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For any real separable Hilbert space $H$, any orthonormal basis $(e_j)$ and independent standard normal random variables $(Z_j)$, the displayed series converges almost surely for each fixed $h\in H$ and in L2 space. Indeed, the Parseval identity for a Hilbertian basis gives ${\sum }_j⟨h,e_j{⟩}^2=\Vert h{\Vert }^2$, so the partial sums form an L2 space-bounded martingale. The martingale convergence theorem supplies both limits. Every finite linear combination has the normal distribution with variance $\Vert {\sum }_j{u}_j{h}_j{\Vert }^2$. Thus its finite-dimensional characteristic function is $\exp (-\Vert {\sum }_j{u}_j{h}_j{\Vert }^2/2)$, independent of the chosen orthonormal basis. This proves equality of the finite-dimensional distributions of the resulting isonormal Gaussian processes; pathwise equality needs a coupling.

For every $c>0$, the stochastic process ${\tilde{B}}_t=c^{-1/2}{B}_{ct}$ is standard Brownian motion. Its Brownian increments remain independent and have variance $c^{-1}c(t-s)=t-s$; continuity and the zero starting value are preserved. Equivalently,Equality here is of whole-process distributions, including every finite-dimensional distribution. It is not an equality of the two expressions along a particular realization. With $c=4$, speeding time up fourfold requires dividing displacement by two to recover the standard law. The figure applies exactly these coordinate changes to four finite samples; the shared colors indicate reused samples, while the dashed lines show pointwise standard deviations, not bounds on entire sample paths.

The same square-root normalization explains how a simple random walk approaches Brownian motion. Let $S_k={\sum }_{j=1}^k{\xi }_j$, where the independent steps take values $+1$ and $-1$ with equal probability. Then $R_n(t)=S_{\lfloor nt\rfloor }/\sqrt{n}$ has variance $\lfloor nt\rfloor /n$. Its linear interpolation isand ${\sup }_{0\le t\le 1}|Y_n(t)-R_n(t)|\le 1/\sqrt{n}$. Donsker's theorem says $Y_n\Rightarrow B$ in $C[0,1]$ with the uniform norm: this is convergence in distribution of random paths. The interpolation bound vanishes, but the theorem does not assert pointwise convergence of an arbitrarily fixed walk to an independently chosen Brownian motion. The usefulness is that many-step discrete motion has a universal continuous limit.^[ref]

For a real-valued stochastic process $(X_t)$ with finite second moments, write $m(t)=\mathbb{E}{X}_t$. Its covariance function isThe diagonal gives the variance at each time; off-diagonal values record linear dependence between times. For the independent standard normal random variables above, $C_X(m,n)=1_{\{m=n\}}$, whereas the shared Gaussian process $Y_n=Z$ has $C_Y(m,n)=1$ for all $m,n$. For a fair simple random walk on the integer line, $C(m,n)=\min (m,n)$; Brownian motion has the continuous-time counterpart $C(s,t)=\min (s,t)$. These formulas quantify how much random history the observations share.

Every covariance function is a positive-semidefinite kernel. Indeed, for any real coefficients $a_1,\dots ,a_k$,Thus a proposed covariance function must make every finite covariance matrix positive semidefinite. For a Gaussian process, the mean function and the covariance function determine every finite-dimensional distribution, including degenerate ones such as the shared process. Conversely, any real positive-semidefinite kernel satisfying $C(s,t)=C(t,s)$, together with a specified mean function, defines consistent multivariate normal distributions; the Kolmogorov extension theorem then constructs a Gaussian process. For a general stochastic process, covariance alone does not determine the full law. The exceptional completeness of this description is what makes Gaussian process regression possible with a mean and a covariance model.

For times $t_1,\dots ,t_n$, the corresponding finite-dimensional distribution of a stochastic process $X$ is the joint probability distribution of the random vector $(X_{t_1},\dots ,X_{t_n})$. The collection of these distributions records every finite set of coordinates of the process.

Knowing only the one-time marginal distributions is insufficient. Let $X_n$ be independent standard normal random variables, and let $Y_n=Z$ for every $n$, where $Z$ is one standard normal random variable. Every $X_n$ and $Y_n$ has the same standard normal distribution. Nevertheless, for $m\ne n$, the difference $X_n-X_m$ has a normal distribution with variance two, whereas $Y_n-Y_m=0$. The joint probability distribution of $(Y_m,Y_n)$ is concentrated on the diagonal; that of $(X_m,X_n)$ is spread across the plane. A model for tomorrow's value therefore needs the relation between observations as well as their separate distributions.

