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Standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ on a probability space with three defining properties: $B_0=0$ almost surely; its sample paths $t\mapsto B_t$ are continuous almost surely; and its Brownian increments over any finite collection of disjoint time intervals are independent, withHere $N(m,v)$ denotes the normal distribution with expectation $m$ and variance $v$. The alternative symbol $W_t$ denotes the same kind of stochastic process. Continuity is a simultaneous statement about an entire sample path, rather than a separate assertion at each fixed time. Relative to its natural Brownian filtration, a future Brownian increment is independent of all observations up to its starting time.

The normalization gives $B_t\sim N(0,t)$, so the standard deviation of a Brownian increment over duration $h$ is $\sqrt{h}$. For $h=1$, $0.01$, and $0.0001$, the corresponding standard deviations are $1$, $0.1$, and $0.01$. Time becoming one hundred times shorter makes typical displacement ten times smaller. Brownian scaling expresses this rule for the whole stochastic process, while the Brownian covariance kernel describes how positions at different times remain related despite independent increments.

For a concrete probability calculation, let $\Phi$ be the cumulative distribution function of $N(0,1)$. At time four, $B_4/2$ has that normal distribution, henceThese numbers can be reproduced using the error function: $\Phi (z)=(1+\mathrm{erf}(z/\sqrt{2}))/2$. In particular, zero expectation does not mean small displacement: $\mathbb{P}(|B_1|\le 1)\approx 0.6826894921$, but the same fixed interval contains only about $38.3\%$ of the probability at time four.

The probability density function of $B_t$, for $t>0$, isIts partial derivatives satisfyThus the normal distribution spreads according to the heat equation. Starting at a fixed position $x_0$ replaces $x$ by $x-x_0$. This is a useful bridge between individual random trajectories and the deterministic evolution of their probability density function: the same Brownian motion provides both descriptions.

Physical diffusion uses its own scale. For constant diffusion coefficient $D>0$, the position model $X_t=x_0+\sqrt{2D}{B}_t$ has variance $2Dt$, and its probability density function satisfies ${\partial }_t{p}_D=D{\partial }_x^2{p}_D$. For example, $D=0.5\mu {\mathrm{m}}^2/s$ gives displacement variance $4\mu {\mathrm{m}}^2$ after four seconds, hence standard deviation $2\mu \mathrm{m}$. Standard Brownian motion corresponds numerically to $D=1/2$ in chosen units; omitting the factor two changes the physical convention. Einstein's 1905 paper relates diffusion to the mean squared displacement in precisely this convention. ^[ref]

Continuity does not supply a smooth physical velocity. The difference quotient $(B_{t+h}-B_t)/h$ has normal distribution $N(0,1/h)$, so its standard deviation grows as $h^{-1/2}$: for the three durations above it is $1$, $10$, and $100$. This calculation motivates the nowhere differentiability of Brownian motion, but it alone is not a proof of the simultaneous assertion at every time. The theorem says that, with probability one, no finite derivative exists anywhere along the sample path. A drawn polygon only displays a finite grid; its straight segments cannot establish this theorem. See Theorem 7.1.6 in Probability: Theory and Examples (fifth edition) by Rick Durrett.

For physical particles, Brownian motion is a model of position on suitable observation scales. It does not claim that microscopic velocity is a derivative of this idealized path. Einstein explicitly restricts his assumption of independent increments to intervals that are not too short. ^[ref]

Another way to measure the roughness is quadratic variation. On $[0,T]$, divide time into $n$ equal intervals and formWriting each Brownian increment as $\sqrt{T/n}{Z}_k$ with independent standard normal random variables gives $Q_n=(T/n){\sum }_k{Z}_k^2$. Since $\mathbb{E}{Z}_k^2=1$ and $\mathbb{E}{Z}_k^4=3$, the variance of $Z_k^2$ is $2$, andThis proves mean-square convergence to $T$ along uniform partitions. For $T=2$ and $n=100$, the expectation is $2$, the variance is $0.08$, and the root mean square of the error is $\sqrt{0.08}\approx 0.2828427125$. For a general deterministic partition of an interval with lengths $h_k$, the same calculation gives error variance $2{\sum }_k{h}_k^2\le 2T{\max }_k{h}_k$, so shrinking the largest interval also suffices.

