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For a discrete-time Markov chain and a set $A$ of absorbing states, its absorption time is the hitting time $T_A=\inf \{m\ge 0:X_m\in A\}$. It is zero when the initial state already lies in $A$, and infinite on a sample path that never enters $A$. Finiteness almost surely need not imply a finite expected value. For a finite-state model, a uniform positive probability of absorption within a fixed number of steps gives a geometric series bound on the survival probabilities and proves that the mean is finite. The mean absorption time of a lazy symmetric random walk is an explicit example.

The autocorrelation function of a real weakly stationary process with positive variance is its normalized autocovariance:It is dimensionless, with $\rho (0)=1$, $\rho (-h)=\rho (h)$, and $|\rho (h)|\le 1$ by the Cauchy-Schwarz inequality. It is unchanged by replacing $X_t$ with $a{X}_t+b$ for fixed $a\ne 0$. This makes autocorrelation useful when comparing records measured on different scales. The ACF is a property of the model; the sample autocorrelation function estimates it and will generally not reproduce theoretical zeros exactly.

An AR(1) with coefficient $\varphi =0.7$ has autocorrelations $0.7,0.49,0.343$ at lags one through three. With $\varphi =-0.7$ these become $-0.7,0.49,-0.343$: the same diminishing magnitude, alternating in sign. An MA(1) with $\theta =0.7$ instead has lag-one autocorrelation $0.7/1.49=70/149\simeq 0.4698$ and zero at every higher lag. Equal numerical coefficients in an autoregressive model and a moving-average model do not mean equal dependence. See Bartlett, lecture 3: theoretical ACFs and prediction.

The bars show exact model autocorrelations, not estimates from the illustrated sample paths in time series. The AR(1) retains a diminishing trace of a disturbance at arbitrarily large lags; the MA(1) stops sharing driving noise after lag one. A zero autocorrelation means uncorrelated random variables, and does not generally imply independent random variables. For a Gaussian process it does imply independence of the two coordinates.

There is a direct prediction interpretation. Among affine predictions based on just $X_t$, the best linear predictor of $X_{t+h}$ is $\mu +\rho (h)(X_t-\mu )$, with mean squared prediction error $\gamma (0)[1-\rho (h{)}^2]$. Thus negative autocorrelation is also useful predictive information. A prediction using the whole observed past can do better; best linear prediction from a finite past describes that problem. Interpreting the affine prediction as a conditional expectation requires additional assumptions, such as a joint normal distribution.

A biased random walk on the integer line takes unit steps right and left with unequal probabilities $p$ and $1-p$, using independent and identically distributed random variables as increments. For $S_0=0$, the expected value is $n(2p-1)$: positive for $p>1/2$ and negative for $p<1/2$. Changing $p$ to $1-p$ reverses the preferred direction. The exact endpoint probability mass function is given under simple random walk on the integer line.

For example, with $p=0.65$ each step has expected value $0.30$ and variance $0.91$. After $80$ steps,The displacement of the expected value from zero is about $24/8.532\approx 2.813$ standard deviations, yet individual sample paths still move left as well as right, as the figure under simple random walk illustrates. Bias lets this elementary stochastic process model an average directional tendency while retaining fluctuations. Boundary and return questions have their own treatments under gambler's ruin and the specialized children below.

Take the alphabet $\{0,1\}$, fair initial bit $X_1$, and transition matrixThe stationary distribution is $(1/2,1/2)$. Every emitted symbol is a fair bit, while the next symbol flips the previous one with probability $q$. For $q<1/2$, repeating the previous symbol is the better one-step prediction; for $q>1/2$, predicting its opposite is better. The resulting minimum next-symbol error probability is $\min (q,1-q)$, compared with $1/2$ when no history is used.