For a real-valued stochastic process, its finite-dimensional distributions determine its probability law on the product sigma-algebra. With an uncountable index set, a path property such as continuity need not be measurable in that sigma-algebra; specifying a suitable space of sample paths and a regular modification of a stochastic process is a separate step. This distinction is explained in MIT, The basics of stochastic processes, section 1.1.

A centered Gaussian process with almost surely continuous sample paths and covariance $\mathbb{E}[B_s{B}_t]=\min (s,t)$ is standard Brownian motion. Here Gaussian process means that every finite-dimensional distribution is a multivariate normal distribution; having separate normal distributions at each time would not suffice. This characterization is stated in Pitman, Basic Properties of Brownian Motion, Definition 15.2.

To compute the Brownian covariance kernel directly, suppose $0\le s\le t$. Since $B_t=B_s+(B_t-B_s)$ and the added Brownian increment is centered and independent of $B_s$,Symmetry gives $\min (s,t)$ for arbitrary nonnegative times. For instance, $\mathrm{Cov}(B_1,B_4)=1$ and their correlation is $1/\sqrt{1\cdot 4}=1/2$. Positions are therefore related even though disjoint Brownian increments are independent: both positions contain the displacement accumulated during the first unit of time. Conversely, taking differences of $\min (s,t)$ gives zero covariance for disjoint increments; their joint multivariate normal distribution turns this into independence.

For fractional Brownian motion, $H\in (0,1)$ controls the variance $|t-s{|}^{2H}$ of increments and the scaling of its finite-dimensional distributions. Larger $H$ means smaller increment variance on intervals shorter than one and greater sample-path Hölder continuity.

A Markov process is a stochastic process whose present state contains all the information in its observed past needed to predict its future. Precisely, if ${\mathcal{F}}_s$ is the natural filtration generated by observations up to time $s$, then, for $t>s$ and any measurable set $A$ of states,The equality is between conditional probabilities and holds almost surely. It expresses the Markov property: the past and future have conditional independence given the present. It does not assert independence of consecutive states. In the sunny/rainy example below, tomorrow's sunny probability is $0.8$ after sunshine but $0.4$ after rain, so knowing today's state matters.

The memory assumption concerns the chosen state space. If an equipment model's failure probability depends on its age, recording only “working” or “failed” may lose relevant information; recording age as part of the state can retain it. Similarly, a dependence on two previous observations can sometimes be represented by using their ordered pair as the current state. Whether such a description fits observations is a modeling question.

Transition Markov kernels describe the conditional distribution of a state at a later time. Successive kernels and the initial probability distribution determine every finite-dimensional distribution, through repeated conditional probability. The time-homogeneous Markov property is an additional assumption: the kernel depends on elapsed time rather than calendar time. The attraction is compression: prediction can proceed from a current state instead of an entire recorded history. Brownian motion and continuous-time Markov chains illustrate that a Markov process need not evolve in discrete time.

Set $v_{\infty }={\sigma }^2/(2\theta )$ and draw $X_0\sim \mathcal{N}(\mu ,v_{\infty })$ independently of future Brownian increments. The Ornstein-Uhlenbeck process is then a strictly stationary process, withEvery time has the same normal distribution, and the covariance function depends only on the time lag. For a Gaussian process these properties also make all its finite-dimensional distributions invariant under a common time shift. The stationary distribution balances the continual random forcing against the contracting drift.^[ref]

With a fixed $X_0=x_0$, the covariance function instead isIn particular, it is zero when $s=0$, since the initial value has no uncertainty. The second term records the fixed start; it is the general scaling of the zero-start Ornstein-Uhlenbeck covariance. The pure exponential covariance function displayed above assumes stationary initialization. For a fixed start, that function emerges as the late-time limit at fixed lag, while the one-time probability distributions approach $\mathcal{N}(\mu ,v_{\infty })$ by convergence in distribution.

For the stipulated $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, stationary initialization has variance one from the outset. Its autocorrelation is $e^{-|h|}$, so the correlation time is one: a lag of one reduces the autocorrelation to $e^{-1}\approx 0.3679$. Its raw second moment $\mathbb{E}[X_s{X}_t]=4+e^{-|t-s|}$ includes the squared expected value; it must not be confused with the covariance.

