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With finitely many nonzero marks $z_j$ and independent Poisson processes $N^j$ of rates ${\lambda }_j$, the displayed process is a Lévy process with no Gaussian component. Its Lévy measure is ${\sum }_j{\lambda }_j{\delta }_{z_j}$, its mean is $t(\gamma +{\sum }_j{\lambda }_j{z}_j)$, and its variance is $t{\sum }_j{\lambda }_j{z}_j^2$. Its paths have finitely many jumps on every bounded interval and deterministic slope $\gamma$ between jumps. Coincident mark sizes combine their rates; zero marks have no effect. The drift here uses the uncompensated finite-sum convention, which differs from the drift parameter in a compensated Lévy–Khintchine formula.

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.

The standard symmetric Cauchy process is the Lévy process with Lévy characteristic exponent $\Psi (u)=|u|$. Its time-$t$ distribution for $t>0$ is the Cauchy distribution of location zero and scale $t$. It can be constructed by subordination of a Lévy process: evaluate an independent standard Brownian motion at the Brownian first-passage subordinator, whose Laplace exponent is $\sqrt{2\lambda }$.

Use the conventionThen $\Psi$ is the characteristic exponent of the Lévy process. Some authors instead write $e^{t\psi (u)}$, with $\psi =-\Psi$. The sign convention should always be specified. The Lévy–Khintchine formula describes all possible exponents.

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.

A compound Poisson process has the formwhere $N$ is a Poisson process and the independent identically distributed marks $Y_n$ are independent of $N$. It is a Lévy process with a finite Lévy measure.

For a Lévy process, stochastic continuity gives continuity of its characteristic function in time at every fixed frequency. The elementary estimate $|{\phi }_{X_s}(u)-{\phi }_{X_t}(u)|\le |u|\delta +2\mathbb{P}(|X_s-X_t|>\delta )$ proves this by first taking $s\to t$ and then $\delta \downarrow 0$. The argument works for any stochastically continuous stochastic process and needs no finite-moment assumption.

A real Lévy process with continuous paths is Brownian motion with deterministic linear drift: $X_t=bt+\sigma W_t$. Continuity forces the jump measure in the Lévy–Khintchine formula to vanish, leaving characteristic function $\exp (t(ibu-{\sigma }^2{u}^2/2))$. If its law is also invariant under $X_t\mapsto {\lambda }^{-1}{X}_{{\lambda }^{2}t}$ for every $\lambda >0$, then $b=0$. The zero-variance case is included.

At unit jump rate, this is $X_t={\sum }_{k=1}^{N_t}{\xi }_k$, where $N$ is a rate-one Poisson process and the independent marks are uniform signs. Equivalently it is the difference of two independent rate-$1/2$ Poisson processes. It is a Lévy process with characteristic function $\exp \{t(\cos \theta -1)\}$, mean zero, variance $t$, and Lévy characteristic exponent $1-\cos \theta$ in the negative-exponent convention. A general total jump rate rescales time.

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.

Independent increments and stationary increments make a Lévy process's characteristic functions multiplicative in time. Together with continuity of Lévy characteristic functions and the value one at time zero, this forces an exponential. A continuous local complex logarithm is additive: its failure of additivity would be a continuous integer multiple of $2\pi i$, hence zero. The Cauchy functional equation and subdivision then give the exponential at every time. This construction of the characteristic exponent of a Lévy process precedes the Lévy–Khintchine formula describing the exponent's possible form.

A probability distribution is infinitely divisible if, for every positive integer $n$, it is the distribution of a sum of $n$ independent and identically distributed random variables. Their common distribution may depend on $n$. A normal distribution has this property by dividing its mean and variance by $n$; a Poisson distribution has it by dividing its rate. The unit-time law of a Lévy process is infinitely divisible because it is the sum of its increments on $n$ equal time intervals.

The Lévy measure $\nu$ of a Lévy process records the intensity of its jumps: for every measurable set $A$ bounded away from zero, $\nu (A)$ is the expected number per unit time of jumps whose sizes lie in $A$. It satisfies

For $\alpha >0$, the Lévy process with characteristic function $\mathbb{E}{e}^{iu{X}_t}=(1+u^2/{\alpha }^2{)}^{-t}$ has zero Gaussian coefficient, zero uncompensated drift, and the displayed Lévy measure. Indeed, the symmetric jump integral is $2{\int }_0^{\infty }(\cos (uy)-1)e^{-\alpha y}dy/y$. Differentiating in $u$ gives $-2u/({\alpha }^2+u^2)$, and integrating back from zero gives $-\log (1+u^2/{\alpha }^2)$. The time-one law is the Laplace distribution of rate $\alpha$. The jump activity is infinite, since $K$ has infinite mass near zero, but the jump variation is finite because $\int (1\wedge |y|)K(dy)<\infty$.

By the usual definition of a Lévy process, $X_0=0$ almost surely. The two martingale assumptions giveFor a continuous Lévy process, the jump measure in the Lévy–Khintchine formula is zero, and the process has the form of a Brownian Gaussian component plus a deterministic drift. Thus its characteristic function isThe first moment makes $m=0$, and the second moment makes ${\sigma }^2=1$. ConsequentlyEquivalently, continuity and the martingale identity $X_t^2-t$ identify the quadratic variation as $t$, so the Lévy characterization of Brownian motion gives the same conclusion. A continuous nonzero drift is excluded by the mean-zero martingale assumption.

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