Given either state, the next-symbol information entropy is $h_2(q)$. ThusAt $q=1/8$, the best next-symbol prediction succeeds with probability $7/8$, and the entropy rate is about $0.54356444$ bits per symbol. The adjacent symbols share mutual information $I(X_1;X_2)=1-h_2(q)\simeq 0.45643556$ bits. This is useful structure even though the single-symbol counts reveal no bias.

At $q=0$, the random initial bit is repeated forever; at $q=1$, it alternates forever. Both have $H(X_1,\dots ,X_n)=1$ and zero entropy rate. At $q=1/2$, the symbols are independent and identically distributed fair bits, giving entropy rate one. Zero rate therefore need not mean that the initial choice was deterministic.

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 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.

Consider the chosen finite minimization problemOnly adjacent integers are neighbors. At each interior state propose left or right with probability $1/2$ each; at an endpoint propose its sole neighbor with probability $1/2$ and itself with probability $1/2$. This proposal distribution is symmetric, including at the endpoints. The local minimum at $1$ is relative to these allowed moves; $4$ is the unique global minimum.

At $1$, improving-only local search rejects both neighbors because $E(0)=2$ and $E(2)=4$ exceed $E(1)=1$. For simulated annealing, choose $T_{\ast }=1/\ln 2$, using the natural logarithm. Conditional on proposing $2$ from $1$, the acceptance probability isThe one-step transition probability is instead $P_{T_{\ast }}(1,2)=(1/2)(1/8)=1/16$. Both subsequent rightward proposals are downhill, so the conditional probability of the precise next three states being $2,3,4$, given $X_0=1$ and fixed $T_{\ast }$, isThis is one possible escape route, not the probability of eventually reaching $4$. Rejections, excursions toward $0$, and repeated attempts give many other sample paths. The calculation establishes possibility, without claiming a general performance advantage from a selected sample path.

For comparison, halving $T_{\ast }$ makes the uphill acceptance probability $1/64$; doubling it makes that probability $1/(2\sqrt{2})$. Cooling makes it harder to leave this local minimum, even though the stationary distribution increasingly favors the global minimum. That tension is the central scheduling problem.

Take a birth-death chain on $\{0,1,2,3,4\}$. At each interior state, move up one state with probability $p=1/2$ and down one state with probability $q=1/2$. The endpoints are absorbing states: $P_{00}=P_{44}=1$. This is a small gambler's ruin model in which the state can represent capital, and reaching $4$ is success.

Let $T_a=\inf \{n\ge 0:X_n=a\}$ be the hitting time of endpoint $a$, and let $T=\min (T_0,T_4)$ be the first endpoint hitting time. The desired hitting probability is $h_i={\mathbb{P}}_i(T_4<T_0)$. First-step analysis and the Markov property giveThe interior equation says adjacent differences are equal. Four equal increments must bridge the boundary values $0$ and $1$, soFor example, from state $2$ the hitting probability is $1/2$, and its one-step average is $\frac{1}{2}(1/4)+\frac{1}{2}(3/4)=1/2$. Individual sample paths can wander back and forth, but the same averaging rule already accounts for all of them.

Now ask how long it takes to reach either endpoint. The expected hitting time $t_i={\mathbb{E}}_{i}T$ counts steps, including the transition that first reaches an endpoint. First-step analysis givesThis is a discrete Poisson equation, $(I-P)t=1$ in the interior. The extra $1$ counts the step just taken. The quadratic valuessatisfy the equations; from state $2$, $t_2=1+\frac{1}{2}(3)+\frac{1}{2}(3)=4$ steps. The hitting probability is harmonic, whereas the expected hitting time has a nonzero source term. These are special cases of expected duration of symmetric gambler's ruin; see Hao Wu's MIT lecture 16, gambler's ruin, slide 9.