The upper panel shows three simulated sequences with independent stationary initial values and exact transitions at spacing $0.05$. Joining their points only guides the eye; it does not simulate a continuous sample path between those points. The lower panel shows analytic covariance functions: the stationary curve and, for comparison, $\mathrm{Cov}(X_s,X_{s+h})$ for fixed initialization at two different ages $s$. These are simulated samples and model curves, not measured trajectories or empirical covariance estimates.

For any real orthonormal basis $(G_i)$, the argument of (b)(ii) gives almost sure convergence and convergence in L2 of $\tilde{W}h$. Finite linear combinations have normal distributions, and the Parseval identity for a Hilbertian basis givesThus $W$ and $\tilde{W}$ are centered Gaussian processes with the same covariance function. More explicitly, their finite-dimensional characteristic functions are bothThis proves the basis invariance of an isonormal Gaussian process: equality of all finite-dimensional distributions, and hence equality in law of the indexed collections.

For the specified cosine orthonormal basis, put ${\lambda }_i=\pi (i-\frac{1}{2})$. Its coefficient against an interval indicator isConsequently the Brownian half-integer sine expansion, a Karhunen–Loève expansion, isFor every fixed $t$, the sum converges almost surely and in L2 space because the coefficient squares sum to $\Vert 1_{[0,t]}{\Vert }^2=t$. Its continuous modification is a standard Brownian motion by (b)(v). The expansion also holds in $L^2(\Omega \times [0,1])$: the expectation of the squared tail is the integral of the coefficient-square tail, which tends to zero by the dominated convergence theorem, with bound $t$.

To realize this expansion for the original $B$, choose $b_i=W{G}_i$. These are Jointly Gaussian with covariance matrix equal to the identity, hence are independent standard normal random variables. The finite orthogonal projections of $1_{[0,t]}$ onto $G_1,\dots ,G_N$ converge in $\mathcal{H}$, and the identity $\mathbb{E}|Wh{|}^2=\Vert h{\Vert }^2$ transfers that convergence to the displayed expansion for $B_t$. With an independently chosen sequence $(b_i)$ the resulting Brownian motion has the same law; a pathwise identification uses the coupling just given.

Use the exponent convention $\mathbb{E}{e}^{iu{X}_t}=e^{t\psi (u)}$. A real Lévy process starts at $0$, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Its characteristic exponent of a Lévy process satisfies $\psi (0)=0$ and is continuous. This convention has the opposite sign from the alternative $e^{-t\Psi (u)}$, so $\psi =-\Psi$.

For real $u$, the exponential martingale of a Lévy process is integrable because $|M_t^u|=e^{-t\mathrm{Re}\psi (u)}$ is deterministic and finite. For $s\le t$, its conditional expectation with respect to the natural filtration isThe second equality uses both independent increments and stationary increments. Thus this is a complex martingale, meaning that its real and imaginary parts are martingales.

A probability distribution is an infinitely divisible distribution when, for every positive integer $n$, it is the distribution of a sum of $n$ independent and identically distributed random variables. Their common distribution is allowed to depend on $n$. For a Lévy process,and these summands are independent with the common law of $X_{1/n}$. This proves the required infinite divisibility directly.

The Lévy–Khintchine theorem states that a real infinitely divisible distribution has characteristic function $e^{\psi (u)}$, where, for a unique triplet with the following fixed truncation,with $b\in \mathbb{R}$, $a\ge 0$, $K$ a positive measure, $K(\{0\})=0$, and $\int (1\wedge y^2)K(dy)<\infty$. Conversely every such triplet gives an infinitely divisible distribution and a Lévy process, unique in law, whose time-$t$ characteristic function is $e^{t\psi (u)}$. The Lévy measure $K$ need not be finite near zero. Taylor expansion makes the compensated integrand $O(y^2)$ there, while the measure is finite away from zero, so the integral is well defined. The drift parameter depends on the truncation: using $|y|\le 1$ instead can alter $b$ by the contribution of atoms at $\pm 1$. Here $a$ denotes Gaussian variance, not drift.

