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Construct a rate-$\lambda$ Poisson process, for $\lambda >0$, from independent waiting times $E_j$ with exponential distribution $\mathrm{Exp}(\lambda )$. Put $S_k=E_1+\cdots +E_k$, $S_0=0$, and $N_t=\max \{k:S_k\le t\}$. The strong law of large numbers gives $S_k/k\to 1/\lambda$, so there are only finitely many arrivals on each finite interval. Consequently the counting paths are càdlàg, and $N_0=0$ almost surely.

At any deterministic time $s$, the memorylessness of the exponential distribution says that the residual waiting time has again exponential distribution $\mathrm{Exp}(\lambda )$ and is independent of the observed history. Subsequent waiting times are fresh independent copies. Thus the Poisson process restarts independently at $s$, proving independent increments and stationary increments. Integrating the joint waiting-time densities over $0<s_1<\cdots <s_k\le t<s_{k+1}$ givesso its time values have the expected Poisson distribution.

For $h\downarrow 0$,Together with stationary increments, this gives stochastic continuity at every time, from either side where applicable. All the Lévy process requirements hold, andFor $\lambda =0$, the identically zero process gives the degenerate case.

Use the intrinsic definition of a Lévy process: $X_0=0$ almost surely, its increments over disjoint time intervals are independent, their probability distributions depend only on interval length, and $X$ has stochastic continuity. In symbols,with times restricted to the half-line. Such a stochastic process has a càdlàg modification, and is usually represented by that version. Requiring càdlàg paths in the definition is a common equivalent convention at the level of modifications; it is important to distinguish this from a claim about the paths of an arbitrary supplied version.

For the characteristic function, writeHere $\psi (0)=0$ and $\psi$ is the positive-time-sign characteristic exponent of a Lévy process. The independent increments and stationary increments give ${\varphi }_{s+t}(\theta )={\varphi }_s(\theta ){\varphi }_t(\theta )$; stochastic continuity gives continuity in time and ${\varphi }_0=1$. This continuous multiplicative semigroup has the stated exponential form. Its exponent has the Lévy–Khintchine formulawhere $b\in \mathbb{R}$, ${\sigma }^2\ge 0$, and the Lévy measure $\nu$ satisfies $\int (1\wedge y^2)\nu (dy)<\infty$. The truncation convention fixes the drift coefficient $b$; the displayed sign convention agrees with $\Psi =-\psi$.

A standard real Brownian motion relative to a filtration $({\mathcal{F}}_t)$ is an adapted stochastic process with $B_0=0$ almost surely, continuous paths almost surely, andEquivalently, it has independent increments and stationary increments, each with a normal distribution of mean zero and variance equal to the elapsed time, together with continuous paths and the stated initial value. The filtration may be its natural filtration. A Brownian motion started at $z$ is $z+W_t$ for a standard process $W$; multidimensional standard Brownian motion has independent standard real coordinate processes. The normalization is variance $t$ at time $t$.

A real Lévy process is a stochastic process $(X_t{)}_{t\ge 0}$ with $X_0=0$ almost surely, independent increments, stationary increments, and stochastic continuity. The latter means $X_t-X_s\to 0$ in probability as $t\to s$. Its usual path convention is càdlàg: paths are right-continuous and have left limits. Equivalently one starts with the increment and continuity-in-probability conditions and then chooses the càdlàg version.

The Lévy characteristic exponent is specified here bySome conventions use $e^{t\tilde{\psi }(\theta )}$ instead, with $\tilde{\psi }=-\psi$. The sign convention must be fixed before computing the exponent. The zero process satisfies this definition; excluding it requires an additional nondegeneracy condition.

Both Poisson processes start at zero and have càdlàg paths. On disjoint time intervals their increments are independent, and independence of the two processes makes the difference increments independent as well. Their increment laws depend only on interval length, so $X$ has stationary increments. For a time interval of length $h$, the probability that either Poisson process has a jump is $1-e^{-h}\to 0$. This proves stochastic continuity. Consequently the difference is a Lévy process, equivalently a continuous-time symmetric simple random walk with total jump rate one and jumps $+1$ and $-1$ each chosen with probability $1/2$.

The characteristic function of a Poisson variable of mean $\lambda$ is $\exp \{\lambda (e^{i\theta }-1)\}$. Independence therefore givesThus, in the negative-exponent convention of part (a),In the positive-exponent convention the answer is $\tilde{\psi }(\theta )=\cos \theta -1$. The Compound Poisson process has mean zero and variance $t$, consistent with the expansion $\psi (\theta )={\theta }^2/2+O({\theta }^4)$.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.

A standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ satisfying $B_0=0$ almost surely, having continuous paths almost surely, and having independent increments withHere $N(0,t-s)$ is the normal distribution with zero expected value and variance $t-s$. Thus the increments are also stationary increments.

If Brownian motion is defined relative to a specified filtration $({\mathcal{F}}_t)$, additionally require that it is an adapted process and that $B_t-B_s$ is independent of ${\mathcal{F}}_s$. For its natural filtration, this follows from independent increments. These are the defining properties of standard Brownian motion.

The Loewner local growth property gives a continuous Loewner driving function, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The Conformal Markov property of SLE therefore gives stationary increments and independent increments.

Consequently $U$ is a continuous Lévy process. In the Lévy–Khintchine formula, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and Brownian motion:Conformal invariance under dilations gives $U_{r^{2}t}/r\stackrel{d}{=}{U}_t$. Comparing the expectations of these Gaussian random variables yields $\mu rt=\mu t$ for every $r>0$, hence $\mu =0$. The Brownian term already has the required law by Brownian scaling. Therefore the characterization of the SLE driving function isThe case $\kappa =0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.

The Conformal Markov property of SLE states that, conditional on the hull $K_t$, the image under $g_t-U_t$ of the future hull has the same law as the original hull and is independent of the past.

For a Loewner chain with continuous driver $U$, this property says that $U_{t+s}-U_t$ is independent of the past and has the same distribution as $U_s$. Thus $U$ has stationary increments and independent increments. Every continuous process with those properties is a Brownian motion with drift, so $U_t=at+\sigma B_t$. Conformal scale invariance giveswhich forces $a=0$. Writing $\kappa ={\sigma }^2$ yields

The Conformal Markov property of SLE says that, conditionally on the curve through time $t$, mapping out the initial hull by $g_t-U_t$ turns the future into an independent Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. Since $U_t=\sqrt{\kappa }{B}_t$, the stationary increments and independent increments of Brownian motion give the asserted independence and equality in law.

For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.

Part (c), iterated over disjoint nested annuli, gives independent increments, and the law $X_s-X_t\stackrel{d}{=}{X}_{s-t}$ gives stationary increments. Every finite vector is jointly Gaussian by the definition of the Zero-boundary Gaussian free field, and a continuous version was assumed. Moreover $X_0=0$, because the field has zero boundary values.

Thus $X$ is a continuous centered Gaussian process with stationary increments and independent increments. Its variance is a continuous additive function on the nonnegative real numbers, so $\mathrm{Var}(X_t)={\sigma }^{2}t$ for some ${\sigma }^2\ge 0$. The Gaussian-process characterization of Brownian motion now givesfor standard Brownian motion $B$. This is the Circle-average process of the Gaussian free field is Brownian motion.

Stationary increments give $N_5-N_3\sim \mathrm{Poisson}(2\lambda )$. ThereforeThere is zero probability of an arrival at either deterministic endpoint, so endpoint inclusion does not alter the result.

A homogeneous Poisson process of rate $\lambda >0$ is an integer-valued counting process with $N_0=0$ almost surely, nondecreasing right-continuous paths with unit jumps, and independent stationary increments satisfyingIncrements over disjoint time intervals are independent. In particular,The finite Poisson counts ensure that there is no explosion in finite time. The increment formula also gives stochastic continuity, since the probability of any arrival in an interval of length $h$ is $1-e^{-\lambda h}\to 0$.

Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.

The unit-time increments $D_j=N_j-N_{j-1}$ are independent and identically distributed random variables, by independent increments and stationary increments of the Poisson process. Each has the Poisson distribution of parameter one, hence expected value one and variance one. The central limit theorem therefore gives

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

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

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

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

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

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

If $X$ is a Lévy process and $T$ an independent subordinator, then $Y_t=X_{T_t}$ is a Lévy process. Conditional on the clock, increments of $X$ over disjoint clock intervals are independent; averaging over the independent stationary increments of the clock gives the same properties for $Y$. If $\mathbb{E}{e}^{iu{X}_s}=e^{-s\Psi (u)}$, its characteristic function is $\mathbb{E}{e}^{-T_t\Psi (u)}$. For standard Brownian motion, $\Psi (u)=u^2/2$, so this is the clock's Laplace transform at $u^2/2$.