Compute a probability distribution function related to Poisson process.

Question

Let $(N_{t})_{t\geq 0}$ be a standard Poisson process. I'd like to compute the probability $$\mathbb{P}\Big((-1)^{N_{t}-N_{t_{s}}}=\dfrac{m}{m_{s}}\Big),$$ with $m, m_{s}\in \{-1,1\}$.

I understand that a random variable $N$ has Poisson$(\lambda)$ if $$\mathbb{P}(N=n)=\dfrac{\lambda^{n}}{n!}e^{-\lambda},\ n=0,1,2,\cdots$$

But what should I do to apply this result?

Thank you!

Answer 1

If $\frac m {m_s}=1$ then $P((-1)^{N_t-N_{ts}}=\frac m {m_s})$ is the probability that $N_t-N_{ts}$ is even. Since $N_t-N_{ts}$ is Poisson random variable with parameter $\lambda (ts-t)$ this probability is $\sum_{n \, \text {even}} e^{-\lambda (ts-t)} \frac {(\lambda (ts-t))^{n}} {n!}$. The case when $\frac m {m_s}=-1$ is similar.