We now prove that continuity forces $K=0$, using the suggested truncation rather than merely quoting a path classification. For $\epsilon >0$, ${\lambda }_{\epsilon }=K(|y|\ge \epsilon )$ is finite. The truncated exponent is that of the Compound Poisson processwhere $N$ has rate ${\lambda }_{\epsilon }$ and the marks have law $K(dy)1_{\{|y|\ge \epsilon \}}/{\lambda }_{\epsilon }$ when the rate is positive. The remaining exponent $\psi -{\psi }_{\epsilon }$ also has an admissible Lévy–Khintchine formula triplet. Hence construct independently a Lévy process $Z^{\epsilon }$ with that exponent. The sum $Y^{\epsilon }+Z^{\epsilon }$ has the same finite-dimensional distributions as $X$, and hence the same law on càdlàg path space. Continuity is therefore an almost-sure property of the sum if it is one of $X$.

But independent compound Poisson jumps cannot cancel: conditional on the path of $Z^{\epsilon }$, its jump times are countable; the independent Poisson jump times have continuous distributions and avoid them almost surely. At each jump of $Y^{\epsilon }$, the sum has the same nonzero jump, of size at least $\epsilon$. If ${\lambda }_{\epsilon }>0$, there is positive probability of such a jump on any fixed positive-length time interval, contradicting almost-sure continuity. Thus $K(|y|\ge \epsilon )=0$ for every $\epsilon >0$, and taking $\epsilon =1/n$ gives $K=0$.

The characteristic function is now $\exp [t(ibu-a{u}^2/2)]$. If $a>0$, the process $B_t=(X_t-bt)/\sqrt{a}$ has continuous paths and independent centered normal increments of variance $t-s$, so it is standard Brownian motion on the original probability space. If $a=0$, the increments are deterministic, and continuity together with equality at all rational times gives $X_t=bt$ simultaneously for all $t$ almost surely. A Brownian motion on an enlarged space can be used in the zero-variance representation if necessary. Therefore

Standard one-dimensional Brownian motion is a real process with $B_0=0$, continuous paths almost surely, and increments $B_t-B_s\sim N(0,t-s)$ independent of the past at time $s$, for $0\le s<t$. These conditions determine its entire finite-dimensional distribution family.

The Wiener theorem asserts that such a process exists, equivalently that there is a unique Wiener measure on $C_0([0,\infty ),\mathbb{R})$ under which the coordinate process has these properties. To construct it, prescribe centered multivariate normal laws with covariance $\mathbb{E}{B}_s{B}_t=\min (s,t)$. This covariance is positive semidefinite becauseThe laws are consistent under removing coordinates, so the Kolmogorov extension theorem constructs the process. Normal increments have fourth moment $\mathbb{E}|B_t-B_s{|}^4=3|t-s{|}^2$. The Kolmogorov continuity theorem therefore supplies a continuous modification, retaining all the prescribed finite-dimensional laws and hence the independent normal increments. Apply this on every compact time interval, taking consistent versions; continuity and equality at rational times make them agree on overlaps. The resulting law on continuous-path space is unique because evaluations at rational times generate its Borel sigma-algebra. This is a proof sketch of the existence theorem, independent of the random-walk limit in Question 5.

Write $m_l=\lfloor N{t}_l\rfloor -\lfloor N{t}_{l-1}\rfloor$ and $\varphi (v)=\mathbb{E}{e}^{iv\xi }$. Since $\mathbb{E}\xi =0$ and $\mathbb{E}{\xi }^2=1$, the characteristic function expansion at zero is $\varphi (v)=1-v^2/2+o(v^2)$. Indeed, $|e^{ix}-1-ix|\le x^2/2$ for real $x$, and $(e^{iv\xi }-1-iv\xi )/v^2\to -{\xi }^2/2$. The dominated convergence theorem, with dominating variable ${\xi }^2/2$, proves the expansion after taking expectations. Initially omit the endpoint interpolation terms. The integer-time increments involve disjoint sets of independent steps, so their joint characteristic function isHere $m_l/N\to t_l-t_{l-1}$. Taking the logarithm near $1$ givesThis proves the claimed limit for the integer-time approximation.

