Lévy process

A Lévy process starts at zero, has independent stationary increments, is stochastically continuous, and is conventionally taken to have càdlàg paths.

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.

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.

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.

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 $\sqrt{2\lambda }$.

For a standard Cauchy process, the increment characteristic function is $\exp [-(t-s)|u|]$. Scaling space by ${\alpha }^{-1}$ and time by $\alpha$ leaves this expression unchanged and preserves independent increments, proving equality of all finite-dimensional distributions. Equivalently, $\alpha X_{t/\alpha }$ has the original process law. Scaling both space and time by $\alpha$ instead changes the exponent by ${\alpha }^2$ and is not an invariance unless $\alpha =1$.

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.

With the convention $\mathbb{E}{e}^{iu{X}_t}=e^{t\psi (u)}$, this complex process is a martingale for each real $u$. Its absolute value is the deterministic finite number $e^{-t\mathrm{Re}\psi (u)}$. Conditioning on the past at time $s$ and using independent increments multiplies $e^{iu{X}_s-t\psi (u)}$ by $e^{(t-s)\psi (u)}$, giving $M_s^u$. Under the opposite exponent convention the sign of the compensating term reverses.

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.

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.

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 $u^2/2$.

A subordinator is a Lévy process with nondecreasing paths. Its nonnegative stationary increments have a Laplace transform of the formwhere $\Phi$ is called its Laplace exponent. Its standard path convention is càdlàg. Subordinators provide random clocks for subordination of a Lévy process.

For a finite-valued subordinator $T$ starting at zero, the Laplace exponent of a subordinator is the function $\Phi$ determined byIts Lévy–Khintchine formula takes the formwhere $d$ is the deterministic drift coefficient and $\nu$ is the Lévy measure. This representation separates steady motion of the clock from its positive jumps.^[ref]

For a deterministic clock $T_t=dt$, $\Phi (\lambda )=d\lambda$. For a Poisson process of rate $a$ with unit jumps, $\Phi (\lambda )=a(1-e^{-\lambda })$. For a sum of independent subordinators, the Laplace exponent of a subordinator is the sum of the individual functions, because the Laplace transforms multiply. The transform at $\lambda =0$ gives $\Phi (0)=0$.

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.

The Brownian first-passage subordinator has a countable dense set of jump levels almost surely, and is continuous at every other level. Its paths are finite nondecreasing functions, so on a bounded level interval there can be only finitely many jumps larger than any fixed positive size; a countable union proves countability. For density, suppose its path were continuous throughout an open level interval. It is strictly increasing, since the original continuous Brownian path takes different values at different passage times. Its continuous inverse on that interval would parametrize a positive-length time interval on which the Brownian path increases monotonically. This contradicts nowhere monotonicity of Brownian motion. Thus every level interval contains a jump. In particular dense discontinuities do not mean that the path is nowhere continuous: it is continuous at every level outside a countable random set.

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

The jump activity of a Lévy process describes whether it has finitely or infinitely many jumps in a bounded time interval. Its Lévy measure $\nu$ decides this: finite activity corresponds to $\nu ({\mathbb{R}}^d\setminus \{0\})<\infty$, in which case the jump component is a Compound Poisson process. Infinite mass gives infinitely many small jumps in every positive-length time interval, almost surely. For every $\epsilon >0$, however, jumps larger than $\epsilon$ still have a finite rate.

Activity and total variation are different. With zero Brownian component, sample paths have finite variation on compact time intervals exactly whenFor example, $\nu (dx)=c{x}^{-1-\alpha }dx$ on $(0,\infty )$, with $c>0$ and $0<\alpha <1$, has infinite total mass but a finite integral of $x$ near zero. It therefore permits infinite jump activity with finite variation. Conversely, Brownian motion has no jumps but infinite total variation on every positive-length time interval, almost surely.^[ref]

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

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.

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 nondecreasing compound Poisson input $U$, let $W={\sup }_{t\ge 0}(U(t)-Ct)$ and $W_{\delta }={\sup }_{k\ge 0}(U(k\delta )-Ck\delta )$. The second supremum is bounded above by the first. For each $t$, the next grid point is at most $\delta$ later, its input is at least $U(t)$, and it offers at most an extra $C\delta$ service. Hence $W\le W_{\delta }+C\delta$. When the workload is finite almost surely, these inequalities give stationary-tail brackets. If its logarithmic exponential tail rate exists, adding the bounded shift $C\delta$ does not change that rate.

Let $Y_t={\sum }_{j=1}^{N_t}{X}_j$ for a rate-$\lambda$ Poisson process and independent and identically distributed nonnegative marks independent of that process, with $\varphi (q)=\mathbb{E}{e}^{-q{X}_1}$. The Laplace transform of a nonnegative random variable for the increment over $[s,t]$ is $\exp [\lambda (t-s)(\varphi (q)-1)]$. That increment is independent of the past of $Y$, so the displayed nonnegative process has mean one and satisfies $\mathbb{E}[Z_t\mid \sigma (Y_u:u\le s)]=Z_s$. No finite first moment of the marks is needed for this martingale when $q\ge 0$.

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.

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 the unit-rate continuous-time symmetric simple random walk, the processes $n^{-\alpha }{X}_{nt}$ have three finite-dimensional regimes. For $0<\alpha <1/2$ no probability limit is possible, because the one-time characteristic functions tend to zero off the origin and are discontinuous there. At $\alpha =1/2$ the limit is standard Brownian motion. For $\alpha >1/2$ it is the identically zero Lévy process. This follows from $n(1-\cos (n^{-\alpha }\theta ))\sim ({\theta }^2/2)n^{1-2\alpha }$ and independent increments. Thus a nonzero limit uniquely determines the exponent, whereas merely asking for a Lévy limit does not.

