Infinite Divisibility and the Lévy-Khintchine Formula

Splitting a law into n identical pieces

Start with a question that sounds almost childish but turns out to organise a whole theory. Given a distribution mu, can I write a random variable X with law mu as a sum X = Y_1 + ... + Y_n of n independent, identically distributed pieces — and can I do this for every n at once? A law for which this is always possible is called infinitely divisible. The Gaussian passes instantly: a normal with variance sigma^2 is the sum of n independent normals each of variance sigma^2/n. The Poisson passes too: Poisson(lambda) is the sum of n independent Poisson(lambda/n). So does any Gamma law, and the Cauchy. But a uniform or a bounded random variable like a single coin flip fails — you cannot cut a coin flip into independent identical halves.

The cleanest place to test divisibility is the characteristic function phi(t) = E[e^(i t X)], because independence turns convolution of laws into a product of characteristic functions. Writing X as a sum of n iid pieces is exactly the statement that phi is an n-th power: phi(t) = (phi_n(t))^n for some characteristic function phi_n, and this must hold for every n. So infinite divisibility says phi has an n-th root that is itself a characteristic function, for all n. That single algebraic demand is surprisingly rigid, and pinning down its consequences is the whole game.

From the CLT problem to a new limit class

Why should we care about this class beyond curiosity? Because it is exactly the class of possible limits in a generalised central limit problem. Recall the ordinary CLT: sums of iid variables with finite variance, centred and scaled by sqrt(n), converge to a Gaussian. Now relax the setup to a triangular array — row n has its own row variables X_{n,1}, ..., X_{n,k_n}, independent within a row, each individually negligible (the 'uniform asymptotic negligibility' condition that no single summand dominates). The fundamental theorem of the general central limit problem says: the possible non-degenerate weak limits of such row sums are precisely the infinitely divisible laws — no more, no less.

This is the structural reason infinite divisibility deserves a formula of its own. The Gaussian is the special limit you get when the array has finite variances and the Lindeberg condition holds; the Poisson is the limit of the law of rare events, a triangular array of tiny Bernoullis. Both are infinitely divisible, and the generalized CLT for heavy tails produces the stable laws, which are infinitely divisible too. So a single inventory of infinitely divisible laws will simultaneously contain the Gaussian world, the Poisson world, and the heavy-tailed stable world. The Lévy-Khintchine formula is that inventory.

The Lévy-Khintchine formula

Here is the theorem in one breath. A probability law mu on R is infinitely divisible if and only if its characteristic function can be written as phi(t) = e^(psi(t)), where the characteristic exponent psi has the form psi(t) = i b t - (1/2) sigma^2 t^2 + integral over R of ( e^(i t x) - 1 - i t x 1_{|x| < 1} ) nu(dx). Here b in R is a drift, sigma^2 >= 0 is a Gaussian variance, and nu is a measure on R\{0} — the Lévy measure — satisfying the single integrability condition integral of min(1, x^2) nu(dx) < infinity. The triple (b, sigma^2, nu) is uniquely determined by mu, and conversely every such triple gives an infinitely divisible law. This is the Lévy-Khintchine formula.

Read the three pieces as three physical mechanisms. The term i b t is a deterministic drift — a constant velocity. The term -(1/2) sigma^2 t^2 is the characteristic exponent of a Gaussian, the continuous 'diffusion' part. The integral is the genuinely new object: it superposes infinitely many independent jumps, where nu(dx) is the rate (per unit time, once we make this a process) of jumps of size near x. The exponent psi is additive under convolution exactly because log of a product is a sum, which is why an infinitely divisible law's logarithm splits so cleanly into drift + Gaussian + jumps.

Why every piece must look like this

The cleanest way to see where the three pieces come from is to start from the easy direction and build infinitely divisible laws by hand, then argue that nothing else can occur. The building blocks are drift, Gaussian, and the compound Poisson mechanism; the formula is what you get by superposing them and taking a careful limit of the small jumps. Walking the construction makes the integrability condition feel inevitable rather than imposed.

Compound Poisson is infinitely divisible. Fix a finite jump measure nu (finite total mass). The compound Poisson law — Poisson-many iid jumps drawn from nu/nu(R) — has exponent integral of (e^(i t x) - 1) nu(dx). It is infinitely divisible because splitting the Poisson rate into n equal parts splits the law into n iid pieces. This single block already produces all 'finite-activity' jump laws.
Superpose with drift and Gaussian. Independent infinitely divisible pieces multiply their characteristic functions, hence add their exponents. So drift i b t, Gaussian -(1/2) sigma^2 t^2, and a compound Poisson exponent combine into i b t - (1/2) sigma^2 t^2 + integral of (e^(i t x) - 1) nu(dx). Every finite-activity infinitely divisible law on R has exactly this shape.
Let the small jumps proliferate — but compensate. Allow nu to have infinite mass near 0. The raw sum of infinitely many tiny jumps need not converge, so subtract their mean: replace e^(i t x) - 1 by e^(i t x) - 1 - i t x for small x. The compensator i t x is the centring that turns the small-jump sum into a convergent martingale-like limit, and integral of x^2 nu(dx) < infinity near 0 is precisely the L^2 control that makes it work.
Converse: nothing else can occur. Starting from an arbitrary infinitely divisible phi, take its continuous logarithm psi (possible because phi never vanishes for such laws), and analyse psi as the limit of the exponents of the n-th roots. A compactness/tightness argument on the approximating jump measures extracts a limiting (b, sigma^2, nu) and shows psi must have the stated form. Uniqueness then follows from the uniqueness of the characteristic function.

Reading the formula: examples and the bridge to processes

The triple (b, sigma^2, nu) is a dial you can read off for every classical law. Gaussian: nu = 0, only sigma^2 survives — pure diffusion, no jumps. Poisson(lambda): sigma^2 = 0, nu = lambda times a point mass at 1 — one jump size, finite rate. Gamma: sigma^2 = 0, nu(dx) = (c/x) e^(-x/theta) dx on x > 0 — infinitely many small positive jumps, infinite activity but finite variation. alpha-stable (0 < alpha < 2): sigma^2 = 0, nu(dx) proportional to |x|^(-1-alpha) dx — a scale-free power law of jumps, which is why these laws have infinite variance for alpha < 2 and even infinite mean for alpha <= 1.

Levy-Khintchine exponent  psi(t) = log E[e^{i t X}]:

  psi(t) = i b t  -  (1/2) sigma^2 t^2
           + INT_{R\{0}} ( e^{i t x} - 1 - i t x * 1{|x|<1} ) nu(dx)

  with   INT min(1, x^2) nu(dx) < infinity      (Levy measure)

  triple (b, sigma^2, nu):   drift / Gaussian / jumps

  Gaussian   :  nu = 0
  Poisson(L) :  sigma^2 = 0,  nu = L * delta_1
  alpha-stable: sigma^2 = 0,  nu(dx) ~ |x|^{-1-alpha} dx

The characteristic exponent and the triple, with three readings.

Why phrase everything through the exponent psi rather than phi? Because psi is additive, and additivity is the seed of a stochastic process. If we define a process by phi at time t equal to e^(t psi(t-argument)) — that is, the law at time t has characteristic function e^(t psi) — then increments over disjoint time intervals are independent and stationary, the hallmark of a Lévy process. The next guide builds exactly that process and shows its paths are càdlàg (right-continuous with left limits), the jumps coming straight from nu. Infinite divisibility of the time-1 law is precisely what lets the law be 'spread continuously over time', so a Lévy process exists for every infinitely divisible law and for no other.