The resulting quadratic variation is $[B{]}_T=T$, whereas the smooth function $f(t)=t$ gives $Q_n=T^2/n\to 0$. This difference explains why the ordinary rules of calculus need the second-order correction in the Itô formula. In the figure, coarser sums reuse the same finest-grid Brownian increments; their individual values can oscillate with refinement even though the error tends to zero in mean-square convergence. Refining a fixed interpolating polygon forever would instead give zero quadratic variation, so the finite-grid illustration must not be confused with the limiting Brownian motion.

Historically, Brown's 1828 report describes microscopic observations made in 1827, starting with particles released from pollen and extending to inorganic materials; this challenged an explanation based solely on living matter. His observations and the later mathematical model are distinct contributions. ^[ref] Einstein's 1905 paper derived a quantitative connection between kinetic theory, diffusion, and displacement. ^[ref] Wiener's 1923 Differential-Space developed integration on spaces of paths, a foundation for Wiener measure; it also acknowledges earlier work and his own previous papers. These dates mark documented contributions rather than an assertion that each subject had a single discoverer. ^[ref]

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 Brownian motion and deterministic partitions of an interval $[0,t]$ whose mesh $|\pi |$ tends to zero, the displayed limit holds in L2 space. Split the error into $U_{\pi }={\sum }_j{B}_{t_j}((\Delta B_j{)}^2-\Delta t_j)$ and the Riemann sum error $V_{\pi }={\sum }_j{B}_{t_j}\Delta t_j-{\int }_0^t{B}_{s}ds$. The summands of $U_{\pi }$ are martingale differences, hence orthogonal in L2 space, and the Gaussian fourth moment givesThe Cauchy-Schwarz inequality and the Brownian increment variance give $\mathbb{E}{V}_{\pi }^2\le (t/2){\sum }_j(\Delta t_j{)}^2\le t^2|\pi |/2$. Both vanish. This proves the limit directly without a stochastic change-of-variables formula.

For deterministic square-integrable $h$, approximate by step functions. Their Itô integrals are sums of independent normal Brownian increments. The Itô isometry gives convergence in $L^2$, and the variances converge to $\int h^2$; convergence of characteristic functions proves the normal limit. Each linear combination of finitely many such integrals is another deterministic stochastic integral, so they are jointly Gaussian.

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.

The Lévy construction builds Brownian motion on $[0,1]$ by successive independent midpoint displacements on dyadic intervals. Start with $B_0=0$ and $B_1=Z_0$, where $Z_0$ has the standard normal distribution. On each dyadic interval $[u,v]$, choose a fresh independent standard normal random variable $Z$ and use the displayed midpoint rule. Conditional on the endpoints, this is the correct Brownian bridge midpoint law.

To verify the increment law, if $D=B_v-B_u\sim N(0,\ell )$ with $\ell =v-u$, the new increments are $D/2\pm \sqrt{\ell }Z/2$. They are Jointly Gaussian, each has variance $\ell /2$, and their covariance is zero, hence they are independent. Induction gives all dyadic Brownian increments. The change between consecutive linear interpolants has uniform norm $2^{-n/2-1}{\max }_{k<2^n}|Z_{n,k}|$. The Gaussian tail bound and union bound bound the chance that this maximum exceeds $n$ by $2^{n+1}{e}^{-n^2/2}$. The First Borel-Cantelli lemma and ${\sum }_{n}n{2}^{-n/2}<\infty$ prove uniform convergence almost surely. The continuous limit has the required Brownian increments at arbitrary times by dyadic approximation.

