Let $N(t)$ be a Poisson random process, i.e., $N(t)$ is the number of arrivals between $[0,t]$, let $t_1, t_2$ be two times, then we claim:
$$\mathbb{E}[(-1)^{N(t_1) + N(t_2)}] = \mathbb{E}[(-1)^{|N(t_1) - N(t_2)|}] = \mathbb{E}[(-1)^{N(|t_1 - t_2|)}]$$
Does anyone know how to justify the equalities?
For the first one, $\mathbb{E}[(-1)^{N(t_1) + N(t_2)}] = \mathbb{E}[(-1)^{|N(t_1) - N(t_2)|}]$, I am guessing that the sign of $(-1)^{N(t_1) + N(t_2)}$ is equal to the sign of $(-1)^{|N(t_1) - N(t_2)|}$.
The only clue I have about the second one is that the Poisson process satisfies a so-called stationary increment property. That is, given time $t_2 > t_1$, $\Pr[N(t_2) - N(t_1) = n] = \Pr[N(t_2 - t_1) = n]$. But I do not see how this definition carries over to the expectation as well.
Yes, for part 1, it's literally the fact that $(-1)^{a+b} = (-1)^{|a-b|}$ for any natural numbers $a$ and $b$ which can easily be shown by case analysis on whether $a$ and $b$ are even or odd.
For part two you're much further along than you think. You've shown that $N(t_2)-N(t_1)$ has the same distribution as $N(t_2-t_1)$ . This immediately implies that $$E[f(N(t_2)-N(t_1))] = E[f(N(t_2-t_1))] $$ for any function $f$ for which the expectation is defined. The expectation of (a function of) a variable is entirely determined by its distribution. Proof: you compute $E(f(X))$ by doing the sum $$ \sum_{n=0}^\infty f(n)P(X=n). $$