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Let a probability space have two sample points, one of probability zero and one of probability one, and give it the complete power-set sigma-algebra at every time. Let $A\subseteq [0,1]$ be non-Borel. Assign the stochastic process the path $1_A(t)$ at the null sample point and the zero path at the other point. Each fixed-time variable is measurable and the paths are càdlàg almost surely, but the stochastic process is not progressively measurable: a section in the time variable of a product-measurable function must be Borel measurable at every sample point. The zero stochastic process is an indistinguishable progressively measurable version. Completing the filtration does not complete the time-product sigma-algebra or remove this distinction.

For standard Brownian motion started at zero, the strict first-passage times form a subordinator in the level parameter $a$. The Strong Markov property gives independent increments and stationary increments; the strict inverse of the continuous running maximum has càdlàg paths. The Brownian first-passage Laplace transform gives Laplace exponent $\Phi (\lambda )=\sqrt{2\lambda }$, so the process is strictly stable of index $1/2$. Choosing the non-strict hitting times preserves each fixed-level law but generally loses right continuity at random levels, as in the fixed-level versus simultaneous Brownian passage-time equality.

A càdlàg modification of a stochastic process is a modification of a stochastic process whose paths are càdlàg on one event of probability one. Equality almost surely at every separately fixed time does not imply indistinguishability of stochastic processes when the original process lacks path regularity. A stochastic process starting at zero, with independent increments, stationary increments, and stochastic continuity, has a càdlàg modification and is a Lévy process under the intrinsic definition.

If a càdlàg function has finite variation on compact intervals, its total-variation process is finite, nondecreasing, and càdlàg. Left limits follow from monotonicity. For right continuity, approximate the variation on $[t,s]$ by one finite partition. Changing its first endpoint from $t$ to $u\downarrow t$ changes the sum by at most $|X_u-X_t|$, which tends to zero. Additivity of variation then bounds $V(u)-V(t)$ by an arbitrarily small error. Its jump is the absolute jump of $X$. Without locally finite variation, right continuity can fail even as an extended-valued assertion: $X_t=t\sin (1/t)$ with $X_0=0$ has infinite variation on every interval starting at zero.

If two versions of a stochastic process both have càdlàg paths on an event of probability one, equality at every rational time holds simultaneously outside one null event. Approaching an arbitrary time by rationals from the right then gives equality at every time on the same event. Thus the stochastic processes are indistinguishable. Right continuity suffices; left limits are unnecessary for this conclusion.

Suppose Lévy processes $X^n$ converge to $X$ in probability at every fixed time, andThen $X$ starts at zero, has independent increments and stationary increments, and has stochastic continuity. Finite increment vectors inherit their independent laws by convergence in distribution; the near-zero estimate supplies continuity at zero. Hence $X$ has a càdlàg modification that is a Lévy process. The hypotheses alone cannot assert that the original version has càdlàg paths.

For independent Poisson processes with rates $r_{+},r_{-}\ge 0$, their difference is a Lévy process with characteristic function $\exp (t[r_{+}(e^{iu}-1)+r_{-}(e^{-iu}-1)])$. When the combined rate is positive, its jumps occur at rate $r_{+}+r_{-}$, with direction probabilities proportional to the two rates. Equal unit rates give $\exp (2t(\cos u-1))$ and a rate-two Compound Poisson process with Rademacher distribution jumps. Its paths are integer-valued, càdlàg, and of finite variation on compact intervals.

Let $Y$ be a Compound Poisson process with nonzero jump marks and positive rate, independent of a càdlàg process $Z$. Conditional on the path of $Z$, its discontinuities form a countable set. The Poisson jump times have continuous distributions and avoid that set almost surely. At every jump of $Y$, the sum $Y+Z$ therefore has the same nonzero jump. In particular a continuous Lévy process cannot have a positive-rate independent compound Poisson component. Applying this to each truncation of its Lévy measure proves that its jump measure is zero.

For a semimartingale with càdlàg paths $X$ and $f\in C^2$, the Itô formula uses the left-limit gradient in $\int f^{\prime }(X_{s-})d{X}_s$, the continuous quadratic variation in its second-order integral, and the jump correction ${\sum }_{s\le t}[f(X_s)-f(X_{s-})-f^{\prime }(X_{s-})\Delta X_s]$. In several dimensions use the corresponding gradient, Hessian and continuous bracket matrix. The correction is the part of each jump not captured by first-order linearization.

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

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.

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.

