= Solution
Take two independent rate-one <Poisson processes> $N^+$ and $N^-$, and let
$$
X_t=N_t^+-N_t^-.
$$
The <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 $1-e^{-2h}\to0$, 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> gives
$$
\boxed{\mathbb E e^{iuX_t}
=\exp\bigl(t(e^{iu}-1)+t(e^{-iu}-1)\bigr)
=e^{2t(\cos u-1)}.}
$$
The <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>.
Back to article page
