The Lévy–Khintchine Formula and Triplets

From Increments to a Triplet

A Lévy process is described by the law of each increment: its increments are stationary and independent, and it is continuous in probability. The Lévy–Khintchine formula packages these increment laws into three ingredients: a drift parameter, a Gaussian variance and a measure describing jumps. The formula also specifies how small jumps are compensated, so its drift parameter must always be read together with its truncation convention.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. When a Gaussian component is present, let $B_t$ denote a standard Brownian motion on this space. A Lévy process $X_t$ starts at $X_0=0$, has stationary independent increments, and is continuous in probability. For a real random variable $Y$, its characteristic function at $u\in\mathbb{R}$ is $\mathbb{E}[e^{iuY}]$. The characteristic exponent $\psi$ of $X$ is defined by $\mathbb{E}[e^{iuX_t}]=e^{t\psi(u)}$.

In the formula below, the integral is a deterministic Lebesgue integral against a measure, not a stochastic integral. A Lévy measure $\nu$ is a measure on $\mathbb{R}\setminus\{0\}$ satisfying $\int_{\mathbb{R}\setminus\{0\}}(1\wedge x^2)\,\nu(dx)<\infty$. This condition permits infinitely many small jumps while ensuring that jumps bounded away from zero have finite measure.

Theorem: (Lévy–Khintchine formula.) Every Lévy process has a characteristic exponent of the form $$ \psi(u)=i\gamma u-\frac{1}{2}\sigma^2u^2+ \int_{\mathbb{R}\setminus\{0\}}\left(e^{iux}-1-iux\mathbf{1}_{\{|x|\le1\}}\right)\nu(dx), $$ where $\gamma\in\mathbb{R}$, $\sigma^2\ge0$, and $\nu$ is a Lévy measure. Conversely, every such choice defines the characteristic exponent of a Lévy process. For the fixed truncation $x\mathbf{1}_{\{|x|\le1\}}$, the triplet $(\gamma,\sigma^2,\nu)$ is unique. This representation, converse and uniqueness are the standard Lévy–Khintchine theorem; its proof is beyond this tutorial.

The integral in the formula is well-defined. For $|x|\le1$, Taylor's estimate gives $|e^{iux}-1-iux|\le \tfrac12u^2x^2$. For $|x|>1$, the truncation term is zero and $|e^{iux}-1|\le2$; the Lévy-measure condition ensures integrability on both regions. The indicator $\mathbf{1}_{\{|x|\le1\}}$ includes the endpoint $x=1$ and $x=-1$.

Reading the Three Components

The parameter $\gamma$ is the drift under the chosen truncation convention, $\sigma^2$ is the variance rate of the Gaussian component, and $\nu$ is the jump measure. In particular, $\nu(A)$ describes the rate of jumps with sizes in a set $A$ bounded away from zero. The subtraction of $iux$ for $|x|\le1$ centers the contribution of small jumps in the exponent; it is why $\gamma$ should not automatically be read as the mean rate of $X_t$.

For example, if $\int_{|x|>1}|x|\,\nu(dx)<\infty$, then $X_t$ has a finite mean and $\mathbb{E}[X_t]=t\left(\gamma+\int_{|x|>1}x\,\nu(dx)\right)$. The compensated small-jump contribution has mean zero; the displayed large-jump integral supplies the additional mean. This identity requires the stated large-jump first-moment condition.

The triplet depends on the truncation used to write the exponent, although the process and its characteristic exponent do not. Here the chosen truncation is $h_0(x)=x\mathbf{1}_{\{|x|\le1\}}$. A different truncation $h$, equal to $x$ near zero and bounded, writes the same exponent as $$ \psi(u)=i\gamma_hu-\frac12\sigma^2u^2+ \int_{\mathbb{R}\setminus\{0\}}\left(e^{iux}-1-iuh(x)\right)\nu(dx). $$ The integral here is again a deterministic integral against $\nu$.

Proposition: If $h$ is bounded and agrees with $x$ on some neighbourhood of zero, then the drift parameter for truncation $h$ is $$ \gamma_h=\gamma+\int_{\mathbb{R}\setminus\{0\}}\big(h(x)-h_0(x)\big)\,\nu(dx). $$ The integral is finite, and $\sigma^2$ and $\nu$ do not change.

Proof. Since $h=h_0$ in a neighbourhood of zero, their difference vanishes there. Outside that neighbourhood, the Lévy measure has finite mass: for any $\varepsilon>0$, $\nu(\{|x|\ge\varepsilon\})<\infty$, by the Lévy-measure condition. The function $h-h_0$ is bounded, so its integral is finite. Substitute the proposed $\gamma_h$ into the exponent using truncation $h$. The resulting drift and integral terms are $$ i\gamma u+i u\int(h-h_0)\,d\nu+\int(e^{iux}-1-iuh(x))\,\nu(dx) =i\gamma u+\int(e^{iux}-1-iuh_0(x))\,\nu(dx). $$ This is exactly the exponent using $h_0$. Thus the process exponent is unchanged, while the Gaussian coefficient and jump measure in the expression are the same. $\square$

Worked Triplets

Worked Example: Brownian Motion with Drift

Let $X_t=mt+sB_t$, where $m\in\mathbb{R}$, $s\ge0$, and $B_t$ is standard Brownian motion. Identify its triplet under the truncation $h_0$.