An Ornstein-Uhlenbeck process combines mean reversion with continuing random fluctuations. In the scalar model, its stochastic differential equation iswhere $W$ is a standard Brownian motion. The drift points towards $\mu$: it is negative above $\mu$ and positive below it. The coefficient $\theta$ sets the speed of mean reversion, while $\sigma$ sets the strength of the random forcing. We consider a fixed or normal initial value independent of future Brownian increments, giving a Gaussian process and a Markov process.

For a stipulated example, use $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$ and the fixed initial value $X_0=6$, in arbitrary consistent state and time units. Then the expected value relaxes as $2+4e^{-t}$ and the variance grows as $1-e^{-2t}$. At time one they are about $3.4715$ and $0.8647$; the limiting variance is one. The sample paths keep fluctuating even after the expected value is close to two. The Ornstein-Uhlenbeck transition distribution describes this evolving uncertainty; Ornstein-Uhlenbeck stationary initialization instead starts with that limiting normal distribution already in place.

A physical motivation is the velocity of a particle undergoing Brownian motion: linear drag removes a persistent initial velocity, while random forcing continually renews it. The original calculation gives a normal distribution for the velocity after subtracting its exponentially decaying initial contribution, with decay rate equal to the friction coefficient divided by the particle's mass.^[ref] A position coordinate in an overdamped harmonic optical trap gives another interpretation of the same scalar equation, with $\theta =k/\zeta$ for spring constant $k$ and drag coefficient $\zeta$. Its physical assumptions and connection to measurements are developed in thermal calibration of an optical trap.

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.

In the Lévy construction of Brownian motion, start with $B_0=0$ and $B_1=Z_0$, where $Z_0$ has the standard normal distribution. Suppose values are defined at the endpoints $u=k{2}^{-n}$ and $v=(k+1)2^{-n}$. Setusing fresh independent standard normal random variables $Z_{n,k}$, also independent of everything previously used. Join the values by straight segments to obtain a continuous piecewise linear process $B^{(n)}$ on the dyadic grid.

To verify the Brownian increments, write $D=B_v-B_u$ and $\ell =v-u$. The two new increments are $D/2+\sqrt{\ell }{Z}_{n,k}/2$ and $D/2-\sqrt{\ell }{Z}_{n,k}/2$. If $D\sim N(0,\ell )$, their variances are $\ell /2$ and their covariance is zero. They are Jointly Gaussian, so they are independent. Induction proves the required Brownian increments on every dyadic grid.

For convergence, successive interpolants differ by tents of disjoint interiors, soThe Gaussian tail bound and the union bound giveThese probabilities are summable. The First Borel-Cantelli lemma and ${\sum }_{n}n{2}^{-n/2}<\infty$ therefore prove uniform convergence of the interpolants almost surely. Their limit has continuous sample paths. Approximating any finite set of times by dyadic times passes the Jointly Gaussian increment laws to the limit, giving a standard Brownian motion on $[0,1]$.

Equivalently, with ${\psi }_{n,k}(t)={\int }_0^t{H}_{n,k}(s)ds$, these tents give the Lévy-Ciesielski expansion

On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, a standard one-dimensional Brownian motion is a real-valued stochastic process $B=(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous sample paths almost surely, and independent increments, such thatHere the second parameter is the variance. Thus its Brownian increments have the normal distribution with mean zero and variance equal to elapsed time. Equivalently, relative to a specified filtration $({\mathcal{F}}_t)$, require $B$ to be an adapted process and $B_t-B_s$ to be independent of ${\mathcal{F}}_s$ with this normal distribution. Taking the natural Brownian filtration recovers the first definition.

Fix $t>0$, a deterministic partition of an interval $\pi$, and write $\eta (s)=t_j$ on $[t_j,t_{j+1})$. Its mesh is $|\pi |={\max }_j(t_{j+1}-t_j)$. The Brownian increment variance givesThus the left-endpoint sums ${\sum }_j{B}_{t_j}\Delta B_j$ converge in L2 space to the Itô integral. Although these step coefficients are unbounded, clipping them at finite levels first makes them bounded simple predictable processes; the Itô isometry and their finite second moments remove the clipping.