Take a complete, right-continuous filtration $({\mathcal{F}}_t)$, an increasing family of sigma-algebras. Adapted means that $X_t$ is ${\mathcal{F}}_t$-measurable at each $t$; càdlàg means right-continuous with left limits. A local martingale is an adapted càdlàg process $X$, with integrable initial value, for which there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $X_t^{{\tau }_n}=X_{t\wedge {\tau }_n}$ is a true martingale. Such a sequence is a localizing sequence. A true martingale is integrable at each time and satisfies $\mathbb{E}(X_t\mid {\mathcal{F}}_s)=X_s$ for $s\le t$.

The necessary and sufficient condition is that $X$ is a Class DL process:A family $\mathcal{Z}$ is uniformly integrable if its members are integrable and ${\sup }_{Z\in \mathcal{Z}}\mathbb{E}(|Z|1_{\{|Z|>K\}})\to 0$ as $K\to \infty$. A stopping time $\tau$ satisfies $\{\tau \le t\}\in {\mathcal{F}}_t$ for all $t$.

For necessity, on $[0,T]$ the optional stopping theorem gives $X_{\tau }=\mathbb{E}(X_T\mid {\mathcal{F}}_{\tau })$; conditional expectations of one integrable variable form a uniformly integrable family. For sufficiency, $X_{t\wedge {\tau }_n}\to X_t$ almost surely, and Class DL upgrades this convergence to $L^1$ on each finite horizon. Pass to the limit in $\mathbb{E}(X_{t\wedge {\tau }_n}\mid {\mathcal{F}}_s)=X_{s\wedge {\tau }_n}$ to obtain the martingale identity. Requiring uniform integrability of all times on the entire half-line would be stronger than necessary; Brownian motion is already a counterexample to that stronger requirement.

For càdlàg semimartingales, the semimartingale integration by parts formula isFor continuous processes the left limits can be replaced by the values. The bracket includes the jump products in the general formula.

For a continuous ${\mathbb{R}}^d$-valued semimartingale $X$ and $f\in C^2({\mathbb{R}}^d)$, the Itô formula isThe formula is localized when the derivatives are unbounded; continuity of $X$ makes them bounded after stopping on compact sets.

To prove the one-dimensional formula for polynomials without assuming that formula, use induction. It is immediate for constants and $X$. SupposeWrite $X=X_0+M+A$ with $M$ a continuous local martingale and $A$ finite variation. The induction hypothesis identifies the martingale part of $X^k$ as $k{X}^{k-1}\cdot M$. The stochastic-integral covariation identity, obtained by polarization from the quadratic-variation identity, givesApply semimartingale integration by parts to $X^{k}X$. Its first-order terms combine to $(k+1)X^{k}dX$, and its bracket terms combine toThis proves the induction step. Taking linear combinations proves the polynomial Itô formula from integration by parts for every polynomial $f$.

A Markov jump process relative to $({\mathcal{F}}_t)$ is an adapted càdlàg process with piecewise-constant paths and finitely many jumps on each bounded time interval, whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. For a time-homogeneous process on a countable state space, write $q(x,y)\ge 0$ for its transition rates and $q(x)={\sum }_{y\ne x}q(x,y)<\infty$. In state $x$, the holding time is exponential of rate $q(x)$ and the next state is $y$ with probability $q(x,y)/q(x)$; a zero rate makes the state absorbing. The generator is $\mathcal{A}f(x)={\sum }_{y\ne x}q(x,y)(f(y)-f(x))$. Nonexplosion is included here so that the process is defined at every finite time.

Here is a finite-jump, bounded-rate form of the Kurtz fluid limit theorem. For each $N$, let $Z^N$ be a chain on a countable state space $E_N\subseteq {\mathbb{R}}^d$ with jumps $\ell /N$ at rates $N{\beta }_{\ell }(z)$, with $\ell$ in a fixed finite subset $\mathcal{J}\subset {\mathbb{R}}^d$. Assume the nonnegative ${\beta }_{\ell }$ are uniformly bounded, admissible jumps remain in the state space, andis globally Lipschitz with constant $L$. Suppose $Z_0^N\to z_0$ in probability, and let $z$ solve $\dot{z}=F(z)$, $z(0)=z_0$. Then for every finite $T$,Bounded total jump rates ensure nonexplosion; the globally Lipschitz drift gives a unique global ODE solution. These explicit hypotheses suffice for the population example, without needing a more general local version of the theorem.

To prove the result, compensate the jumps of this density-dependent Markov jump process. The drift and the coordinate martingales satisfyHere $⟨M^{N,i}⟩$ is predictable quadratic variation, not the sum of realized squared jumps. The compensation identity follows because each possible increment $\ell /N$ has intensity $N{\beta }_{\ell }$, and its squared coordinate increment contributes ${\ell }_i^2{\beta }_{\ell }/N$ per unit time. Bounded rates and jump sizes make these martingales square-integrable.

