Consider a process $(X_t)_{t\ge 0}$ with stationary increments. Then it is true that for $s<t$, we have
$\mathbb{P}(|X_t - X_s| \le \varepsilon) = \mathbb{P}(|X_{t-s}| \le \varepsilon)$,
however we cannot (am I right?) say the same thing about its supremum:
$\mathbb{P}(\sup_{a < s < t < b} |X_t - X_s| \le \varepsilon) \neq \mathbb{P}(\sup_{a < s < t < b}|X_{t-s}| \le \varepsilon)$.
But how about the Poisson process $(N_t)_{t \ge 0}$ or any (almost surely) monotone process? Can we write that
$\sup_{a < s < t < b} |N_t - N_s| = \sup_{a < s < t < b} N_t - N_s = \sup_{a < s < t < b} N_t - \inf_{a < s < t < b} N_s = N_{b^-} - N_{a^+} \sim N_{b^- - a}$ a.s.?
And as a consequence: $\mathbb{P}(\sup_{a < s < t < b} |N_t - N_s| \le \varepsilon) = \mathbb{P}(N_{b^- - a} \le \varepsilon) \le \mathbb{P}(N_{b - a} \le \varepsilon)$? Generally, what I am asking is if my thinking is correct and if not, how can I deal with supremum of increment of Poisson process?