At a fixed time $t$, the omitted endpoint term is $R_N(t)=N^{-1/2}(Nt-\lfloor Nt\rfloor ){\xi }_{\lfloor Nt\rfloor +1}$, and $\mathbb{E}|R_N(t){|}^2\le 1/N$. A fixed linear combination of the errors in the $k$ increments therefore tends to zero in $L^1$, by the triangle inequality and Cauchy-Schwarz inequality, even when two endpoint terms share a step. Since $|e^{ix}-e^{iy}|\le |x-y|$, restoring interpolation changes the joint characteristic function by a quantity tending to zero. HenceThe limit is the joint characteristic function of independent centered normal increments with the indicated variances. By Lévy continuity theorem, the increment vectors converge in distribution to Brownian increment vectors. The value vector is obtained by the fixed invertible triangular map of cumulative sums; thus the continuous mapping theorem gives convergence of $(S_{t_1}^N,\dots ,S_{t_k}^N)$ to $(B_{t_1},\dots ,B_{t_k})$. This proves every required finite-dimensional distribution limit. The finite fourth moment is not needed for this part; it will give tightness in part (b).

We prove fourth-moment tightness of polygonal random walks uniformly, including intervals shorter than a mesh cell or crossing a mesh boundary. For $s<t$, let $c_j$ be the length of the overlap of $(Ns,Nt)$ with $(j-1,j)$. The interpolated increment isPut $m_4=\mathbb{E}{\xi }^4<\infty$. Expanding the fourth power and using centered independent steps, only the terms with four equal indices or two equal pairs survive. ConsequentlyIf $m_4\le 3$, the last correction is nonpositive. If $m_4>3$, use $\sum c_j^4\le (\sum c_j^2{)}^2$. In either case the expression is at most $\max (3,m_4)N^{-2}(\sum c_j^2{)}^2$. Since $c_j^2\le c_j$,Also $S_0^N=0$ and the paths are continuous. The stated tightness criterion applies with $\gamma =4$, $\alpha =1$ and the displayed constant, so the laws ${\mu }_N$ are tight on $C_0([0,1],\mathbb{R})$ with its uniform topology. Part (a) established all the Brownian finite-dimensional distribution limits. The supplied implication from tightness plus those limits now yields the Donsker invariance principle:where $\mu$ is Wiener measure. No estimate restricted to grid endpoints would suffice by itself for the printed uniform criterion; the overlap coefficients handle every pair of times.

With the inner boundary specified in the PDF, the Ising boundary condition pins $\omega {|}_{\partial \Lambda }=\zeta$. Explicitly ${\mu }_{\Lambda }^{\zeta }$ is ${\mu }_{\Lambda }$ conditioned on those Ising spins. Its partition function sums over unpinned interior Ising spins only; edges between two pinned sites contribute a constant that cancels on normalization.

Pinning coordinates preserves the FKG lattice condition, since meet in a lattice and join in a lattice still have the same pinned values. It also gives monotonicity in the pins. For any increasing observable $f$, positive association of random variables with an unpinned Ising spin $\omega (z)$ givesBoth conditional probabilities are positive. The same inequality holds after other Ising spins have been pinned. Changing pins one at a time proves stochastic domination of the measures as $\zeta$ increases.

Now let $\Lambda \subset {\Lambda }^{\prime }$ and take an increasing observable depending on a fixed set inside the smaller box. Given the Ising spins on $\partial \Lambda$, the larger-box measure induces precisely the smaller-box interior law with those boundary values: nearest-neighbor interactions cannot see farther once this boundary is fixed. Its conditional law is therefore between the all-zero and all-one smaller-box laws. Averaging yieldsThese are bounded monotone sequences along any increasing box exhaustion.

For every finite set $S$, the event that all Ising spins in $S$ are one is increasing, so its probability converges in both boundary sequences. Probabilities of arbitrary finite zero/one patterns are finite inclusion-exclusion combinations of these all-one probabilities, and hence converge too. The limits are consistent finite-dimensional distributions; they define measures on $\{0,1{\}}^{{\mathbb{Z}}^d}$ and give weak convergence of probability measures in the product topology. Thus both extremal infinite-volume Ising measures ${\mu }^0,{\mu }^1$ exist, with ${\mu }^0$ below ${\mu }^1$ in stochastic domination. This argument proves existence even when the two limits are different.

Disjoint time strips of the Poisson random measure are independent, and its intensity is translation invariant in time. The construction and its limits therefore give independent increments and stationary increments, withFor the scaling actually printed in the original PDF,For example, $\alpha =2$, $t=1$, $u=1$ gives $e^{-4}$ rather than $e^{-1}$. Thus the requested equality of process laws is false unless $\alpha =1$.

