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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.

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.

The coordinates of a vector-valued Compound Poisson process are independent exactly when its Lévy measure is supported on the union of the coordinate axes. For marks $(g_1(U),g_2(U))$ with $U$ uniform on $[0,1]$ and continuous $g_i$, this is equivalent tofor every $y\in [0,1]$.

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

When paths move only by jumps of size $\pm 1$, there is neither a Gaussian component nor a continuous drift. The Lévy measure is supported on $\{-1,1\}$ and has finite mass because these points are bounded away from zero. The process is therefore a Compound Poisson process, expressible as a difference of independent Poisson processeswhere the rates are ${\lambda }_{+},{\lambda }_{-}$ respectively. The martingale assumptions implyHence $\boxed{{\lambda }_{+}={\lambda }_{-}=1/2}$. The embedded jump chain is a simple symmetric random walk, and the total jump rate is one. In particular,This is the symmetric Skellam distribution, with the explicit mass functionFor $k\ge 0$, sum the joint probabilities of $N_{+}=j+k,N_{-}=j$; for $k<0$, interchange the two variables. The characteristic function is $\mathbb{E}{e}^{iu{X}_1}=\exp (\cos u-1)$, consistent with the rate-one symmetric Poisson difference process. The condition that paths move only by jumps is essential to exclude a separate continuous component.

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

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,

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 $\Psi =-\psi$.

The joint process is a two-dimensional Compound Poisson process whose Lévy measure isTwo coordinates of a Lévy process are independent exactly when its Lévy measure charges only the coordinate axes and its Gaussian covariance has no cross term. Here there is no Gaussian part, so independence is equivalent toSince $g_1{g}_2$ is continuous, this is equivalent to pointwise vanishing. Conversely, when the product vanishes, the mark sets where $g_1$ and $g_2$ are nonzero are disjoint; independent thinning of the Poisson random measure gives independent coordinate processes. Hence

For a Poisson random measure on $E$ with finite intensity $\nu (E)$, its integral of a finite-valued measurable mark function is the finite sum of the marks at its random atoms. Its characteristic function is $\mathbb{E}\exp (iu\int fdN)=\exp (\int (e^{iuf}-1)d\nu )$. On a time-mark space with intensity $ds\nu (dz)$ and finite mark intensity, restriction to each bounded time interval has finite intensity and defines a Compound Poisson process. Zero-valued marks do not affect the integral and need not appear in its Lévy measure.

If a Lévy measure satisfies ${\int }_{|y|\le 1}|y|K(dy)<\infty$, the jump integral in the Lévy–Khintchine formula can be written without compensation: $\mathbb{E}{e}^{iu{X}_t}=e^{t\psi (u)}$, with the displayed exponent. Its integral is absolutely convergent because $|e^{iuy}-1|\le |u||y|$ near zero and is at most $2$ elsewhere. If the compensated formula uses drift $\gamma$, the uncompensated drift is $b=\gamma -{\int }_{|y|\le 1}yK(dy)$. The Gaussian coefficient is still ${\sigma }^2\ge 0$.