The elementary identity $B_{t_{j+1}}^2-B_{t_j}^2=2B_{t_j}\Delta B_j+(\Delta B_j{)}^2$ givesThe squared Brownian increments are independent, with means $\Delta t_j$ and variances $2(\Delta t_j{)}^2$, since a centered normal random variable of variance $v$ has fourth moment $3v^2$. ThereforePassing to the convergence in L2 limits yieldsThis holds for each fixed $t$; continuous sample paths of both sides extend equality from rational times to all $t$ on one event of probability one.

Summing the exact cubic increment identity givesThe first sum tends to zero in L2 space. Indeed, odd moments of centered normal random variables vanish, a centered normal random variable of variance $v$ has sixth moment $15v^3$, and independence of the Brownian increments givesThe second sum converges in L2 space to $3{\int }_0^t{B}_{s}ds$ by (i).

For the last sum, first note that $\mathbb{E}{\int }_0^t{B}_s^{4}ds={\int }_0^{t}3s^{2}ds=t^3$, so $B_s^2$ is a square-integrable stochastic integrand. If $s=t_j+u$ and $D=B_s-B_j$, then $D$ is independent of $B_j$, with $D\sim N(0,u)$. HenceIt follows thatThe Itô isometry therefore makes ${\sum }_j{B}_j^2\Delta B_j$ converge in L2 space to ${\int }_0^t{B}_s^{2}d{B}_s$, with bounded truncations supplying the initial simple predictable processes if needed. Taking limits in the cubic identity provesThe proof uses only increment algebra, moments of normal random variables and the definition of the Itô integral.

To establish the Brownian weighted quadratic variation in L2 for the deterministic partition of an interval, decomposewhere $U_{\pi }={\sum }_j{B}_j((\Delta B_j{)}^2-\Delta t_j)$ and $V_{\pi }={\sum }_j{B}_j\Delta t_j-{\int }_0^t{B}_{s}ds$.

The summands in $U_{\pi }$ are martingale differences: $B_j$ is ${\mathcal{F}}_{t_j}$-measurable, and the following Brownian increment is independent of ${\mathcal{F}}_{t_j}$. Their conditional means are zero, so distinct summands are orthogonal in L2 space. Since $\mathbb{E}{B}_j^2=t_j$ and $\mathbb{E}[((\Delta B_j{)}^2-\Delta t_j{)}^2]=2(\Delta t_j{)}^2$,For the Riemann sum error, the Cauchy-Schwarz inequality givesThus the triangle inequality in L2 space provesThe time integration variable is $s$.

For a positive starting value, Itô's formula applied to $\log S_u$ gives $d\log S_u=(\mu -{\sigma }^2/2)du+\sigma d{W}_u$. Hence the geometric Brownian motion isThe future Brownian increment $W_T-W_t$ is independent of ${\mathcal{F}}_t$ and has normal distribution $N(0,\tau )$. If ${\tilde{W}}_v=W_{t+v}-W_t$, then $\tilde{W}$ is a fresh Brownian motion and the increment is exactly ${\tilde{W}}_{\tau }$. This specifies the meaning of the paper's abbreviated $W_{\tau }$; the original process value $W_{T-t}$ is not generally that future increment. The stochastic exponential solution also verifies positivity for $S>0$.

For $t>0$, put $M_t={\sup }_{0\le s\le t}{B}_s$. For $\lambda >0$, reflect the Brownian path after its first hit of $\lambda$. The Strong Markov property and symmetry of subsequent Brownian increments make this reflection preserve the path distribution. It exchanges the events $\{M_t\ge \lambda ,B_t<\lambda \}$ and $\{B_t>\lambda \}$. Since $B_t$ has no atom at $\lambda$, the Brownian reflection principle givesThe case $\lambda =0$ follows by decreasing positive levels to zero: the probability is one. At $t=0$, $M_0=0$ and the probability is zero for every $\lambda \ge 0$.