The increment $X_t-X_0=mt+sB_t$ is normal with mean $mt$ and variance $s^2t$. Its characteristic function is $$ \mathbb{E}[e^{iuX_t}]=e^{iumt-\frac12u^2s^2t} =\exp\left(t\left(imu-\frac12s^2u^2\right)\right). $$ There is no jump contribution, so $\nu=0$. Comparing with the formula gives $\gamma=m$ and $\sigma^2=s^2$, hence the triplet is $(m,s^2,0)$.

Check: substituting this triplet into the formula gives $\psi(u)=imu-\tfrac12s^2u^2$, the exponent just computed. If $s=0$, the process is deterministic, $X_t=mt$, and the same formula reduces to $\psi(u)=imu$.

Worked Example: A Compound Poisson Process with Small Jumps

Let $N_t$ be a Poisson process of rate $4$, and set $X_t=\tfrac12N_t$. Identify the triplet using the specified truncation.

For a Poisson random variable with mean $4t$, summing its probability mass function gives $$ \mathbb{E}[e^{iuX_t}] =\sum_{n=0}^{\infty}e^{iu n/2}e^{-4t}\frac{(4t)^n}{n!} =e^{-4t}\exp(4t e^{iu/2}) =\exp\left(4t(e^{iu/2}-1)\right). $$ Thus $\psi(u)=4(e^{iu/2}-1)$. A jump of size $\tfrac12$ lies in $\{|x|\le1\}$, so $\nu=4\delta_{1/2}$, where $\delta_a$ denotes unit mass at $a$. In the Lévy–Khintchine formula this measure contributes $4(e^{iu/2}-1-iu/2)$. Taking $\gamma=2$ adds $i2u$, and $$ i2u+4(e^{iu/2}-1-iu/2)=4(e^{iu/2}-1). $$ The triplet is therefore $(2,0,4\delta_{1/2})$.

Check: the mean of $X_t$ is $\tfrac12(4t)=2t$. The triplet's mean formula gives $t\left(2+\int_{|x|>1}x\,\nu(dx)\right)=2t$, since $4\delta_{1/2}$ places no mass above $1$. The nonzero drift parameter is the compensation required by the small-jump truncation, even though $X_t$ was defined as a nonnegative jump sum.

Worked Example: The Compensated Poisson Process

Let $N_t$ have rate $\tfrac32$, and let $X_t=2(N_t-\tfrac32t)$. Identify its triplet, and check its mean and variance.

The characteristic function of $2N_t$, computed from the Poisson mass function, is $\exp(\tfrac32t(e^{i2u}-1))$. The deterministic subtraction contributes the factor $e^{-i3ut}$, so $$ \mathbb{E}[e^{iuX_t}] =\exp\left(t\left(\frac32(e^{i2u}-1)-i3u\right)\right). $$ A jump has size $2$, which is outside $\{|x|\le1\}$, so the Lévy integral contributes $\tfrac32(e^{i2u}-1)$ without a truncation correction. The triplet is $(-3,0,\tfrac32\delta_2)$.

The Poisson mean and variance are both $\tfrac32t$, so $\mathbb{E}[X_t]=2(\tfrac32t)-3t=0$ and $\operatorname{Var}(X_t)=2^2(\tfrac32t)=6t$. The triplet mean formula also gives $-3t+t\int_{|x|>1}x\,\nu(dx)=-3t+2(\tfrac32)t=0$. The variance check agrees with the jump-size-squared rate $2^2(\tfrac32)=6$.

Why the Truncation Convention Matters

The drift parameter is not, by itself, a description of the visible linear trend in every jump process. With the fixed truncation, jumps of size at most $1$ are centered inside the integral, while jumps larger than $1$ are not. Changing the truncation shifts the drift to compensate. The Gaussian variance and jump measure remain the same under the change described above.

Common mistake. For a Poisson process $N_t$ of rate $2$ with unit jumps, setting $\gamma=0$ because the process has no separately written continuous drift gives the exponent $$ 2(e^{iu}-1-iu), $$ since unit jumps satisfy $|x|\le1$. This is the exponent of $N_t-2t$, not $N_t$: the missing $i2u$ term changes its mean from $2t$ to $0$. The correct triplet for $N_t$ is $(2,0,2\delta_1)$, because $$ i2u+2(e^{iu}-1-iu)=2(e^{iu}-1). $$ The correction is to match the entire characteristic exponent, not to infer $\gamma$ from the process's informal description.

The triplet is useful because it separates three sources of randomness and makes independent components easy to combine. Independent Lévy processes have exponents that add; correspondingly, their drift and Gaussian variance add, and their Lévy measures add. The triplet also records which compensation convention is being used, an essential detail when moving from characteristic exponents to pathwise descriptions of jumps.

Check Your Understanding

Use the fixed truncation $x\mathbf{1}_{\{|x|\le1\}}$ unless a question states otherwise.

What triplet corresponds to $X_t=3t+2B_t$? Verify it by writing its characteristic exponent.
For a compound Poisson process with rate $5$ and jump size $\tfrac14$, find $\nu$ and $\gamma$, then check the resulting exponent.
Why does a unit-jump Poisson process require a nonzero drift parameter under the stated truncation?
State the Lévy-measure integrability condition and explain what it implies about jumps bounded away from zero.
If a new truncation agrees with $x$ near zero, which triplet components can change, and by what formula does the drift change?

The Lévy–Khintchine Formula and Triplets

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