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For specified exposure periods with constant death hazard functions $h_k$, the integrated hazard through the endpoint is ${\sum }_k{h}_k\Delta t_k$. The displayed event probability follows from the survivor function identity. For different latent exposure schedules mixed with probabilities $w_j$, average the schedule-specific probabilities, ${\sum }_j{w}_j(1-e^{-H_j})$, rather than exponentiating the average integrated hazard. When the hazards are small, $1-e^{-H_j}\simeq H_j$ gives a person-time approximation. A comparison must specify the entire custody or treatment schedule; later changes to exposure cannot be inferred from an endpoint rate alone.

A plotting position assigns a probability level to an order statistic for a Q-Q plot. The displayed formula is one convention, not a universal definition. R's ppoints uses $p_i=(i-a)/(n+1-2a)$, with $a=1/2$ by default for $n>10$, which gives our choice at $n=19$; for $n\le 10$ its default is $a=3/8$. NIST's normal probability plot instead describes an approximation to uniform distribution order statistic medians. State the convention when comparing plots, especially their extreme quantiles. ^[ref][ref]

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].

The sum of independent Bernoulli random variables with possibly unequal success probabilities $p_i$ has a Poisson binomial distribution. Its probability generating function is $G_K(z)={\prod }_i(1-p_i+p_{i}z)$: expand the product to read off the probability mass function. Linearity of expectation and additivity of variance for independent random variables give $\mathbb{E}K={\sum }_i{p}_i$ and $\mathrm{Var}(K)={\sum }_i{p}_i(1-p_i)$. Equal $p_i=p$ recover the binomial distribution.

For three trials with success probabilities $1/10,3/10,4/5$, expanding $(9/10+z/10)(7/10+3z/10)(1/5+4z/5)$ gives masses $0.126,0.572,0.278,0.024$ at $0,1,2,3$. Their expected value is $1.2$ and variance is $0.46$. Replacing all three success probabilities by their average $0.4$ preserves the expected value but gives binomial distribution variance $0.72$, so it does not preserve the law.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

The normalized coherent state has number amplitude $e^{-|\alpha {|}^2/2}{\alpha }^n/\sqrt{n!}$, hence the displayed Poisson law with mean and variance $|\alpha {|}^2$. A product of coherent modes has factorized number probabilities.

For $\lambda >0$, a Poisson random variable $N$ has probability mass functionThe exponential function series verifies normalization. At $\lambda =0$, define $N=0$ with certainty. The equality of mean and variance is a distinguishing model property, though it does not by itself characterize the law.

When counting events in a homogeneous Poisson process, distinguish its rate $r$ from the distribution parameter: an interval of duration $t$ has $\lambda =rt$. A rate measured per minute times a duration measured in minutes gives a dimensionless expected count. This process assumes independent increments, a constant rate, and a negligible chance of multiple events relative to interval length in a sufficiently short interval. A rate alone does not establish a Poisson distribution. See Bertsekas and Tsitsiklis, chapter 6 summary, pages 46-47.

For a constructed arrival model with $r=2$ events per minute over $t=1.5$ minutes, $\lambda =3$. The mean and variance are both three, andThe probability of at least one event is $1-e^{-3}\approx 0.9502129$. These results describe the specified process, not every arrival mechanism with the same average rate.

The Poisson limit theorem, often called the law of rare events, connects this law to the binomial distribution. Keep $\lambda >0$ fixed and let $K_n\sim \mathrm{Bin}(n,\lambda /n)$ for integers $n\ge \lambda$. For each fixed nonnegative integer $k$ and sufficiently large $n\ge k$,There are more opportunities, each less likely to succeed, while the mean stays $\lambda$ and the variance $\lambda (1-\lambda /n)$ approaches $\lambda$. For $\lambda =3$, the probability of three successes is $0.2668279$ at $n=10$, $0.2310569$ at $n=50$, and $0.2257398$ at $n=200$, approaching the Poisson distribution value $0.2240418$.

The panels show the same count scale and the same Poisson distribution reference. At $n=10$, $p=0.3$ is not especially small and the difference is visible. No universal trial-count cutoff guarantees an adequate approximation for every event.

The independence assumption matters even when every success is rare. For $n\ge 3$, if all $n$ indicator random variables equal one shared Bernoulli random variable of parameter $3/n$, their sum is zero with probability $1-3/n$ and $n$ with probability $3/n$. Its mean is three, yet its zero-count probability tends to one, not $e^{-3}$. Some dependent counts still admit approximation through the Stein-Chen method, but this requires additional control of dependence.

For historical orientation, the BnF records Poisson's original probability treatise, Recherches sur la probabilité des jugements, as published in Paris in 1837. This bibliographic fact alone does not establish when the modern distribution terminology arose. See BnF record of Poisson's 1837 treatise.

A Poisson germ-grain model assigns a measurable region to each point of a Poisson point process and takes the union of those regions. A location remains uncovered exactly when no Poisson point has a region containing it, so its coverage probability is determined by the intensity of the corresponding set of germs and marks.

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.