The displayed tail probabilities are exactly those of $|B_t|$, so $M_t$ has the half-normal distribution of $\sqrt{t}|Z|$ with $Z\sim N(0,1)$. ThusFor $A_t={\sup }_{s\le t}|B_s|$, the pointwise inequality $A_t\ge M_t$ gives the lower non-strict bound. To prove strictness when $t>0$, fix $\lambda >0$. Reflection at $\lambda$ also givesSubtracting this from $\mathbb{P}(B_t<-\lambda )=\mathbb{P}(B_t>\lambda )$ givesOn this event $A_t\ge |B_t|>\lambda >M_t$, so $A_t^2-M_t^2$ is strictly positive with positive probability. Finally, $|B_t|$ is a continuous nonnegative submartingale by the conditional Jensen inequality, and part (a) gives $\Vert A_t{\Vert }_2\le 2\Vert B_t{\Vert }_2=2\sqrt{t}$. Its square is therefore integrable, and the strict pointwise comparison on a positive-probability event givesThis is the strict comparison of one-sided and absolute Brownian maxima. The strict lower inequality printed with $t\ge 0$ needs the qualification $t>0$: at $t=0$ both maxima vanish and all three quantities are zero.

We use dyadic quadratic variation of Brownian motion. For fixed rational $0\le u<v$, set $\ell =v-u$ and let $Q_n$ be the sum of the squares of the $2^n$ Brownian increments on its equal-length partition. Independence and the normal fourth moment giveFor every $\eta >0$, Chebyshev's inequality bounds $\mathbb{P}(|Q_n-\ell |>\eta )$ by $2{\ell }^2{\eta }^{-2}{2}^{-n}$, a summable sequence. The Borel-Cantelli lemma implies $Q_n\to \ell$ almost surely. Intersect over positive rational $\eta$ and all rational intervals to obtain one probability-one event on which every such interval has this positive quadratic-variation limit.

If a path were Hölder of exponent $\alpha >1/2$ on a nonempty open interval, choose a compact rational subinterval $[u,v]$ within it and a finite Hölder constant $C$ there. Every partition increment would have absolute value at most $C(\ell 2^{-n}{)}^{\alpha }$, soThis contradicts $Q_n\to \ell >0$. The same pathwise event excludes every exponent greater than $1/2$, without needing an uncountable intersection of events. ThereforeThis is the quadratic variation obstruction to Hölder regularity. It rules out a continuously differentiable trajectory on any interval, since a continuous derivative is bounded on smaller compact intervals and gives a Lipschitz bound. By itself it does not rule out differentiability at an isolated point. The stronger nowhere differentiability of Brownian motion can also be seen directly as follows.

On a mesh of width $h=2^{-n}$ in $[0,M]$, the probability that any specified three consecutive increments all have absolute value at most $Kh$ is at most $C{K}^3{h}^{3/2}$: each increment is $N(0,h)$ and the three are independent. There are at most $M/h$ possible blocks, so the probability of any such block is at most $CM{K}^3{h}^{1/2}$, summable in $n$. By Borel-Cantelli lemma, for every pair of positive integers $M,K$ there are eventually no such blocks. But a finite derivative at some time $t<M$ would imply $|B_s-B_t|\le C_t|s-t|$ near $t$. The three mesh increments beginning at the mesh point immediately to the left of $t$ would then all be at most $6C_{t}h$ for all sufficiently fine meshes. Choosing an integer $K\ge 6C_t$ contradicts the preceding event. Thus finite derivatives do not exist anywhere almost surely, including a finite right derivative at $0$. The Hölder estimates below $1/2$ express continuity with a roughness scale; they do not imply differentiability.

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).

Write $X_t=x+W_t+\mu t$ with $W$ a zero-starting Brownian motion. A unit-time increment exceeds $2b$ with probabilityIf the process has not exited before an integer time $k$, it is in $(-b,b)$; an increment exceeding $2b$ forces it outside at time $k+1$ and hence forces an exit by then. Independence of Brownian increments gives $\mathbb{P}(T>k)\le (1-p{)}^k$. Thus the geometric tail bound from a uniform escape probability yieldsIn particular $T$ is finite almost surely.

