Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 / 2

A real Lévy process is a real-valued stochastic process $(X_t{)}_{t\ge 0}$ with the following properties:

One convention also includes càdlàg paths in the definition. Equivalently, under the intrinsic definition above one chooses the càdlàg modification, which exists for such a stochastic process. Thus the usual working version of a Lévy process has right-continuous paths with left limits. There is no assumption of finite moments or continuous paths.

Fix $u\in \mathbb{R}$ and let $s\to t$. For any $\delta >0$, use $|e^{ia}-e^{ib}|\le \min (2,|a-b|)$ to obtainThe probability tends to zero by stochastic continuity. Taking the limit superior and then letting $\delta \downarrow 0$ provesThis proves continuity of Lévy characteristic functions from the elementary estimate, without requiring moments or replacing convergence in probability by an unjustified almost sure limit. At $t=0$, time approaches from the right. If stochastic continuity is formulated only at zero, stationary increments give the same argument at every $t$: the absolute value of $X_s-X_t$ has the law of $|X_{|s-t|}|$. For $u=0$ the characteristic function is identically one.

For fixed $u$, write $h(t)={\phi }_{X_t}(u)$. Independent increments and stationary increments giveThe preceding part gives continuity. Also $h$ never vanishes: if $h(t_0)=0$ with $t_0>0$, then $h(t_0/m{)}^m=0$ for every positive integer $m$, contradicting $h(t_0/m)\to 1$.

Here is a direct proof of the exponential form of Lévy characteristic functions. Choose $a>0$ small enough that $|h(t)-1|<1/2$ on $[0,a]$. The principal complex logarithm gives a continuous $L(t)=\log h(t)$ there, with $L(0)=0$. For $s,t\ge 0$ with $s+t\le a$, the multiplicative identity impliesThis difference is continuous on the connected triangle of allowed $(s,t)$ and equals zero at $(0,0)$, so it is identically zero. Thus $L$ satisfies the additive Cauchy functional equation locally. Subdivision gives $L(a/m)=L(a)/m$ and $L(ka/m)=kL(a)/m$; continuity then gives $L(t)=tL(a)/a$ for every $t\in [0,a]$.

Define $\eta (u)=L(a)/a$. For any $t\ge 0$, choose an integer $m$ with $t/m\le a$; thenThe coefficient is unique: if two coefficients give the same exponential for every $t\ge 0$, their derivatives at zero agree. In particular $\eta (0)=0$, and $\mathrm{Re}\eta (u)\le 0$ follows from $|h(t)|\le 1$. The characteristic exponent of a Lévy process has therefore been obtained from first principles, without invoking the Lévy–Khintchine formula.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.