A Poisson Metropolis chain samples a Poisson distribution by proposing an upward or downward unit step with equal probability and reflecting a downward proposal at zero into a self-proposal. Its upward and downward transition probabilities are the displayed $u_n,d_n$, with $d_0=0$, and the diagonal entry is $1-u_n-d_n$. For ${\pi }_n=e^{-\lambda }{\lambda }^n/n!$, adjacent stationary fluxes equal $\frac{1}{2}\min ({\pi }_n,{\pi }_{n+1})$. Hence detailed balance holds. Positive adjacent probabilities imply an irreducible Markov chain; $P(0,0)\ge 1/2$ implies aperiodicity. The normalized invariant Poisson distribution implies positive recurrence, so the countable-state Markov chain convergence theorem applies. The boundary proposal probability is essential: forcing every proposal at zero to go upward would require a different acceptance ratio.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

For $n\ge 3$ independent lengths with the uniform distribution on $[0,1]$, the probability of forming a nondegenerate polygon is $1-1/(n-1)!$. The necessary and sufficient condition is that every length is less than the sum of the others. The failure events are disjoint except on sets of probability zero. The sum of $n-1$ lengths has probability density function $s^{n-2}/(n-2)!$ on $0<s<1$, by repeated convolution of probability densities. A specified rod is too long with probability ${\int }_0^1(1-s)s^{n-2}/(n-2)!ds=1/n!$; summing over the $n$ rods proves the formula. For three and four rods the probabilities are $1/2$ and $5/6$.

At $p=1/2$ for independent bond percolation on the square lattice, there exist $C<\infty$ and $\alpha >0$ with ${\mathbb{P}}_{1/2}(0\leftrightarrow \partial [-n,n{]}^2)\le C{n}^{-\alpha }$. The Russo-Seymour-Welsh theorem and the Harris-FKG inequality give a uniform positive probability of a closed dual graph cycle around each geometrically spaced square ring. The independent annular barriers for percolation give the polynomial bound and imply zero critical percolation probability. This does not determine the exact critical exponent.

Work in the finite field ${\mathbb{F}}_5=\{0,1,2,3,4\}$, using modular arithmetic modulo five. Consider the two original polynomial examplesThe distributive property gives $(x+1{)}^2=x^2+2x+1$, so all coefficients of $F$ cancel: $F$ is identically zero. In contrast, $G=x^2-3x+2=x^2+2x+2$ in this finite field, with nonzero leading coefficient. Its roots of a polynomial are exactly one and two: a product in a field is zero only if a factor is zero.

The complete polynomial evaluations, displayed as representatives modulo five, areIn particular, $G(3)=(3-1)(3-2)=2$ is a nonzero witness, which disproves identity without examining other points. At one or two, a zero value fails to reveal that $G$ is nonzero. A random variable $a$ with a uniform distribution on a finite set ${\mathbb{F}}_5$ hits those two polynomial roots with probability $2/5$.

A crucial limit appears with $H(x)=x^5-x$. Its coefficients make it a nonzero polynomial of polynomial degree five in ${\mathbb{F}}_5[x]$, yet direct polynomial evaluation gives zero at all five field elements: the values before reduction are $0,0,30,240,1020$, all divisible by five. Thus $H$ induces the same polynomial function as the zero polynomial. Any number of trials restricted to ${\mathbb{F}}_5$ will falsely accept this identity. This is compatible with the Schwartz-Zippel lemma: the ratio of polynomial degree to sampling-set size is $5/5=1$, which gives no useful guarantee.

When a polynomial's representation permits polynomial evaluation over a larger field extension, a larger sampling set can restore the useful degree ratio. Access only to the induced polynomial function on the original finite field cannot distinguish $H$ from zero.^[ref]

After $T$ input-bit queries, every probability amplitude is a polynomial of degree at most $T$, because the bit-query matrix entries are affine in the input bits. An acceptance probability is a sum of squared absolute amplitudes and has polynomial degree at most $2T$. Multilinear reduction on the Boolean cube does not increase degree. Exact acceptance therefore implies $Q_E(f)\ge \lceil \mathrm{deg}(f)/2\rceil$. For bounded error the acceptance polynomial only approximates the Boolean function; its exact representing degree is not the relevant bound.

An ideal position measurement of a normalized wavefunction $\psi$ assigns a probability to each spatial region $A$ through the Born rule: $P(A)={\int }_A|\psi (x){|}^{2}dx$. This specifies outcome statistics; it does not by itself specify the post-measurement state. A physical detector has finite spatial resolution, which must be included in its measurement model.

Let $S,T$ be independent random variables with exponential distributions of respective positive rates $\alpha ,\beta$. Condition on $T=t$ and use the survival function of $S$:At $u=0$ this is $P(S>T)$. Dividing by that probability proves that the positive difference conditional on $S>T$ again has an exponential distribution with rate $\alpha$. The memorylessness of the exponential distribution persists after an independent competing event has already occurred. Replacing the difference by a negative sentinel when $S\le T$ adds an atom of a measure of mass $\alpha /(\alpha +\beta )$ at that sentinel without changing the positive tail.