Say k > 0, t $\geq$ 0 and $N_{t}$ is a Poisson Process with rate $\lambda$ >0 ;
Is $$M(t) = \dfrac{N_{t+k^2} - N_{t}}{k}$$ a Poisson Process?
The answer is clearly no since the above process is not a counting process. However I don't exactly understand what would happen if k were to be 1. Would m(t) be a PP for k=1? Intuitively, I feel like for k=1, the given M(t) should be a PP but some very good questions were raised by my friend who asked how we would try to satisfy the property of independent increments for a Poisson process. A detailed explanation would be appreciated.
Edit: For clarification, can we say that if $N_{t}$ satisfies independent increments, $M(t)=M(t)\big|_{k=1}$ will have independent increments?
$M$ does not have independent increments. Let $s < t$. Then
$$ M(t) - M(s) = [N(t+1) - N(s+1)] - [N(t) - N(s)]. $$
From this, we note that
$$ \mathbf{Cov}(M(2) - M(1), M(1) - M(0)) = -\mathbf{Var}(N(2) - N(1)) = -\lambda. $$
So $M(2) - M(1)$ and $M(1) - M(0)$ are not independent.
That $M$ is not a Poisson process can be verified by other means:
$M(t)\sim\operatorname{Poisson}(\lambda)$ does not depend on $t$, which is contrary to the Poisson processes.
The sample paths of $M$ are not monotone increasing almost surely. On the other hand, Poisson sample paths are monotone increasing with probability one.