Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 26 / 6

A real Lévy process starts at zero with probability one, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Stationarity means $X_{t+s}-X_s$ has the law of $X_t$; independence means increments over disjoint ordered time intervals are independent. Stochastic continuity means $X_s\to X_t$ in probability as $s\to t$. One can equivalently impose starting at zero, stationary independent increments and stochastic continuity first, and then take a càdlàg modification.

A Poisson random measure with sigma-finite intensity $\nu$ on a measurable space $(E,\mathcal{E})$ is a countably additive integer-valued random measure $N$ such that $N(A)$ has Poisson distribution with parameter $\nu (A)$ whenever $\nu (A)<\infty$, and counts on disjoint measurable sets are independent. A set of infinite intensity has infinite count with probability one. For jump processes the space is often time times a mark space, with intensity $dt\nu (dz)$. These conditions specify both the marginal count laws and their joint independence.

Let $Z=X+Y$, where the two Lévy processes are independent as processes. For any disjoint ordered time intervals, the increment vectors of $X$ and $Y$ are independent of one another, and each vector has independent coordinates. Thus the pairs of corresponding increments are independent across intervals, and so are their sums. The law of each summed increment is the convolution of the two increment laws, depending only on the interval length. This proves independent increments and stationary increments for $Z$.

Also $Z_0=0$, and for every $\epsilon >0$,Thus $Z$ is stochastically continuous. The sum of two càdlàg functions is càdlàg. All defining properties hold, so $X+Y$ is a Lévy process. Independence of the entire two processes, not merely equality of some one-time laws, is what supplies independent summed increments.

For $t>0$, use the continuous logarithm of the characteristic function, normalized to zero at $u=0$, to writeThus the deterministic drift is $+c$. In particular the last contribution is not the negative drift of a compensated unit-jump process.

Take a Poisson random measure $N$ on $(0,\infty )\times \{+,-,p\}$ with intensityDefine the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the Poisson stochastic integral with finite intensitywhere the three counts are independent Poisson processes of rates $a/2,a/2,c$. Their characteristic functions multiply towhich is exactly the given expression. The constructed process is a Lévy process, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.

Equivalently, using its nonzero-jump measure, the Lévy measure isand the unmarked representation is $X_t=ct+{\int }_{(0,t]\times (\mathbb{R}\setminus \{0\})}zN(ds,dz)$ with intensity $ds\nu (dz)$. The finite-jump case of the Lévy–Itô decomposition realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.

The atomic compound Poisson process with drift has càdlàg paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the Lévy measure; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,