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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 measurement in 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.

A positive operator-valued measure is a family of positive operators $E_i$ satisfying ${\sum }_i{E}_i=I$. Each $E_i$ is called an effect and determines a probability $p_i=\mathrm{Tr}(E_i\rho )$ for input density operator $\rho$. The effects need not be orthogonal projections or commute. This permits measurements with more outcomes than the dimension of the Hilbert space, such as the trine qubit POVM. Coupling to a quantum ancilla and measuring the larger system supplies a physical route to generalized measurements. ^[ref].

For an elementary unsharp $Z$ example on a qubit, takeOn the eigenstate $|0⟩$, $p_{\pm }=(1\pm \eta )/2$; at $\eta =1/2$ they are $3/4$ and $1/4$. At $\eta =0$ the outcome is a fair coin independent of the input, and at $\eta =1$ these effects specify a sharp $Z$ projective measurement. Choosing $K_{\pm }=\sqrt{E_{\pm }}$ specifies one quantum instrument, but $U_{\pm }\sqrt{E_{\pm }}$ for outcome-dependent unitary operators has the same effects and different state updates. Thus POVM does not determine the post-measurement state.

When every entry of a payoff matrix $B$ is positive, its game value $v$ is positive. Setting $x=p/v$ converts the row player's probability vector and guaranteed value into the displayed linear program, whose optimum is $1/v$. Conversely normalizing a feasible $x$ gives a mixed strategy guaranteeing $1/(1^{T}x)$. The dual maximizes $1^{T}y$ under $By\le 1$, $y\ge 0$ and recovers the column strategy by normalization. Adding a constant to every payoff preserves Nash equilibria and lets this positive-payoff formulation handle general finite zero-sum games.

For an integrable finite-dimensional random vector $Y$, let $L=\{h:h\cdot Y=0\text{ almost surely}{\}}^{\perp }$. The displayed convex cone is open relative to $L$. Given a bounded strictly positive $Z$, the perturbation $Z_{\epsilon }=Z(1+\epsilon \cdot Y/(1+\Vert Y\Vert ))$, $\Vert \epsilon \Vert <1$, changes its weighted expectation by $A_Z\epsilon$, whereFor $0\ne v\in L$, $v^T{A}_{Z}v>0$: otherwise $v\cdot Y=0$ almost surely. Thus $A_Z$ is invertible on $L$, and these perturbations give a neighborhood of every point of the cone. The hyperplane separation theorem consequently gives the alternative: either zero is a strictly positively weighted expectation, or a nonzero direction in $L$ has a nonnegative dot product with $Y$ almost surely and a positive dot product with positive probability. To obtain the latter assertion, test the separating direction with $Z=1_{\{h\cdot Y<0\}}+\epsilon$ and let $\epsilon \downarrow 0$.

Let $N$ have the positive-integer geometric distribution with success parameter $p\in (0,1)$, and conditionally toss $N$ independent coins with head probability $\theta \in (0,1)$. Exactly one head has conditional probability $n\theta (1-\theta {)}^{n-1}$. Combining this with the prior by Bayes' theorem gives the displayed posterior with $r=(1-p)(1-\theta )$; its normalization uses ${\sum }_{n\ge 1}n{r}^{n-1}=(1-r{)}^{-2}$. For $p=\theta =1/2$, the posterior is $9n/4^{n+1}$. It is also a shifted negative binomial distribution, with $N-1$ equal in law to the number of failures before two successes of probability $1-r$.

For a mixed Poisson distribution with known exposure $n>0$, multiply the unnormalized posterior density ${\theta }^s{e}^{-n\theta }\pi (\theta )$ by $\theta$ and compare its integral with the next probability mass. The factors $n^s/s!$ give the displayed identity whenever $p_n(s)>0$. Both mass probabilities have the same exposure $n$. For a future conditionally independent unit-exposure Poisson distribution count, this is also its posterior predictive expected value. It holds for arbitrary proper mixing distributions, with integrals taken against the prior measure when there is no density.

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.

After the quantum quench, hold $H_f$ fixed and evolve without an intervening energy measurement. The unitary time evolution isProjection onto the original $|0⟩$ gives probability amplitude $c^2{e}^{+i{\Delta }_{f}t/(2\hslash )}+s^2{e}^{-i{\Delta }_{f}t/(2\hslash )}$. Squaring its modulus yields the two-level quantum return probabilityTheir sum is one. For $\theta =\pi /3$, the first minimum of $P_0$ is $1/4$ at $t=\pi \hslash /{\Delta }_f$, and the return is complete at $T=2\pi \hslash /{\Delta }_f$. At $\theta =0$ or $\pi$ there is no oscillation, because the initial quantum state is already a final energy eigenstate.

Changing measurement basis reveals the evolving relative quantum phase. Measuring the final energy first instead destroys this quantum coherence between final energy eigenstates in the unconditioned ensemble: its later original-basis return probability is $c^4+s^4=1-\frac{1}{2}{\sin }^2\theta$, equal to $5/8$ for this example. That constant differs from the oscillation of the unmeasured pure state. The work protocol and the oscillation protocol therefore use separate preparations.

Postselection conditions a quantum measurement on a specified outcome. If its branch acts by a linear operator $K$ on a normalized input $|b⟩$, its probability is $\Vert K|b⟩{\Vert }^2$, and its normalized output is $K|b⟩/\Vert K|b⟩\Vert$. This conditional state is defined only when the probability is nonzero. Physically unsuccessful trials are discarded; postselection is not a deterministic implementation of the nonlinear normalization map.

An AlphaFold confidence score for local structural accuracy, usually reported on a 0–100 scale. It is not a probability that a complete structure is correct.

For a success index $J$ with geometric distribution of parameter $q=2^{-n}$, write $J-1=2^{n}Q+R$, $0\le R<2^n$. Encode $Q$ by $Q$ ones followed by zero, then encode $R$ in exactly $n$ bits. This prefix code has length $n+Q+1$. Since $\mathbb{P}(Q\ge k)=(1-q{)}^{k{2}^n}$, its expected length is the displayed expression, at most $n+e/(e-1)$. Thus a very rare success can be communicated in approximately the logarithm of its inverse probability, rather than one bit per trial.

// Target: quantum-theory.bigb