For a strictly positive jump rate $\lambda$, a Compound Poisson process $X_t={\sum }_{j=1}^{N_t}{Y}_j$ is a martingale for its natural filtration exactly when $\mathbb{E}|Y_1|<\infty$ and $\mathbb{E}{Y}_1=0$. The marks are iid and independent of the rate-$\lambda$ Poisson process.

Necessity of mark integrability follows from the positive-probability event $N_t=1$: on that event $X_t$ has the mark law, so an infinite first absolute mark moment would make $X_t$ nonintegrable. With integrable marks, independent increments giveThe martingale condition is therefore precisely zero mark mean. For arbitrary integrable marks, $X_t-\lambda t\mathbb{E}{Y}_1$ is the compensated martingale.

Symmetric unit-sign marks give a martingale of variance $\lambda t$, whereas deterministic unit marks give $N_t$, whose compensated version is $N_t-\lambda t$. The strict-rate assumption avoids a degenerate exception: at $\lambda =0$ the process is identically zero and is a martingale regardless of a hypothetical unused mark law.

For a vector-valued Compound Poisson process, its coordinate processes are independent exactly when its Lévy measure is supported on the union of the coordinate axes. A jump can then change only one coordinate. Splitting the marked Poisson process by the axis of its mark gives independent coordinate jump processes. Conversely, independent finite-activity coordinates have no simultaneous nonzero jumps almost surely, excluding Lévy mass away from the axes.

For a positive rate and two-dimensional marks $(g_1(U),g_2(U))$, with $U$ uniform on $[0,1]$, the condition is $g_1(U)g_2(U)=0$ almost surely. If both functions are continuous, it is equivalent toA nonzero value of the continuous product would persist on an interval of positive uniform probability. Without continuity only the almost-everywhere condition is necessary.

For example, marks $((1/2-U{)}^{+},(U-1/2{)}^{+})$ give independent coordinates by thinning into the two half-intervals. Marks $(U,1-U)$ produce common jumps in both coordinates almost surely, so the coordinates are dependent even though a single scalar uniform generates the marks. The positive-rate qualification is needed: a zero-rate process has independent constant coordinates irrespective of its unused mark functions.

Every real Lévy process has a unique triplet $(a,b,K)$, with $a\in \mathbb{R}$, $b\ge 0$ and a Lévy measure $K$ satisfying $\int (1\wedge x^2)K(dx)<\infty$ such that

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 ${\sigma }^2\ge 0$.

The Lévy-Itô decomposition builds a real Lévy process from drift, continuous Gaussian noise and Poisson jumps. For the triplet $(a,b,K)$ in the Lévy-Khintchine formula with small-jump cutoff one, take independent standard Brownian motion $B$ and Poisson random measure $N$ of intensity $dsK(dx)$. ThenHere $b\ge 0$, $K$ is a Lévy measure and $\tilde{N}=N-dsK(dx)$ denotes compensation. The large-jump term has finitely many jumps on each bounded time interval because $K(|x|>1)<\infty$.

The small-jump integral is defined by limits, rather than by subtracting two potentially infinite sums. For $\epsilon >0$, let $J_t^{\epsilon }={\int }_0^t{\int }_{\epsilon <|x|\le 1}x\tilde{N}(ds,dx)$. These are square-integrable martingales. The compensated Poisson isometry givesas $0<\delta <\epsilon \downarrow 0$. Doob's L2 maximal inequality also controls the supremum on $[0,T]$, so the limit exists locally uniformly in probability, indeed in the L2 norm of that supremum. This uses exactly the small-jump second-moment condition of a Lévy measure.^[ref]

The components have independent increments; multiplying their characteristic functions recovers the Lévy-Khintchine exponent. The decomposition separates continuous fluctuations from large jumps and possibly infinitely many small jumps. A finite small-jump Second moment does not require a finite first absolute jump moment, which explains why compensation and its limiting construction are essential.

For the small-jump cutoff convention in the Lévy-Khintchine formula, let $(a,b,K)$ be the triplet of a real Lévy process. Its paths are continuous almost surely exactly when $K=0$. They are everywhere differentiable almost surely exactly when $K=0$ and $b=0$, giving the deterministic path $X_t=at$. A compound Poisson path can have a derivative between its jumps without being an everywhere differentiable path.

finite variation on compact time intervals has the different criterionUnder these conditions the uncompensated Lévy-Khintchine formula gives a drift plus an absolutely summable small-jump process and finitely many large jumps. A nonzero Brownian component precludes finite variation, while an infinite absolute small-jump sum already forces infinite variation.

Integrability requires only the large-jump first moment:The small compensated jumps are square integrable and hence integrable. When this condition holds, $\mathbb{E}{X}_t=t[a+{\int }_{|x|>1}xK(dx)]$; the small-jump compensation convention is why no small-jump mean appears.

Likewise a finite second moment is equivalent to $\int x^{2}K(dx)<\infty$, and thenThe Lévy-Itô decomposition proves these moment statements by independent Brownian, compensated-small-jump and compound-Poisson-large-jump parts. The large-jump one-event case also shows why the tail moment conditions are necessary.

If a Lévy process has mean zero and $\mathrm{Var}(X_1)={\sigma }^2<\infty$, then $\mathrm{Var}(X_t)=t{\sigma }^2$ and $X_t^2-t{\sigma }^2$ is a martingale.