The intended self-similarity of a Cauchy process rescales space and time in opposite directions:Indeed, for $0\le s<t$,The transformed process retains independent increments. For any ordered times, the joint characteristic function of its successive increments is therefore the same product as for $X$; cumulative summation proves equality of all finite-dimensional distributions. Both constructions have càdlàg paths, so these distributions also determine their laws on the usual path space. This derives the corrected invariance and explicitly diagnoses the false printed assertion.

A Lévy measure $K$ on the real line is a nonnegative Borel measure satisfyingIt records the intensity per unit time of jumps of a Lévy process. Under the additional small-jump first-moment condition, the uncompensated Lévy–Khintchine formula iswhere $b\in \mathbb{R}$ and ${\sigma }^2\ge 0$. The integral is absolutely convergent: near zero its integrand has absolute value at most $|u||y|$, while outside a neighborhood of zero it is at most $2$ and $K$ has finite mass. If the usual compensated Lévy–Khintchine formula uses drift $\gamma$, this convention has $b=\gamma -{\int }_{|y|\le 1}yK(dy)$.

A Lévy process starts at zero and has stationary independent increments and stochastic continuity. Its characteristic function at time $t$ is $e^{t\psi (u)}$, with $\psi (0)=0$ and $\psi$ continuous. The given time-one characteristic function is strictly positive. It therefore has the unique continuous logarithm vanishing at zero,Any other continuous exponent would differ by an integer multiple of $2\pi i$; continuity on the real line and the value at zero make that integer identically zero. This determines all increment laws. More explicitly, for $0=t_0<t_1<\cdots <t_m$,Thus it determines every finite-dimensional distribution. For the standard càdlàg version of a Lévy process, coordinates at rational times determine the path, so these laws determine the entire process law on its canonical path space as well.

To find the Lévy measure, try the symmetric density $e^{-\alpha |y|}/|y|$. Its jump exponent isDifferentiation under the integral is justified by the integrable bound $2e^{-\alpha y}$. The elementary Fourier sine transform givesSince $J_{\alpha }(0)=0$, integration in $u$ yields $J_{\alpha }(u)=-\log (1+u^2/{\alpha }^2)$. Choosing $\alpha =\sqrt{2}$ matches the prescribed exponent exactly. This is the Lévy measure of a symmetric Laplace time-one law:The candidate satisfies the Lévy measure condition because its density is of order $1/|y|$ near zero and decays exponentially at infinity; moreover ${\int }_{|y|\le 1}|y|K(dy)=2{\int }_0^1{e}^{-\sqrt{2}y}dy<\infty$. Uniqueness of the triplet in the Lévy–Khintchine formula proves that these are the required coefficients. The infinite mass near zero means infinitely many small jumps, not a Gaussian component; their total variation is finite on bounded time intervals because $\int (1\wedge |y|)K(dy)<\infty$.

Fix times $t_1,\dots ,t_d\in [0,1]$, and setThese are iid bounded centred random vectors, andTheir covariance matrix isWe prove the required central limit theorem calculation explicitly. For $u\in {\mathbb{R}}^d$, the scalar $u\cdot V_1$ is bounded, has mean zero and second moment $u^{T}Cu$. The Taylor expansion of the exponential, with its uniformly bounded cubic remainder, givesIndependence implies that the characteristic function of the sum vector is the $n$th power of this expression, so it tends to $\exp (-u^{T}Cu/2)$. This is the characteristic function of a centred multivariate normal distribution with covariance $C$. The Lévy continuity theorem states that convergence of characteristic functions to a characteristic function continuous at zero gives convergence in distribution; apply it here.

By part (ii), the corresponding Brownian bridge vector has exactly this covariance and mean. ThereforeRepeated times or the endpoint times may make $C$ singular; the characteristic-function argument still applies. This proves every finite-dimensional distribution convergence requested, the uniform case of the Finite-dimensional Brownian bridge limit of an empirical distribution process.

A real Gaussian process $(Z_t{)}_{t\in T}$ is a family of random variables such that every finite vector $(Z_{t_1},\dots ,Z_{t_d})$ has a multivariate normal distribution, allowing degenerate distributions. Equivalently every finite real linear combination is a Gaussian random variable. Its mean and covariance kernel are $m(t)=\mathbb{E}{Z}_t$ and $C(s,t)=\mathbb{E}[(Z_s-m(s))(Z_t-m(t))]$.

