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When paths move only by jumps of size $\pm 1$, there is neither a Gaussian component nor a continuous drift. The Lévy measure is supported on $\{-1,1\}$ and has finite mass because these points are bounded away from zero. The process is therefore a Compound Poisson process, expressible as a difference of independent Poisson processeswhere the rates are ${\lambda }_{+},{\lambda }_{-}$ respectively. The martingale assumptions implyHence $\boxed{{\lambda }_{+}={\lambda }_{-}=1/2}$. The embedded jump chain is a simple symmetric random walk, and the total jump rate is one. In particular,This is the symmetric Skellam distribution, with the explicit mass functionFor $k\ge 0$, sum the joint probabilities of $N_{+}=j+k,N_{-}=j$; for $k<0$, interchange the two variables. The characteristic function is $\mathbb{E}{e}^{iu{X}_1}=\exp (\cos u-1)$, consistent with the rate-one symmetric Poisson difference process. The condition that paths move only by jumps is essential to exclude a separate continuous component.

For the exit law, the two identified processes have the same argument. Starting at zero, neither can skip $-a$ or $b$: Brownian motion is continuous, and the symmetric Poisson difference process has unit jumps with integer boundary levels. Therefore $|X_{t\wedge T}|\le K=\max (a,b)$, and if $T$ is finite, $X_T\in \{-a,b\}$.

To justify finiteness rather than presume it, apply the optional stopping theorem to the martingale $X_t^2-t$ at the bounded stopping time $t\wedge T$:The first boundary hitting time is a stopping time in both cases: in the continuous case use the continuous-path hitting criterion, and in the jump case inspect the successive jump times. Monotone convergence now gives $\mathbb{E}T\le K^2$, so $T<\infty$ almost surely. The stopped position tends to $X_T$ as $t\to \infty$ and is bounded, making dominated convergence applicable. Stopping $X$ at $t\wedge T$ and taking this limit yields $\mathbb{E}{X}_T=0$.

If $\pi =\mathbb{P}(X_T=-a)$, then $0=-a\pi +b(1-\pi )$. Thus in both cases the entire distribution isThis is the same balanced gambler's ruin law despite the different one-time distributions. As a further consistency check, passing to the limit in the stopped square-martingale identity givesAll uses of an unbounded exit time were obtained by limits of bounded-time sampling with an explicit bound on the stopped position.

Let $X_t=N_t^{+}-N_t^{-}$, where $N^{+}$ and $N^{-}$ are independent Poisson processes of rate $\lambda$. This symmetric Poisson difference process is centered, has jumps $\pm 1$, and has variance rate ${\sigma }^2=2\lambda$. If $a,b$ are positive integers, then $X_T\in \{-a,b\}$ exactly. Optional sampling of the martingale $X_t$ givesConsequentlyPart (d) now gives the mean exit time

The distribution of $N_{+}-N_{-}$ for independent Poisson random variables with means ${\lambda }_{+},{\lambda }_{-}$ is the Skellam distribution. Its characteristic function is $\exp ({\lambda }_{+}(e^{iu}-1)+{\lambda }_{-}(e^{-iu}-1))$, its mean is ${\lambda }_{+}-{\lambda }_{-}$ and its variance is ${\lambda }_{+}+{\lambda }_{-}$. For equal means $\lambda >0$,This is the fixed-time law of a symmetric Poisson difference process.

// Target: markov-process.bigb