We use the Doob L2 maximal inequality in the explicit form $\mathbb{E}{\sup }_{s\le T}|M_s{|}^2\le 4\mathbb{E}|M_T{|}^2$ for a scalar square-integrable martingale starting at zero. Applying it coordinatewise and setting $K={\sup }_z{\sum }_{\ell }|\ell {|}^2{\beta }_{\ell }(z)<\infty$ givesSubtract the ODE integral equation. Lipschitz continuity and Gronwall's inequality yieldThe first term tends to zero in probability by hypothesis, and the second by its maximal second-moment estimate. More quantitatively, for $\delta >0$,This proves the stated uniform finite-horizon fluid limit.

For the cells, division replaces one cell by two and increases population by one. At population $k\le N$, the total rate is the sum of the $k$ identical cell rates, namelyThe state $N$ is absorbing. Set $Z_t^N={\xi }_t/N$; its jump is $1/N$ with rate $N\beta (z)$, where $\beta (z)=z(1-z)$ on $[0,1]$. Extending this function by zero outside $[0,1]$ makes it bounded, nonnegative and globally Lipschitz, with Lipschitz constant one. Starting from $Z_0^N=p$, or from $\lfloor pN\rfloor /N$ when integer rounding is necessary, the limiting ODE is the logistic differential equationSeparation of variables, or differentiating $z/(1-z)$, gives its solution. The logistic cell-division process therefore satisfiesHere $K\le 1/4$, so the maximal martingale second moment is at most $T/N$. With exact initial density $p$, the preceding pathwise bound even gives $\mathbb{E}{\sup }_{s\le T}|Z_s^N-z(s){|}^2\le T{e}^{2T}/N$. With integer rounding, the bound $\mathbb{E}{\sup }_{s\le T}|Z_s^N-z(s){|}^2\le 2e^{2T}(T/N+1/N^2)$ follows from $|Z_0^N-p|\le 1/N$ and $(a+b{)}^2\le 2a^2+2b^2$. Thus the deterministic logistic curve approximates the density with fluctuations of order $N^{-1/2}$ on fixed time intervals. The convergence statement is on finite horizons; it does not interchange the limits $t\to \infty$ and $N\to \infty$.

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 $\int (1\wedge |y|)K(dy)<\infty$.

A martingale describes an observable after its predictable drift has been removed. For a Markov chain the drift is the change produced by its transition operator; for Brownian motion it is one half of the Laplacian. Requiring the resulting processes to be martingales for a sufficiently rich class of observables determines the transition law itself. The same identities, combined with optional stopping, determine hitting probabilities, expected exit times and recurrence or transience. The duplicated Q5 paragraph in the PDF is a repeated printing of this one essay request.

For a finite or countable-state Markov chain with transition matrix $P$, define $(Pf)(x)={\sum }_{y}P(x,y)f(y)$ for a bounded test function $f$. The Markov property says $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=(Pf)(X_n)$. Consequentlyis a martingale, since each increment has conditional expected value zero. Conversely, if these processes are martingales for every bounded $f$, taking the conditional expectation of an increment recovers that identity. With $f=1_{\{y\}}$ it gives $\mathbb{P}(X_{n+1}=y\mid {\mathcal{F}}_n)=P(X_n,y)$ for every state $y$. Thus, given the initial law, these identities characterize the Markov chain, not merely a few of its moments. This is the discrete-time Markov chain martingale characterization.

The drift operator is $P-I$. If $(P-I)h=0$ away from a target set, $h(X_{n\wedge T})$ is a stopped martingale, where $T$ is the first hit of that set. Under boundedness or suitable uniform integrability, the optional stopping theorem expresses $h$ in terms of boundary hitting probabilities. Similarly, if $(P-I)v=-1$ away from the target and $v=0$ on it, then $v(X_{n\wedge T})+(n\wedge T)$ is a martingale, giving $v(x)={\mathbb{E}}_{x}T$ when the terminal limit is justified.

For a concrete calculation, take a simple symmetric random walk on $\{0,1,\dots ,N\}$ stopped at $T$, its first hit of $0$ or $N$. In the interior, $h(i)=i$ has zero drift, while $v(i)=i(N-i)$ satisfiesThe stopped identity gives ${\mathbb{E}}_i(T\wedge n)=v(i)-{\mathbb{E}}_{i}v(X_{T\wedge n})\le v(i)$, proving $T$ finite with finite mean. Bounded convergence in both stopped identities now yields the gambler's ruin answersThis illustrates why the integrability justification is part of using a martingale, rather than an automatic consequence of writing an equation.

