Lévy Processes and Characteristic Exponents

From Increment Laws to Characteristic Exponents

Lévy processes extend Brownian motion and Poisson counting processes within one framework: their increments are independent, their increment laws depend only on elapsed time, and the process is continuous in probability. These properties turn the characteristic functions of the increments into a multiplicative family. The corresponding exponential formula is a compact way to encode all one-time distributions and, through independent increments, all finite-dimensional distributions.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. The definitions below do not require a Lévy process to be driven by, or independent of, this Brownian motion. Write $\mathbb{E}$ for expectation under $\mathbb{P}$. No stochastic integrals are used in this tutorial.

Definition: A real-valued process $X=(X_t)_{t\ge0}$ is a Lévy process if $X_0=0$ almost surely, it has paths that are right-continuous with left limits almost surely, it has stationary independent increments, and it is stochastically continuous. Independent increments means that increments over any finite collection of disjoint time intervals are independent. Stationary increments means that the law of $X_{s+t}-X_s$ depends only on $t$, for every $s,t\ge0$. Stochastic continuity means that for every $s\ge0$ and $\varepsilon>0$, $\mathbb{P}(|X_t-X_s|>\varepsilon)\to0$ as $t\to s$, with $t\ge0$.

For $u\in\mathbb{R}$, the characteristic function of $X_t$ is $\phi_t(u)=\mathbb{E}[e^{iuX_t}]$, where $i^2=-1$. It always exists because the integrand has absolute value one. The increment assumptions imply $\phi_{s+t}(u)=\phi_s(u)\phi_t(u)$: write $X_{s+t}=X_s+(X_{s+t}-X_s)$, and use the independence of the two terms and stationarity of the increment. Stochastic continuity ensures that $t\mapsto\phi_t(u)$ is continuous.

Theorem: For each Lévy process $X$, there is a unique function $\Psi:\mathbb{R}\to\mathbb{C}$, called its characteristic exponent, such that $$ \mathbb{E}[e^{iuX_t}]=\phi_t(u)=e^{t\Psi(u)},\qquad t\ge0,\ u\in\mathbb{R}. $$ It satisfies $\Psi(0)=0$. The exponent need not be real: its real part controls the modulus of the characteristic function, and its imaginary part records a phase.

Proof. Fix $u$ and set $f(t)=\phi_t(u)$. Then $f(0)=1$, $f$ is continuous, and $f(s+t)=f(s)f(t)$. First, $f(t)\ne0$ for every $t$: by continuity at zero, $f(t/n)$ is nonzero for all sufficiently large integers $n$, and $f(t)=f(t/n)^n$.

By continuity again, choose $\delta>0$ so that, for $0\le r\le\delta$, $f(r)$ lies in a neighborhood of $1$ on which the continuous logarithm $L(r)$ is defined and $|L(r)|<1/4$. If $s,t,s+t\le\delta$, then $e^{L(s+t)-L(s)-L(t)}=1$. The expression in the exponent has absolute value less than $3/4$, so it cannot be a nonzero integer multiple of $2\pi i$. Thus $L(s+t)=L(s)+L(t)$. Continuity and this local additivity imply $L(r)=cr$ on $[0,\delta]$ for some $c\in\mathbb{C}$: for rational subdivisions this follows by repeated addition, and for every real $r$ it follows by continuity and approximation by such subdivisions. For arbitrary $t\ge0$, choose $n$ with $t/n\le\delta$. Multiplicativity gives $f(t)=f(t/n)^n=e^{ct}$. The value of $c$ is unique, since two such values would satisfy $e^{(c-c')t}=1$ for every $t\ge0$, which, by continuity at $t=0$, forces $c=c'$. Apply this argument for each $u$ to define $\Psi(u)$. At $u=0$, $\phi_t(0)=1$, so $\Psi(0)=0$. $\square$

In the usual terminology, $\Psi$ is continuous in $u$. One way to see this is to use joint continuity of $\phi_t(u)$ and the local logarithm argument above: near each fixed $u$, the same sufficiently small positive time can be used to express $\Psi$ as a continuous logarithm divided by that time. We will use the exponent formula directly; identifying its explicit form for general Lévy processes requires a further representation theorem.

Brownian Motion: A Continuous Example

Worked Example: Drifted Brownian Motion

Let $X_t=at+\sigma B_t$, where $a,\sigma\in\mathbb{R}$ are constants. Identify its Lévy-process properties and calculate its exponent.