For a biased random walk with $p=0.6$ upward and $q=0.4$ downward, keep the same absorbing states. The interior harmonic equation becomes $h_i=p{h}_{i+1}+q{h}_{i-1}$. Using biased gambler's ruin probability with these boundary values givesThe three interior hitting probabilities are approximately $0.415385$, $0.692308$, and $0.876923$. Thus starting at $2$ now succeeds about $69.23\%$ of the time. A direct check is $0.6(57/65)+0.4(27/65)=9/13$. The curved profile reflects the unequal transition weights: its successive increments shrink by the factor $q/p=2/3$. The formula, with the endpoint denoted $T$ there, is verified in Karger and Lynch's MIT lecture 26, sections 2.3-2.4; the five-state numbers here are direct substitutions.

Why does harmonic averaging produce the correct hitting probability? With the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, the Markov property gives the conditional expectation identityFor a function harmonic everywhere, this makes $h(X_n)$ a martingale whenever it is integrable. For a function harmonic only in the interior, $M_n=h(X_{n\wedge T})$ is a stopped martingale. In either five-state example, $0\le M_n\le 1$. Also, from any interior state at most three consecutive upward steps reach $4$, with probability at least $p^3>0$. Repeated three-step blocks give ${\mathbb{P}}_i(T>3k)\le (1-p^3{)}^k$, so $T$ is almost surely finite and has finite expected value; compare finite-interval exit times have geometric tails. Applying optional stopping at $n\wedge T$ and then the bounded convergence theorem givesThe boundedness and finite hitting time justify this passage to the limit. See Steven Lalley's notes, harmonic functions and stopped expectations, theorem 1, and stopped martingale in discrete time for this bounded version of optional stopping.

Uniqueness has a short maximum principle explanation. On a finite state space, the difference of two harmonic extensions with the same boundary conditions is zero on the boundary. If its positive maximum occurs inside, the weighted average forces every successor with positive transition probability to have the same maximum. A path to the boundary propagates that maximum to zero, a contradiction. Applying the argument to the negative of the difference proves uniqueness. Reachability of the boundary is essential: an interior closed communicating class can retain an arbitrary constant.

// Target: markov-process.bigb

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.

A harmonic function assigns a value to each state of a Markov chain such that one transition leaves its expected value unchanged. This turns a question about many possible future sample paths, such as which endpoint is reached first, into a local averaging rule. For a transition matrix $P$, a real-valued function $h$ is harmonic at state $i$ whenHere ${\mathbb{E}}_i$ means expected value conditional on starting at $X_0=i$; on a countable state space the sum must be absolutely convergent. If this holds at every state, write $Ph=h$. At an absorbing state the equation is automatic, so prescribed boundary conditions supply information that the averaging rule cannot.

For a finite Markov chain, choose a boundary set $B$ that can be reached from every state, and assign boundary values $b$. If $T_B$ is its first hitting time, the harmonic extension isChoosing $b$ to be the indicator function of one target within $B$ gives a hitting probability. This is a discrete Dirichlet problem: enforce the averaging rule outside $B$ and fix the values on $B$. See Hao Wu's MIT lecture 8, harmonic extension, slide 5. The construction extends the mean-value idea of a harmonic function; for a random walk on a graph it is a discrete harmonic function.

Consistent probability distributions on finite collections of coordinates in standard Borel spaces determine a unique probability measure on the product sigma-algebra. Consistency means that discarding coordinates by marginalization agrees with the law already assigned to the smaller collection, and relabeling coordinates relabels the law. The theorem turns finite recipes into a complete stochastic process. In particular, products of one fixed probability distribution construct an independent and identically distributed sequence; putting one common random variable in every coordinate constructs the perfectly shared process above. The theorem supplies a law on the coordinate space, but does not by itself ensure continuous sample paths.

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.

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.

Fix an outer radius $R>\rho =|z_0|$. By continuity, reaching zero before radius $R$ requires first reaching every radius $\epsilon \in (0,\rho )$. The planar Brownian annulus hitting probability from (i) therefore impliesThus $\mathbb{P}({\tau }_0<{\tau }_R)=0$ for every fixed $R>\rho$.