For a finite-state continuous-time Markov chain with generator matrix $Q=(q_{ij})$, the corresponding martingale problem for a continuous-time Markov chain usesThese are martingales. Conversely, suppose an adapted càdlàg finite-state process satisfies these identities. Let $P_t=e^{tQ}$, fix $T$, and set $g(s,i)=(P_{T-s}f)(i)$. Then ${\partial }_{s}g+Qg=0$. Apply deterministic integration by parts to the finite collection of indicator martingales to obtain the time-dependent Dynkin formula: $g(s,X_s)-g(0,X_0)-{\int }_0^s({\partial }_{r}g+Qg)(r,X_r)dr$ is a martingale. ThereforeThis recovers the Markov property and the entire semigroup, proving the characterization. On countable spaces one uses nonexplosion and localization to justify the analogous unbounded-rate formulas.

For $d$-dimensional Brownian motion, the Itô formula shows that, for $f\in C_c^{\infty }({\mathbb{R}}^d)$,is a true martingale, since its bounded derivatives make the integral square-integrable on each bounded horizon. This is the martingale problem for Brownian motion with generator $L=\frac{1}{2}\Delta$.

Here is why it also characterizes Brownian motion. Suppose a continuous adapted process $X$, with $X_0=x$, satisfies that martingale problem. Use smooth cutoff test functions equal to $x_i$ and $x_i{x}_j$ inside successively larger balls, and stop before leaving those balls. The coordinate tests show that $X^i-x_i$ are continuous local martingales; the product tests show that $X^i{X}^j-x_i{x}_j-{\delta }_{ij}t$ are local martingales. The Itô product rule consequently givesthe difference between the two proposed compensators is a continuous finite-variation local martingale, hence constant and zero at time zero. For each $u\in {\mathbb{R}}^d$, another application of the Itô formula gives the exponential local martingaleIts modulus is the deterministic quantity $e^{|u{|}^{2}t/2}$, so it is a true martingale on every bounded time interval. Taking its conditional expectation and dividing by its known value at time $s$ givesThe conditional characteristic function identifies a centered multivariate normal distribution of covariance $(t-s)I$, independent of the past. Together with continuity, this is the defining law of Brownian motion. This proves the Lévy characterization of multidimensional Brownian motion in the form needed for the generator characterization.

The analytic use of these martingales parallels the chain case. A harmonic function $h$ satisfies $\Delta h=0$, so it yields a stopped local martingale until the process leaves its domain of harmonicity. Bounded stopped observables permit optional stopping and solve boundary hitting problems. A solution of $\frac{1}{2}\Delta v=-1$ with zero boundary values gives an expected exit time. For example, in the ball of radius $R$, $v(x)=(R^2-|x{|}^2)/d$ gives ${\mathbb{E}}_x{T}_R=(R^2-|x{|}^2)/d$; stopping $|B_t{|}^2-dt$ first at $T_R\wedge n$ proves finiteness and justifies the limiting calculation. The Brownian local time construction in Q4 is another example: the predictable compensators of approximations to $|B_t|$ recover its increasing occupation term.

We now prove the required three-dimensional transience of Brownian motion in dimension at least three, in the strong sense $|B_t|\to \infty$ almost surely. Start first at a point $x$ with $r<|x|<R$, and let $T$ be the first exit from this annulus. The radial Laplacian in dimension three givesThus $|B_{t\wedge T}{|}^{-1}$ is a bounded martingale. The exit time is finite: stopping $|B_t{|}^2-3t$ gives $3\mathbb{E}(t\wedge T)\le R^2-|x{|}^2$. At exit, continuity puts $|B_T|$ at either $r$ or $R$. The optional stopping theorem therefore yieldsand hence the Brownian sphere-hitting probability in dimension threeAs $R\to \infty$, the events $\{{\tau }_r<{\tau }_R\}$ increase to $\{{\tau }_r<\infty \}$: a continuous path is bounded on the finite interval preceding any finite inner hit. This argument assumes no transience. ConsequentlyTo pass from a positive escape probability to almost-sure eventual escape, take Brownian motion started at zero. Let ${\sigma }_R$ be its first hit of the sphere of radius $R$. Stopping the squared-norm martingale proves $\mathbb{E}{\sigma }_R=R^2/3<\infty$. At ${\sigma }_R$, the Strong Markov property and the preceding hitting formula giveThe Strong Markov property here follows from independent Brownian increments by approximating a stopping time from above with dyadic-valued stopping times and using path continuity. Choose $R_m=2^{m}r$. The exit times ${\sigma }_{R_m}$ increase to infinity, since a continuous path is bounded on every compact time interval. The events of a return to the closed ball of radius $r$ after ${\sigma }_{R_m}$ decrease and have probabilities $2^{-m}$. Their intersection has probability zero. Thus almost surely there is some finite exit time after which the path never again enters that ball. Apply this simultaneously to all positive integer $r$. This proves the last visit to a bounded ball for three-dimensional Brownian motion conclusion and therefore

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.