Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 201 / 6

A real Lévy process is a stochastic process $(X_t{)}_{t\ge 0}$ with $X_0=0$ almost surely, independent increments, stationary increments, and stochastic continuity. The latter means $X_t-X_s\to 0$ in probability as $t\to s$. Its usual path convention is càdlàg: paths are right-continuous and have left limits. Equivalently one starts with the increment and continuity-in-probability conditions and then chooses the càdlàg version.

The Lévy characteristic exponent is specified here bySome conventions use $e^{t\tilde{\psi }(\theta )}$ instead, with $\tilde{\psi }=-\psi$. The sign convention must be fixed before computing the exponent. The zero process satisfies this definition; excluding it requires an additional nondegeneracy condition.

Both Poisson processes start at zero and have càdlàg paths. On disjoint time intervals their increments are independent, and independence of the two processes makes the difference increments independent as well. Their increment laws depend only on interval length, so $X$ has stationary increments. For a time interval of length $h$, the probability that either Poisson process has a jump is $1-e^{-h}\to 0$. This proves stochastic continuity. Consequently the difference is a Lévy process, equivalently a continuous-time symmetric simple random walk with total jump rate one and jumps $+1$ and $-1$ each chosen with probability $1/2$.

The characteristic function of a Poisson variable of mean $\lambda$ is $\exp \{\lambda (e^{i\theta }-1)\}$. Independence therefore givesThus, in the negative-exponent convention of part (a),In the positive-exponent convention the answer is $\tilde{\psi }(\theta )=\cos \theta -1$. The Compound Poisson process has mean zero and variance $t$, consistent with the expansion $\psi (\theta )={\theta }^2/2+O({\theta }^4)$.

The printed uniqueness claim omits nondegeneracy of the limit. The zero process is a Lévy process, and every $\alpha >1/2$ yields that limit. We first classify all possibilities under the stated assumptions, then identify the unique nonzero one.

For $t>0$, part (b) givesfor each fixed $\theta \ne 0$. If $\alpha <1/2$, the characteristic functions tend to zero for every nonzero parameter and remain one at zero. This discontinuous function cannot be the characteristic function of a probability law. The characteristic-function convergence theorem therefore rules out even one-dimensional weak convergence of random variables at a positive time.

For joint convergence, take $0=t_0<t_1<\cdots <t_m$ and write $q_k={\sum }_{j=k}^m{\theta }_j$. Independent increments giveAt $\alpha =1/2$, this tends to $\exp \{-\frac{1}{2}{\sum }_k(t_k-t_{k-1})q_k^2\}$. These are exactly the joint laws of standard Brownian motion: increments are independent centred normals with variances $t_k-t_{k-1}$. The assumed limiting Lévy process thus has the Brownian law.

For $\alpha >1/2$, the same joint characteristic functions tend to one. Every limiting vector is zero. The càdlàg path convention then makes the limiting process identically zero almost surely: first intersect the zero-value events at rational times, then use right continuity. The complete scaling classification of a continuous-time symmetric simple random walk isIn particular, if a nonzero limiting Lévy process is intended, the unique exponent is $\alpha =1/2$, andThe finite-dimensional Brownian law, together with the Lévy càdlàg convention, determines the process law and its continuous version. No assertion of path-space convergence of the approximating walks is needed here.