= Solution
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 $1-e^{-h}\to0$. This proves <stochastic continuity>. Consequently \b[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 gives
$$
\mathbb E e^{i\theta X_t}
=\exp\left\{\frac t2(e^{i\theta}-1)+\frac t2(e^{-i\theta}-1)\right\}
=\exp\{t(\cos\theta-1)\}.
$$
Thus, in the negative-exponent convention of part (a),
$$
\boxed{\psi(\theta)=1-\cos\theta.}
$$
In the positive-exponent convention the answer is $\widetilde\psi(\theta)=\cos\theta-1$. The <Compound Poisson process> has mean zero and <variance> $t$, consistent with the expansion $\psi(\theta)=\theta^2/2+O(\theta^4)$.
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