The diffusion generator is $\mathcal{L}=\frac{1}{2}{\partial }_{xx}+\mu {\partial }_x$. The function $f(z)=e^{-2\mu z}$ satisfies $\mathcal{L}f=0$, so the Itô formula gives $df(X_t)=-2\mu f(X_t)d{W}_t$. It is a true martingale, since it is $e^{-2\mu x}$ times the Gaussian exponential martingale $e^{-2\mu W_t-2{\mu }^{2}t}$. The stopped values $f(X_{t\wedge T})$ are uniformly bounded because the stopped path stays in $[-b,b]$. Optional sampling and bounded convergence givewhere continuity ensures that $X_T$ is exactly one of the two endpoints. Solving,This drifted Brownian interval-exit probability tends to $(b-x)/(2b)$ as $\mu \to 0$, as expected.

The printed pathwise conclusion is false without specifying the initial value. We first prove exactly what the hypotheses do imply, by stationary drift saturation.

Stationarity gives $\mathbb{E}{Z}_s^2=1$. The Itô formula yieldsThe stochastic integral is a true square-integrable martingale on every finite horizon, since $\mathbb{E}{\int }_0^t{Z}_s^{2}ds=t$. Also $\mathbb{E}|Z_s{g}_s|\le \lambda$ by the Cauchy-Schwarz inequality. Taking expectations therefore givesEvery integrand is nonnegative, so $\mathbb{E}[Z_s{g}_s]=-\lambda$ for almost every $s$. For those times,It follows that $g_s=-\lambda Z_s$ for almost every time and outcome. Equality of these integrable drift terms gives the exact solutionThus $Z$ solves the same equation, but stationarity has not identified its initial random variable with $B_1$.

For an explicit counterexample, takeThis is adapted to ${\mathcal{G}}_t$. Its independent normal initial value and Brownian noise make it a centered stationary Gaussian process with covariance $e^{-\lambda |t-s|}$, exactly as for $Y$. Hence $\mathbb{E}{Z}_0^2=1$ and $\mathbb{E}{g}_t^2={\lambda }^2$, satisfying every printed hypothesis. Nevertheless $Z_t-Y_t=-2e^{-\lambda t}{B}_1$, which is nonzero almost surely for every finite $t$.

The corrected result is indistinguishability if one adds $Z_0=Y_0$ almost surely. Without that addition, there is equality in law of the stationary Ornstein-Uhlenbeck processes, not necessarily equality of their coupled paths. To see that the initial law is forced, independence of Brownian increments from ${\mathcal{G}}_0$ and stationarity give, for its characteristic function $\phi$,Letting $t\to \infty$ proves $\phi (u)=e^{-u^2/2}$. This is the distinction captured by same-noise stationary Ornstein-Uhlenbeck processes.

After the fixed time $s$, the mapping-out function of the uncentered mapped future isIts expansion is ${\bar{g}}_t(z)=z+2t/z+O(z^{-2})$, so the elapsed time $t$ is still the half-plane-capacity parameterization. Differentiating in $t$ yieldsTherefore the Loewner transform of ${\bar{\gamma }}_t=g_s({\gamma }_{s+t})$ is ${\xi }_{s+t}$, with starting point ${\xi }_s$.

Centering gives ${\tilde{g}}_t(w)={\bar{g}}_t(w+{\xi }_s)-{\xi }_s$, whose driver is ${\tilde{\xi }}_t={\xi }_{s+t}-{\xi }_s$. By independent stationary Brownian increments, $(B_{s+t}-B_s{)}_{t\ge 0}$ is a standard Brownian motion independent of the past through time $s$. HenceThis proves the domain Markov property of a chordal Loewner chain from the composition rule for chordal Loewner driving functions. For possible contacts with the old hull, the mapped trace is understood through the continuous boundary extension or prime ends; on the open half-plane the displayed maps are ordinary conformal maps. The PDF asks first for the driver of the uncentered curve $\bar{\gamma }$; the centered driver is included as well.