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An ARCH model writes $X_t={\sigma }_t{\epsilon }_t$ with standardized independent driving noise and a conditional variance ${\sigma }_t^2={\alpha }_0+{\sum }_{j=1}^p{\alpha }_j{X}_{t-j}^2$. Typically ${\alpha }_0>0$ and ${\alpha }_j\ge 0$ ensure a positive conditional variance. It allows a changing conditional scale even when $X$ is a stationary process.

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

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

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

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

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

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

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

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

Mean reversion is a tendency for a stochastic process to move back towards a reference level. For an Ornstein-Uhlenbeck process, the conditional expected value of a future deviation is the present deviation multiplied by $e^{-\theta h}$ after time $h$. Individual sample paths can move away from that level: mean reversion describes a conditional tendency, not monotone motion or a guaranteed return by a deadline. It also does not by itself imply a stationary process; the initial probability distribution matters.

A Gaussian process has a mean-square derivative at $t$ when its difference quotients converge in L2 space. The mean-square convergence need not give differentiable sample paths. For a centered stationary process with twice continuously differentiable covariance function $K(t-s)$, the difference quotients are Cauchy in L2 space, since their covariance is a rectangular average of $-K^{\prime \prime }$. Their limits form a Gaussian process with covariance function $\mathbb{E}[Y(s)Y(t)]=-K^{\prime \prime }(s-t)$.

Use the Itô product rule with the deterministic factor $e^{-\lambda t}$. Its quadratic covariation with $X$ is zero, givingMultiplication by the integrating factor $e^{\lambda t}$ solves the Ornstein-Uhlenbeck stochastic differential equation:The initial value $B_1$ is standard normal and independent of the future driving Brownian motion. Thus $Y$ is a centered Gaussian process with variance one. For $s\le t$, its independent future noise gives $\mathbb{E}(Y_s{Y}_t)=e^{-\lambda (t-s)}$. Its Gaussian finite-dimensional distributions consequently depend only on time differences, so it is a stationary process. This verifies the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process directly.

Put ${\theta }_0=1$ and define the autoregressive polynomial and moving-average polynomial byFor a minimal, coprime ARMA representation, the usual simultaneous causality and invertibility root criteria for an ARMA model areThe autoregressive condition makes $\Theta /\Phi$ a stable one-sided filter and gives the causal second-order stationary process. The moving-average condition makes its inverse a stable one-sided filter, so the white noise can be reconstructed from present and past observations. If the two polynomials have a common factor, first use common factor cancellation in an ARMA model; necessary-and-sufficient root statements apply to the reduced noise-driven representation. Stationarity here is second-order stationarity; weak white noise alone does not assert strict stationarity of all finite-dimensional laws.

For $p=1$, write $\varphi ={\varphi }_1$, so $|\varphi |<1$. Expanding $(1-\varphi z{)}^{-1}={\sum }_{r\ge 0}{\varphi }^r{z}^r$ gives the Wold representationIn particular $c_0=1$, $c_j=\varphi c_{j-1}+{\theta }_j$ for $1\le j\le q$, and $c_j={\varphi }^{j-q}{c}_q$ for $j\ge q$. The coefficients are square-summable, so this series converges in mean square. Causality and invertibility identify the driver with the linear innovation process, justifying the Wold terminology. For $\varphi =0$, powers with exponent zero equal one and the formula reduces to the finite moving average.

To obtain the covariance equations, subtract $\varphi X_{t-1}$ and take covariance with $X_{t-k}$. Orthogonality of the white noise and the causal expansion implyThus the innovation coefficients of an ARMA(1,q) process give the entire triangular system in one formula:At $k=0$, use ${\gamma }_{-1}={\gamma }_1$, since the autocovariance of a real stationary process is symmetric. The terms are consequently ${\sigma }^2(c_0+{\theta }_1{c}_1+\cdots +{\theta }_q{c}_q)$ at zero, ${\sigma }^2{\theta }_q{c}_0$ at $q$, and the intervening partial sums as required.

For $q=1$, put $\theta ={\theta }_1$ and $c_1=\varphi +\theta$. The equations becomeSubstitution of the second into the first, followed by the geometric recursion, yields the autocovariance of a causal ARMA(1,1) process:This includes $\varphi =0$: only lags $\pm 1$ survive, as for an MA(1) process.

With angular frequency $\lambda$, the periodogram isAt Fourier frequencies $2\pi j/T$ it is the squared modulus of the normalized discrete Fourier transform. If a nonzero mean is unknown, subtract the sample mean first.

For a zero-mean stationary process, expanding the square givesUnder absolute summability of the covariances this converges to the spectral density of a stationary process. Thus the periodogram is generally biased at finite $T$, but asymptotically unbiased under this short-memory time series condition.

