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 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

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 $W\le W_{\delta }+C\delta$. This supplies explicit stationary-tail brackets and shows that time discretization preserves the workload's exponential tail 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.

A compound Poisson process with rate $\lambda$ and mark $Y$ is a martingale exactly when $\mathbb{E}|Y|<\infty$ and $\mathbb{E}Y=0$.

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 $y\in [0,1]$.

Every real Lévy process has a unique triplet $(a,b,K)$ with $b\ge 0$ and $\int (1\wedge x^2)K(dx)<\infty$ 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 ${\sigma }^2\ge 0$.

Given a Brownian motion $B$ and an independent Poisson random measure $N$ with intensity $dtK(dx)$, a Lévy process with triplet $(a,b,K)$ is

For the convention in the Lévy–Khintchine formula, a Lévy process has almost surely differentiable paths exactly when $b=0$ and $K=0$, and continuous paths exactly when $K=0$. It is integrable exactly when ${\int }_{|x|>1}|x|K(dx)<\infty$, and it has a finite second moment exactly when $\int x^{2}K(dx)<\infty$.

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.