaluates incentives for reporting that predictive distribution; it does not supply the unknown probabilities.

Consider an original balanced-readout model, rather than a fit to the experiment. Let $n\in \{0,\dots ,N_{\max }\}$, with a normalized prior probability $p_0(n)\ge 0$ and ${\sum }_n{p}_0(n)=1$. A detected chemical atom has outcome $s\in \{e,g\}$. Define the conditional probabilitiesHere $\chi$ is the accumulated quantum phase per photon, $\varphi$ the controlled analysis quantum phase, and $C$ the fringe contrast. The two probabilities sum to one for every $n$. Actual calibration may also require an offset and asymmetric detection errors; $C$ alone does not specify a quantum instrument.

Assume no photon number change during the record and fresh independently prepared atomic meters. For observed outcomes $s_1,\dots ,s_k$ at recorded settings ${\varphi }_1,\dots ,{\varphi }_k$, the Born rule and Bayes' theorem giveThe posterior probability is defined only for $Z_k>0$. Summing its numerator over $n$ proves normalization. This is also the update of diagonal elements of a field density operator under a measurement in quantum mechanics; it does not assume that an initial quantum superposition was an unknown pre-existing integer. The observed settings are part of the record, especially when selected adaptively.

For an exact finite example take $p_0=(1,1,1,1)/4$, $C=4/5$, and $\chi =\pi /2$. An $e$ outcome at $\varphi =0$ has likelihood $(9,5,1,5)/10$, giving $p_1=(9,5,1,5)/20$. Another $e$ at $\varphi =-\pi /2$ gives $p_2=(9,9,1,1)/20$. A subsequent $g$ at $\varphi =0$ givesThus three atomic records raise the posterior probability of $n=1$ from $1/4$ to $45/68$, while leaving finite uncertainty. No one of these outcomes identifies $n$ with certainty.

The periodic likelihood function imposes an identifiability limit. If $\chi =2\pi /q$ for an integer $q>0$, then $L_s(n+q;\varphi )=L_s(n;\varphi )$ for every setting: the posterior odds between $n$ and $n+q$ never change. Varying $\varphi$ can resolve a cosine reflection ambiguity within a period, but cannot remove this exact period. Changing $\chi$ can supply additional information. A finite cutoff is therefore a physical assumption to justify, not evidence that excluded high photon numbers were absent. At $C=0$ no outcome changes the prior probability; at $0<C<1$ neither binary outcome has an exactly zero likelihood.

The no-loss assumption also has a time limit. In an original zero-temperature loss model, let $\kappa =1/T_c$. A Fock state $|n⟩$ loses its first photon at rate $n\kappa$, so its survival probability is $\exp (-n\kappa t)$ and its mean lifetime is $T_c/n$ for $n>0$. Several records must be acquired within that lifetime to identify a plateau. When losses matter, first predict $p_k^{-}(n)={\sum }_m{T}_{nm}(\Delta t)p_{k-1}(m)$ using a normalized transition matrix, then apply the atomic Bayesian inference update to $p_k^{-}$. Thermal excitation requires upward transitions too. A delayed or broadened inference after a jump need not indicate that an atom caused the jump.

Write $q=P(A)$, $r=P(B\mid A)$ and $s=P(B\mid A^c)$. The law of total probability gives the displayed convex combination. With $q,r$ fixed and $s$ free in $[0,1]$,With $r,s$ fixed and $q$ free in $[0,1]$,The joint cell masses $qr,q(1-r),(1-q)s,(1-q)(1-s)$ are nonnegative and sum to one, proving sharpness in the joint-model interpretation. For ordinary conditional probabilities, a conditioning event must have positive mass. If both $r$ and $s$ are specified by ratios, then $0<q<1$: for $r\ne s$ the last bounds are strict and their endpoints are the infimum and supremum. If $r=s$, the marginal equals that common value for every $q$. Endpoint mixtures remain meaningful if the unused zero-mass conditional is treated as a prescribed conditional model rather than a defined ratio.