For $0\le s<t$, the increment is $X_t-X_s=a(t-s)+\sigma(B_t-B_s)$. Brownian increments on disjoint intervals are independent, and $B_t-B_s$ has law $N(0,t-s)$, so these increments are independent and their laws depend only on interval lengths. For stochastic continuity, the increment has mean $a(t-s)$ and variance $\sigma^2(t-s)$; both tend to zero as $t-s\to0$. Chebyshev's inequality, with the deterministic mean tending to zero as well, gives convergence in probability. The paths are continuous and $X_0=0$.

The characteristic function of a normal variable gives $$ \phi_t(u)=e^{iuat}\,e^{-\sigma^2u^2t/2} =\exp\left(t\left(iua-\frac{\sigma^2u^2}{2}\right)\right). $$ Therefore $\Psi(u)=iua-\tfrac12\sigma^2u^2$.

Check: At $u=0$, the exponent is $0$ and the characteristic function is $1$. The mean and variance computed from $at+\sigma B_t$ are $at$ and $\sigma^2t$, respectively, consistent with the drift and Gaussian spread in the displayed characteristic function. If $\sigma=0$, the process is the deterministic drift $at$, and the formula reduces to $\phi_t(u)=e^{iuat}$.

Compound Poisson Processes: Jumps and a Zero-Jump Term

A compound Poisson process has the form $X_t=\sum_{k=1}^{N_t}Y_k$, where $N_t$ is a Poisson counting process of rate $\lambda>0$ and the independent, identically distributed jump sizes $Y_k$ are independent of $N$. The sum is defined to be zero when $N_t=0$. The Poisson process has stationary independent increments; independent jump sizes then give stationary independent increments for $X$. Stochastic continuity follows because the probability of at least one jump in an interval of length $h$ is $1-e^{-\lambda h}$, which tends to zero as $h\downarrow0$. Its paths are right-continuous with left limits.

Worked Example: A Two-Point Compound Poisson Law

Let $N_t$ have rate $\lambda=\tfrac32$, and let each jump equal $-1$ with probability $\tfrac13$ and $2$ with probability $\tfrac23$. Find the characteristic exponent and evaluate the characteristic function at $u=\pi$.

Condition on $N_t=n$. Independence of the jump sizes gives $\mathbb{E}[e^{iuX_t}\mid N_t=n]=(\mathbb{E}[e^{iuY_1}])^n$, where $\mathbb{E}[e^{iuY_1}]=\tfrac13e^{-iu}+\tfrac23e^{2iu}$. Averaging over the Poisson probabilities yields $$ \begin{aligned} \phi_t(u) &=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{(\lambda t)^n}{n!} \left(\mathbb{E}[e^{iuY_1}]\right)^n\\ &=\exp\left(\lambda t\left(\mathbb{E}[e^{iuY_1}]-1\right)\right). \end{aligned} $$ Thus $\Psi(u)=\tfrac32\left(\tfrac13e^{-iu}+\tfrac23e^{2iu}-1\right)$.

At $u=\pi$, $e^{-i\pi}=-1$ and $e^{2i\pi}=1$. Hence $\mathbb{E}[e^{i\pi Y_1}]=-\tfrac13+\tfrac23=\tfrac13$, so $\Psi(\pi)=\tfrac32(\tfrac13-1)=-1$ and $\phi_t(\pi)=e^{-t}$.

Check: The jump mean is $\mathbb{E}[Y_1]=(-1)(\tfrac13)+2(\tfrac23)=1$; the jump second moment is $\mathbb{E}[Y_1^2]=1(\tfrac13)+4(\tfrac23)=3$. Conditioning on the Poisson count gives $\mathbb{E}[X_t]=\lambda t\,\mathbb{E}[Y_1]=\tfrac32t$ and $\operatorname{Var}(X_t)=\lambda t\,\mathbb{E}[Y_1^2]=\tfrac92t$. These checks also show that a negative jump is permitted in a compound Poisson Lévy process.

Subordinators: Lévy Processes That Only Increase

Definition: A subordinator is a Lévy process whose paths are nondecreasing almost surely. Nondecreasing means $X_s\le X_t$ for all $0\le s\le t$ on an event of probability one. In particular, a subordinator starts at zero and has nonnegative increments. A compound Poisson process with nonnegative jumps is a subordinator; adding a deterministic drift $dt$, with $d\ge0$, preserves this property.

Worked Example: Drift Plus Positive Poisson Jumps

Let $X_t=\tfrac12t+2N_t$, where $N_t$ is a Poisson process of rate $\lambda=\tfrac14$. Calculate the exponent, the mean and variance, and the characteristic function at $u=\tfrac{\pi}{2}$, $t=2$.