For any finite set of indices, these functions give the whole mean vector and covariance matrix. The characteristic function of the vector isThe standard uniqueness theorem for characteristic functions therefore determines every finite-dimensional distribution, including the singular cases. Cylinder events form a generating class for the product sigma-algebra on ${\mathbb{R}}^T$ and are closed under finite intersections, so agreement on all cylinder events gives agreement of the process laws by the uniqueness theorem for probability measures. Thus the mean and covariance determine the law on this coordinate sigma-algebra even when $T$ is uncountable. This statement does not claim uniqueness on an arbitrary finer sigma-algebra of path properties.

Under the path convention used here, a real Lévy process $(L_a{)}_{a\ge 0}$ starts at zero, has independent increments and stationary increments, is stochastically continuous, and has càdlàg paths almost surely. We verify all four requirements for $S$.

The immediate positive excursions proved in part (b) give $S_0=0$. For a deterministic $a\ge 0$, $S_a$ is an almost-surely finite stopping time, and continuity gives $X_{S_a}=a$. The Strong Markov property implies that ${\tilde{X}}_t=X_{S_a+t}-a$ is a fresh Brownian motion independent of the information up to $S_a$. For $b\ge 0$,No crossing of $a+b$ could have occurred before $S_a$. The increment thus has the same law as $S_b$ and is independent of the past. Apply this successively at a finite increasing sequence of deterministic levels to prove stationary independent increments.

For $h>0$ and $\epsilon >0$, part (b) and reflection giveHence $S_h\to 0$ in probability. Stationarity of the increments gives stochastic continuity at every level, from either side where defined.

For path regularity, on a single probability-one event the Brownian running maximum $M_t={\max }_{s\le t}{X}_s$ is continuous, nondecreasing and unbounded, andThis inverse is nondecreasing. If $a_j\downarrow a$, then $S_{a_j}\downarrow L\ge S_a$. For any $t>S_a$, some $u<t$ has $M_u>a$; for all sufficiently large $j$, $M_u>a_j$, so $S_{a_j}\le u<t$. Thus $L=S_a$, proving right continuity at every level simultaneously. On each bounded level interval it is bounded by the finite value at a larger integer level. Monotonicity then gives finite left limits. ThereforeThe Brownian first-passage subordinator is the level-indexed increasing process, rather than a process indexed by the original Brownian time.

For $H$, the inverse relations for a continuous running maximum areFor example, if $b<a$ then $S_b\le H_a$; while for $t<H_a$, $M_t<a$ and any $b\in (M_t,a)$ has $S_b>t$. Taking limits gives $S_{a-}=H_a$. Similarly $H_b\ge S_a$ for $b>a$, and a crossing of any level strictly above $a$ before a given $t>S_a$ bounds $H_b$ from above for $b$ sufficiently close to $a$. This proves $H_{a+}=S_a$. These are the non-strict inverse of a continuous nondecreasing function identities.

At the random level $A$ in part (b), $H_A<S_A=H_{A+}$, so $H$ fails right continuity almost surely. Thus it is not a Lévy process under the stated càdlàg-path convention. It does have the same finite-dimensional distributions and stochastic continuity as $S$, so definitions of a Lévy process that require only the increment and stochastic-continuity properties would call $H$ a Lévy process with a càdlàg modification. The reason for excluding the particular version $H$ here is its path regularity.

A nonstationary process has a statistical law that changes with the time origin. In the weak sense, this includes a time-dependent mean or variance, or a covariance that depends on the two times separately rather than only on their lag. A strictly stationary process requires invariance of every finite-dimensional distribution under a common time shift.

In the preceding plots, series $1$ suggests a stochastic trend or changing level, and series $4$ suggests a deterministic seasonal mean. These are the intended nonstationary examples. A plot alone cannot prove either conclusion: in particular, a stationary random-phase sinusoid can have a seasonal-looking trace and an oscillatory ACF. The interpretation of series $4$ as nonstationary concerns a seasonal mean tied to calendar time.

Three standard responses are to remove a fitted deterministic trend or seasonal mean; to use regular or seasonal differencing for an appropriate trend or seasonal component; and to stabilize a changing variance by a transformation or explicit seasonal scale model. For example, logarithms or a Box–Cox transformation can address level-dependent variance. The operation should match the source of nonstationarity; differencing a varying variance does not generally make it stationary.