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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

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

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.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.

A real Lévy process is a real-valued stochastic process $(X_t{)}_{t\ge 0}$ with the following properties:

One convention also includes càdlàg paths in the definition. Equivalently, under the intrinsic definition above one chooses the càdlàg modification, which exists for such a stochastic process. Thus the usual working version of a Lévy process has right-continuous paths with left limits. There is no assumption of finite moments or continuous paths.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.

A real Lévy process starts at zero with probability one, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Stationarity means $X_{t+s}-X_s$ has the law of $X_t$; independence means increments over disjoint ordered time intervals are independent. Stochastic continuity means $X_s\to X_t$ in probability as $s\to t$. One can equivalently impose starting at zero, stationary independent increments and stochastic continuity first, and then take a càdlàg modification.

A Poisson random measure with sigma-finite intensity $\nu$ on a measurable space $(E,\mathcal{E})$ is a countably additive integer-valued random measure $N$ such that $N(A)$ has Poisson distribution with parameter $\nu (A)$ whenever $\nu (A)<\infty$, and counts on disjoint measurable sets are independent. A set of infinite intensity has infinite count with probability one. For jump processes the space is often time times a mark space, with intensity $dt\nu (dz)$. These conditions specify both the marginal count laws and their joint independence.

Let $Z=X+Y$, where the two Lévy processes are independent as processes. For any disjoint ordered time intervals, the increment vectors of $X$ and $Y$ are independent of one another, and each vector has independent coordinates. Thus the pairs of corresponding increments are independent across intervals, and so are their sums. The law of each summed increment is the convolution of the two increment laws, depending only on the interval length. This proves independent increments and stationary increments for $Z$.

Also $Z_0=0$, and for every $\epsilon >0$,Thus $Z$ is stochastically continuous. The sum of two càdlàg functions is càdlàg. All defining properties hold, so $X+Y$ is a Lévy process. Independence of the entire two processes, not merely equality of some one-time laws, is what supplies independent summed increments.

For $t>0$, use the continuous logarithm of the characteristic function, normalized to zero at $u=0$, to writeThus the deterministic drift is $+c$. In particular the last contribution is not the negative drift of a compensated unit-jump process.

Take a Poisson random measure $N$ on $(0,\infty )\times \{+,-,p\}$ with intensityDefine the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the Poisson stochastic integral with finite intensitywhere the three counts are independent Poisson processes of rates $a/2,a/2,c$. Their characteristic functions multiply towhich is exactly the given expression. The constructed process is a Lévy process, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.

Equivalently, using its nonzero-jump measure, the Lévy measure isand the unmarked representation is $X_t=ct+{\int }_{(0,t]\times (\mathbb{R}\setminus \{0\})}zN(ds,dz)$ with intensity $ds\nu (dz)$. The finite-jump case of the Lévy–Itô decomposition realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.

The atomic compound Poisson process with drift has càdlàg paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the Lévy measure; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,

First, convergence in probability at $t=0$ gives $X_0=0$ almost surely, since every $X_0^n=0$. For any finite set of times, the corresponding vectors converge in probability: the union bound controls the probability that any coordinate differs by more than a fixed tolerance. The same holds for their increment vectors, hence also for their convergence in distribution.

For $0=t_0<t_1<\cdots <t_m$, set $D_j^n=X_{t_j}^n-X_{t_{j-1}}^n$ and $D_j=X_{t_j}-X_{t_{j-1}}$. The characteristic function of a random vector factors for the independent increments of each $X^n$:Pass to the limit using convergence in distribution and bounded continuous functions. The resulting factorization of the characteristic function of a random vector, with the uniqueness theorem for characteristic functions, proves independence of the $D_j$. Similarly $X_{s+t}^n-X_s^n\stackrel{d}{=}{X}_t^n$ passes to the limit, proving stationary increments for $X$.

It remains to prove stochastic continuity. For $\epsilon >0$, the triangle inequality and the union bound imply, for every fixed $n$,The second term vanishes by stochastic continuity of $X^n$. Now let $n\to \infty$ and use the additional near-zero approximation hypothesis. We obtain $X_t\to 0$ in probability as $t\downarrow 0$. The stationary increments transfer this to every time: both $X_{s+h}-X_s$ and $X_s-X_{s-h}$ have the law of $X_h$ for $h>0$ when defined. Thus $X$ has all the intrinsic Lévy process properties, proving closure of Lévy processes under locally controlled convergence in probability.

If one requires the supplied process itself to be càdlàg, the hypotheses justify a càdlàg modification, rather than that stronger pathwise assertion. To see the distinction, take $X^n\equiv 0$, let $U$ have uniform distribution on $(0,1)$, and put $X_t=1_{\{t=U\}}$. At each fixed time $t$, $X_t=0$ almost surely, so both approximation hypotheses hold with zero error probability. Nevertheless every path has an isolated spike at $U$ and is not right-continuous there. Its identically zero modification of a stochastic process is a Lévy process with càdlàg paths. The conclusion is exact under the intrinsic definition, and exact up to modification under the convention requiring càdlàg paths.

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 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.

The printed uniqueness claim omits nondegeneracy of the limit. The zero process is a Lévy process, and every $\alpha >1/2$ yields that limit. We first classify all possibilities under the stated assumptions, then identify the unique nonzero one.

For $t>0$, part (b) givesfor each fixed $\theta \ne 0$. If $\alpha <1/2$, the characteristic functions tend to zero for every nonzero parameter and remain one at zero. This discontinuous function cannot be the characteristic function of a probability law. The characteristic-function convergence theorem therefore rules out even one-dimensional weak convergence of random variables at a positive time.

For joint convergence, take $0=t_0<t_1<\cdots <t_m$ and write $q_k={\sum }_{j=k}^m{\theta }_j$. Independent increments giveAt $\alpha =1/2$, this tends to $\exp \{-\frac{1}{2}{\sum }_k(t_k-t_{k-1})q_k^2\}$. These are exactly the joint laws of standard Brownian motion: increments are independent centred normals with variances $t_k-t_{k-1}$. The assumed limiting Lévy process thus has the Brownian law.

For $\alpha >1/2$, the same joint characteristic functions tend to one. Every limiting vector is zero. The càdlàg path convention then makes the limiting process identically zero almost surely: first intersect the zero-value events at rational times, then use right continuity. The complete scaling classification of a continuous-time symmetric simple random walk isIn particular, if a nonzero limiting Lévy process is intended, the unique exponent is $\alpha =1/2$, andThe finite-dimensional Brownian law, together with the Lévy càdlàg convention, determines the process law and its continuous version. No assertion of path-space convergence of the approximating walks is needed here.