The paths increase at rate $\tfrac12$ between jumps and increase by $2$ at each Poisson jump, so $X$ is a subordinator. Conditioning on $N_t$, or using the compound Poisson computation, gives $$ \phi_t(u)=\exp\left(t\left(\frac{iu}{2}+\frac14(e^{2iu}-1)\right)\right), \qquad \Psi(u)=\frac{iu}{2}+\frac14(e^{2iu}-1). $$ The mean is $\tfrac12t+2\mathbb{E}[N_t]=\tfrac12t+2(\tfrac14t)=t$. The deterministic drift has zero variance, and $2N_t$ has variance $4\operatorname{Var}(N_t)=4(\tfrac14t)=t$.

For $u=\tfrac{\pi}{2}$, $e^{2iu}=e^{i\pi}=-1$. Therefore $\Psi(\tfrac{\pi}{2})=\tfrac{i\pi}{4}+\tfrac14(-2)=-\tfrac12+\tfrac{i\pi}{4}$. At $t=2$, $\phi_2(\tfrac{\pi}{2})=\exp(-1+i\pi/2)=i e^{-1}$.

Check: Directly, $X_2=1+2N_2$, with $N_2$ Poisson of mean $\tfrac12$. Its mean is $1+2(\tfrac12)=2$, and its variance is $4(\tfrac12)=2$, agreeing with the formulas at $t=2$. Also, conditioning on $N_2$ gives $\mathbb{E}[e^{i\pi X_2/2}]=e^{i\pi/2}\mathbb{E}[(-1)^{N_2}] =i\exp(\tfrac12(-1-1))=ie^{-1}$.

A Useful Rule for Independent Sums

Proposition: If $X$ and $Y$ are independent Lévy processes, then $X+Y$ is a Lévy process and its characteristic exponent is $\Psi_X+\Psi_Y$.

Proof. Both processes start at zero and have right-continuous paths with left limits, so their sum does too. For any finite collection of disjoint time intervals, the increments of $X$ are mutually independent, and the increments of $Y$ are mutually independent. Independence of the two processes makes the two increment collections independent of one another. Pairing increments on each interval therefore gives independent increments for the sum. Stationarity follows because the law of each paired increment depends only on the interval length. For stochastic continuity, if $h\to0$, then \[ \mathbb{P}(|(X_t+Y_t)-(X_s+Y_s)|>\varepsilon) \le \mathbb{P}(|X_t-X_s|>\varepsilon/2) +\mathbb{P}(|Y_t-Y_s|>\varepsilon/2), \] and both terms tend to zero. Finally, independence gives $\mathbb{E}[e^{iu(X_t+Y_t)}]=\phi_t^X(u)\phi_t^Y(u) =e^{t\Psi_X(u)}e^{t\Psi_Y(u)} =e^{t(\Psi_X(u)+\Psi_Y(u))}$. Uniqueness of the exponent proves the claim. $\square$

This addition rule is useful when a process is built from independent components. For example, independent Brownian motion and compound Poisson motion have exponents that add. It does not say that every Lévy process can be represented as such a sum; the general structure of exponents is addressed by the Lévy–Khintchine formula.

Common Pitfall: Forgetting the Zero-Jump Contribution

A frequent error is to write the compound Poisson exponent as $\lambda(\mathbb{E}[e^{iuY_1}])$, omitting the $-1$. This incorrect expression would give $\phi_t(0)=e^{\lambda t}$, even though every characteristic function must equal $1$ at zero. The correct exponent is $\lambda(\mathbb{E}[e^{iuY_1}]-1)$. The subtraction accounts for the probability of zero jumps as well as the positive jump counts; it is not an optional centering convention. In particular, the compound Poisson process need not have mean zero.

Check Your Understanding

Use the definitions and exponent formulas above to answer the following.

For $X_t=3t-2B_t$, find the characteristic exponent and check its value at $u=0$.
A compound Poisson process has rate $\lambda$ and fixed jump size $c$. Derive its exponent and state when it is a subordinator.
Explain why stochastic continuity of a compound Poisson process follows from the probability of at least one jump in a short interval.
If two independent Lévy processes have exponents $\Psi_1$ and $\Psi_2$, what is the exponent of their sum, and which independence assumption is needed?
For the process $X_t=\tfrac12t+2N_t$ in the worked example, calculate $\mathbb{E}[X_4]$ and $\operatorname{Var}(X_4)$.