Show that $(-1)^{N_{t}}$ has independent increments, where $(N_{t})_{t>0}$ is a standard Poisson process.

Question

I am working on an exercise in the stochastic process:

Show that $X_{t}:=(-1)^{N_{t}}$ is a Markov Process, where $(N_{t})_{t>0}$ is a standard Poisson process.

I was nearly done, by reading this post: If $N_t$ is a Poisson process and $Y\in\{-1,1\}$, then $X_t = Y(-1)^{N_t}$ is a Markov process

However, I have two questions about the proof in this post.

$(1)$ Why does the fact $(N_{t})$ has independent increments implies $(-1)^{N_{t}}$ has independent increments?

$(2)$ In one step, he used \begin{align*} \mathbb{P}\Big((-1)^{N_{t}}=m, (-1)^{N_{t_{k}}}=m_{k},\cdots&, (-1)^{N_{t_{1}}}=m_{1}\Big)\\ &=\mathbb{P}\Big((-1)^{N_{t}-N_{t_{k}}}=\frac{m}{m_{k}}, \cdots, (-1)^{N_{t_{2}}-N_{t_{1}}}=\frac{m_{2}}{m_{1}}, (-1)^{N_{t_{1}}}=m_{1}\Big) \end{align*} what is the reason behind this equality? stationary?

Thank you!

Answer 1

Observation 1: If $f: \mathbb{R} \to \mathbb{R} \backslash \{0\}$ is a measurable function and $X_1,\ldots,X_n$ any real-valued random variables, then \begin{align*} &\mathbb{P}(f(X_1)=c_1,\ldots,f(X_n)=c_n) \\ &= \mathbb{P} \left( f(X_1)=c_1, \frac{f(X_2)}{f(X_1)} =\frac{c_2}{c_1},\ldots,\frac{f(X_n)}{f(X_{n-1})} = \frac{c_n}{c_{n-1}} \right)\end{align*} for any constants $c_1,\ldots,c_n \in \mathbb{R} \backslash \{0\}$.

Applying this for $f(x)=(-1)^x$ and $X_i = N_{t_i}$ answers your second question.

For the proof of the identity, take $\omega \in A:=\bigcap_{j=1}^n \{f(X_j)=c_j\}$. Then clearly

$$\frac{f(X_j(\omega))}{f(X_{j-1}(\omega))} = \frac{c_j}{c_{j-1}}$$

for all $j=2,\ldots,n$ and so $$\omega \in B:=\{f(X_1)=c_1\} \cap \bigcap_{j=2}^n \left\{\frac{f(X_j)}{f(X_{j-1})} = \frac{c_j}{c_{j-1}} \right\}.$$ Hence, $A \subseteq B$. On the other hand, if $\omega \in B$, then

$$f(X_2(\omega)) = \frac{f(X_2(\omega))}{f(X_1(\omega))} f(X_1(\omega)) = \frac{c_2}{c_1} c_1 = c_2.$$ Iterating the procedure, we get $f(X_k(\omega)) =c_k$ for all $k=3,\ldots,n$, and so $B \subseteq A$. In conclusion, $B=A$ and, in particular, $\mathbb{P}(B)=\mathbb{P}(A)$.

Observation 2: If $X_1,\ldots,X_n$ are indepndent random variables and $f$ a measurable function, then $f(X_1),\ldots,f(X_n)$ are independent.

This is a well-known fact from probability theory. Since the Poisson process $(N_t)_{t \geq 0}$ has independent increments, it follows that $f(N_{t_1}),f(N_{t_2}-N_{t_1}),\ldots,f(N_{t_n}-N_{t_{n-1}})$ are independent for any $0<t_1<\ldots<t_n$. If we choose $f(x)=(-1)^x$ this gives that $(-1)^{N_{t_n}-N_{t_{n-1}}},\ldots,(-1)^{N_{t_2}-N_{t_1}},(-1)^{N_{t_1}}$ are independent. (Note that the linked proof does neither use nor claim that $(-1)^{N_t}$ has independent increments.)