It is nevertheless not a pointwise estimator with statistical consistency unless it is smoothed. For Gaussian white noise, at a nonzero Fourier frequency other than the Nyquist frequency, the real and imaginary Fourier components are independent normal variables. Exactly,where $f={\sigma }^2/(2\pi )$. The variance does not decrease with $T$. Under usual short-memory time series assumptions the same exponential limit is asymptotic for general processes. Averaging nearby frequencies or using a lag-window estimator reduces variance; a frequency bandwidth tending to zero while $T$ times that bandwidth tends to infinity can give statistical consistency. The raw plot remains useful for detecting strong periodic peaks, but increasing the record length alone does not remove its pointwise noise.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

At each point the Gaussian process has a standard normal marginal distribution. Its characteristic function gives the coherent screen fieldIt is constant in $z$, so its Fourier transform is $m\delta (\nu )$. Applying the Fresnel propagator to this zero-transverse-wavenumber component gives the ensemble-averaged reduced spectrumThus the reduced mean is independent of $x$. The physical mean $⟨\psi (x,z)⟩=e^{ikx}{e}^{-{\beta }^2/2}$ still has its carrier phase. The coherent attenuation by a Gaussian phase screen takes place at the screen, with no further coherent attenuation in homogeneous space.

The full spectral acoustic flux depends on the second moment, not the square of this mean. For a stationary process, the generalized cross-spectral correlation isIndeed its propagation factors cancel on $\nu ={\nu }^{\prime }$. This explains why the ensemble-averaged spectral acoustic flux in (b) is also independent of $x$, even for nonzero transverse wavenumbers whose individual complex amplitudes change phase. The coherent atom has weight $|m{|}^2=e^{-{\beta }^2}$; the full stationary phase-screen power spectrum also contains fluctuations. Under the decay assumptions in (b), their integrated weight is $1-e^{-{\beta }^2}$. One must use these weights in a spectral measure of a stationary random field, rather than square a Dirac delta distribution. A phase-only screen preserves the leading-order total acoustic energy flux, although it reduces its coherent part.

Multiplication by the integrating factor $e^{t/2}$ and the Itô product rule give the Ornstein-Uhlenbeck processThe two summands are independent centered normal random variables, because $X_0$ is independent of the Brownian motion and the integrand is deterministic. By the Itô isometry, their variances sum to $e^{-t}+{\int }_0^t{e}^{-(t-u)}du=1$. ThusFor $t\ge s$, split the Itô integral at $s$ to obtain $X_t=e^{-(t-s)/2}{X}_s+{\int }_s^t{e}^{-(t-u)/2}d{B}_u$, where the latter term is centered and independent of $X_s$. ConsequentlyIn particular this initialization makes the Gaussian process a stationary process.

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 population $N(t)$ of a queue counts every admitted customer still inside the selected system at time $t$, including customers receiving service when the server is inside that boundary. Its long-run time average is $L$. The waiting population $N_q(t)$ excludes customers currently receiving service, and its long-run time average is $L_q$. For a single server that serves whenever work is present,where $\rho$ is the server utilization, assuming at most one customer receives service at a time. In a stationary process with ergodicity and finite expected values, these time averages equal $\mathbb{E}N(t)$ and $\mathbb{E}{N}_q(t)$. The workload of a queue instead counts unfinished work, so two queues with the same population can have very different remaining workloads.

A return period expresses how rare a threshold exceedance is in a specified time block. Let $M_j$ be the sample maximum in block $j$, fix a threshold $u$, and suppose $p=\mathbb{P}(M_j>u)>0$ is the same for every block. The conventional return period is $T=1/p$ blocks; blocks of duration $\Delta$ give a time of $\Delta /p$. Thus an annual exceedance probability $p=0.01$ means a 100-year return period. This names a frequency scale, not a schedule or a guarantee of one exceedance in each century. The annual convention and the misleading “once every 100 years” interpretation are discussed in USGS: the 100-year flood.

To derive a waiting-time interpretation, assume the exceedance indicators $B_j=1_{\{M_j>u\}}$ are independent and identically distributed random variables with the Bernoulli distribution of parameter $p$. Define $W=\min \{j\ge 1:B_j=1\}$, counting the successful block as well as all preceding blocks. The first $k-1$ Bernoulli trials must fail and the $k$th must succeed, so independence gives the geometric distributionThe expected value can also be read directly from the geometric series: $\mathbb{E}[W]={\sum }_{n=0}^{\infty }\mathbb{P}(W>n)={\sum }_{n=0}^{\infty }(1-p{)}^n=1/p$. This counts blocks from the starting boundary, not the exact time of an exceedance within its block. Counting only failed blocks instead would give expected value $T-1$. The waiting law and its memorylessness of the geometric distribution are developed in Kyle Siegrist, Random Services: the geometric distribution.

For a horizon of $n$ complete years, at least one exceedance is the complement of no exceedances. Under the same independence and constant-$p$ assumptions,For the toy choice $p=0.01$, this gives $9.6\%$ in 10 years, $26.0\%$ in 30 years, $39.5\%$ in 50 years, and $63.4\%$ in 100 years. Even after 100 years there is a $36.6\%$ chance of no exceedance. These numbers are exact model calculations rounded to one decimal place, not measured frequencies. The median waiting count is 69 years, since $1-0.99^{68}<1/2\le 1-0.99^{69}$; the expected value is 100 years. A mean and a median answer different questions.

