How to prove difference between two independent poisson process is not a poisson process?

Question

It will come under properties of poisson process in some books. The sum of two independent poisson process can be proved as a poisson process using its memoryless property but how to prove difference between two independent poisson process is not a poisson process?

Answer 1

Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ be independent Poisson processes with intensity $\lambda$ and $\mu$, respectively. For $Z_t := X_t-Y_t$ we have

$$\begin{align*} \mathbb{P}(Z_t = -1) &\geq \mathbb{P}(X_t=0, Y_t=1)\\ &= \mathbb{P}(X_t=0) \cdot \mathbb{P}(Y_t = 1) \\ &= e^{-\lambda \cdot t} e^{-\mu \cdot t}t \, \mu > 0 \end{align*}$$ for any $t > 0$. In particular, $Z_t$ is not Poisson distributed. As @AndréNicolas pointed out, this proves that $(Z_t)_{t \geq 0}$ is not a Poisson process. Alternatively, one can show that $(Z_t)_{t \geq 0}$ has (a.s.) negative jumps and this clearly implies that $(Z_t)_{t \geq 0}$ is not a Poisson process.

Note that $(Z_t)_{t \geq 0}$ is nevertheless a Lévy process, i.e. a stochastic process with independent stationary increments and càdlàg sample paths.