Lévy–Khintchine representation

Remarks

The measure $\nu$ is the Lévy measure and $(b, \sigma^2, \nu)$ is the characteristic triple; the value of $b$ depends on the choice of the cut-off $\mathbf{1}_{\set{\abs{x} < 1}}$. The three parts are a drift, a Gaussian part with variance $\sigma^2$, and a jump part. Proof idea for "only if": write $\mu$ as the $n$-fold convolution power of some $\mu_n$. Then $\exp(n(\varphi_n - 1))$ is the characteristic function of a compound Poisson law, and it tends to $\varphi$. The triple is obtained as a limit of the jump measures $n \mu_n$. Example: the normal law with mean $m$ and variance $\sigma^2$ has the triple $(m, \sigma^2, 0)$. Example: the Poisson law with mean $\lambda$ has the triple $(0, 0, \lambda \delta_1)$, since the cut-off term vanishes at $x = 1$ and $\psi(u) = \lambda(e^{iu} - 1)$. For a Lévy process $(X_t)_{t \ge 0}$, the law of $X_1$ is infinitely divisible with exponent $\psi$, and $E[e^{iuX_t}] = e^{t \psi(u)}$ for all $t \ge 0$. The same theorem holds on $\R^d$, with $\sigma^2$ replaced by a positive semidefinite matrix and $ux$ by a dot product.

Going deeper

For a student

The theorem classifies infinitely divisible laws, and through them all Lévy processes. The Lévy–Itô decomposition writes a Lévy process as a Brownian motion with drift plus an independent jump process. The jump process is built from a Poisson random measure with intensity $\nu$. Stable laws with index $\alpha \in (0, 2)$ are the case $\sigma = 0$ and $\nu(dx) = (c_+ \mathbf{1}_{\set{x > 0}} + c_- \mathbf{1}_{\set{x < 0}}) \abs{x}^{-1-\alpha} \, dx$ with $c_+, c_- \ge 0$ not both $0$. The condition $\int \min(1, x^2) \, d\nu < \infty$ allows infinitely many small jumps, as long as they are square-summable.