On any continuous sample path that reaches zero in finite time, the radius has a finite maximum before that time. There is therefore some integer $m>\rho$ for which zero is reached before radius $m$. ConsequentlyEvery event on the right has probability zero, soThis proves that planar Brownian motion avoids a fixed point even though it can reach arbitrarily small positive radii.

For a submartingale $(Z_n)$ and $a<b$, let $U_N(a,b)$ count the completed upcrossings of $[a,b]$ by time $N$. The Doob upcrossing inequality givesFor a martingale $(X_n)$, a useful sharper form isHere $x^{+}=\max (x,0)$ and $x^{-}=\max (-x,0)$. To see the martingale form, use a bounded predictable process $H_k\in \{0,1\}$ that holds one unit after an observation at or below $a$ and releases it on the next observation at or above $b$. Its gain $G_N={\sum }_{k=1}^N{H}_k(X_k-X_{k-1})$ is at least $(b-a)U_N-(X_N-a{)}^{-}$: completed trades earn at least $b-a$, and any unfinished trade loses at most $(X_N-a{)}^{-}$. Since $\mathbb{E}{G}_N=0$, the bound follows. For the submartingale version, apply the same strategy to $(Z_n-a{)}^{+}$, itself a submartingale. Each purchase occurs at value zero, so its gain is at least $(b-a)U_N$. The complementary predictable strategy has nonnegative expected gain, bounding this strategy's expected gain by $\mathbb{E}(Z_N-a{)}^{+}-\mathbb{E}(Z_0-a{)}^{+}$, and in particular by the stated right side.

The almost-sure martingale convergence theorem says that if $C={\sup }_n\mathbb{E}|X_n|<\infty$, then $X_n$ converges almost surely to a finite, integrable random variable $Y$, with $\mathbb{E}|Y|\le C$. Indeed, the Doob upcrossing inequality bounds the expected total number of upcrossings of every rational interval by $(C+|a|)/(b-a)$. By monotone convergence, each such total count is finite almost surely; intersect these probability-one events over all rational $a<b$. If the lower and upper limits of a sample path differed, some rational interval strictly between them would be crossed infinitely often. Thus each remaining path has a limit in the extended real line. The Fatou lemma givesso the limit is finite almost surely and integrable. Almost-sure convergence by itself neither preserves expectations nor supplies this uniform first-moment bound.

For the loss of expectation, let $({\eta }_k)$ be independent variables with the fair Bernoulli distribution and setConditional on the past, survival doubles the value with probability $1/2$ and otherwise sets it to zero, so this is a nonnegative martingale. An infinite string of successes has probability zero. The process therefore eventually vanishes almost surely, giving yes: $X_0=1$ and $Y=0$ almost surely are possible. Its expectations remain one at every finite time; the process is not uniformly integrable.

A nonintegrable limit is also possible for a signed martingale. Take independent ${\xi }_k$ withand let $M_0=1$, $M_n=1+{\sum }_{k=1}^n{\xi }_k$. Every increment has mean zero and a finite absolute expectation, so each $M_n$ is integrable and $(M_n)$ is a martingale. Since ${\sum }_k\mathbb{P}({\xi }_k\ne 0)<\infty$, the Borel-Cantelli lemma makes only finitely many increments nonzero almost surely; hence the finite limit $Y=1+{\sum }_k{\xi }_k$ exists almost surely. To compute its absolute expectation, let $A_k$ be the event that only the $k$th increment is nonzero. Independence givesThe infinite product is positive because its tail logarithms are bounded in absolute value by a constant times the summable series ${\sum }_j{2}^{-j}$. The events $A_k$ are disjoint, and on $A_k$, $|Y|\ge 4^k-1$. ThereforeThus yes: a finite almost-sure martingale limit can have infinite absolute expectation, even with deterministic $M_0=1$. This is almost-sure martingale convergence to a nonintegrable limit; the missing hypothesis is the uniform first-moment bound. For a nonnegative martingale, in contrast, the Fatou lemma would force the limit to be integrable.