The exceedance count $N_n={\sum }_{j=1}^n{B}_j$ has the binomial distribution under independence, with expected value $np$. Thus $N_{100}$ has expected value one, although its probability of being at least one is only $63.4\%$: some realizations contribute several exceedances. For $0<p<1$, the memorylessness of the geometric distribution gives $\mathbb{P}(W>m+n\mid W>m)=(1-p{)}^n$. A long quiet spell does not make the next independent year's exceedance overdue, and an exceedance this year does not force a quiet interval. Two specified consecutive years both exceed with probability $p^2=0.0001$ in this toy model.

A reproducible tabletop experiment makes the assumptions tangible. Put 100 identical slips numbered 1 to 100 in a bag; after thoroughly mixing, draw one and record whether it is 100. Replace the slip and mix again before every draw. Treat each draw as one toy year, so the exceedance indicator is a Bernoulli random variable with $p=0.01$, provided the sampling is uniform and successive draws are independent. Repeat complete 30-draw runs and record the fraction containing at least one 100; separately repeat runs stopped at the first 100 and record their waiting counts. The first fraction targets $1-0.99^{30}$; the waiting-count histogram targets $0.99^{k-1}(0.01)$. Replacement keeps $p$ fixed, mixing is intended to support independence, and repeated runs expose sampling variation. This is a proposed demonstration of the toy model; no experimental results are asserted here.

Stationarity supports a time-invariant annual probability distribution, but does not imply independence. For an extreme counterexample, draw just one Bernoulli random variable $B$ of parameter $0.01$ and set $B_j=B$ for every year. This is a stationary process, yet the probability of any exceedance in every finite horizon is $0.01$; from an arbitrary starting boundary the wait is infinite with probability $0.99$. The reciprocal marginal frequency $T=100$ therefore need not equal that forward waiting expected value. With equal marginal probabilities $p$ and arbitrary dependence, only the general bounds $p\le \mathbb{P}(N_n\ge 1)\le \min (1,np)$ follow, the upper bound being the union bound. Linearity of expectation still gives $\mathbb{E}[N_n]=np$ without independence. Mere zero correlation does not restore the product formula; even pairwise independence does not imply joint independence. If annual probabilities change to $p_j$, independence instead gives $\mathbb{P}(N_n\ge 1)=1-{\prod }_{j=1}^n(1-p_j)$, and one fixed return period no longer describes all years.

Two Ornstein-Uhlenbeck processes with the same driving Brownian motion have the displayed difference. If $Y_0$ is standard normal and independent of future Brownian increments, replacing it by $-Y_0$ produces another stationary process with the same covariance and second moment, but a different pathwise coupling. Equal initial values are necessary for indistinguishability of stochastic processes.

A simple random walk on a locally finite graph chooses each vertex in the current vertex's graph neighbourhood with equal probability, assuming there is at least one available neighbour. On the integer line there are two choices, one unit left or right, so each has probability $1/2$. This is the one-dimensional simple symmetric random walk. Terminology varies: a simple random walk on the integer line is also often allowed unequal step probabilities, with “symmetric” specifying the fair case.

For a simple symmetric random walk starting at $S_0=0$, each increment has the Rademacher distribution, with expected value zero and variance one. Thus linearity of expectation and the variance of a sum of independent random variables giveFor example, after $80$ steps the expected value is zero, the variance is $80$, and the standard deviation is $\sqrt{80}\approx 8.944$. Doubling the typical distance scale takes four times as many steps. This square-root growth explains why a random walk can wander far even when there is no preferred direction.

The colored curves are illustrative sample paths, not averages over all possible walks. The dashed curves show the expected values; the shaded bands show the expected value plus or minus one standard deviation at each time. These bands are a scale of spread, not confidence intervals, and give no guarantee that an entire sample path stays inside them.

Zero expected value does not make this a stationary process: its one-time probability distribution changes with $n$, as the increasing variance already proves. It does have stationary increments and independent increments, because blocks of fresh steps have the same probability distribution at any starting time and disjoint blocks use independent random variables. This separation of position from increment is one reason the random walk is such a useful first example of a stochastic process.

For an adapted stationary process with the displayed bounds, the Itô formula and stationarity give ${\int }_0^t(\mathbb{E}[Z_s{g}_s]+\lambda )ds=0$. The Cauchy-Schwarz inequality makes each integrand nonnegative, so equality holds almost everywhere. Then $\mathbb{E}(g_s+\lambda Z_s{)}^2\le 0$, proving $g=-\lambda Z$ for almost every time and outcome. Thus the process solves the Ornstein-Uhlenbeck stochastic differential equation and its initial value, rather than stationarity alone, determines its coupling with W.

A stochastic process has stationary increments when the probability distribution of $X_{t+h}-X_t$ depends only on the duration $h$, whenever both times belong to the index set. The clock origin then does not change the increment law. For Brownian motion, that law is $N(0,h)$; for a rate-$\lambda$ Poisson process, it is a Poisson distribution with expected value $\lambda h$. Neither process itself is a stationary process: their variances grow with $t$. Stationary increments and a stationary process describe different properties.