Natural filtration of a one-jump counting process Since a one-jump counting process $N(t) := I(T \leq t)$ only attains values in $\{0, 1\}$, am I correct in thinking that its natural filtration $\mathcal{N}_t := \sigma(\{N(s): s \leq t)\}$ does not change in time? I.e. $\mathcal{N}_s = \mathcal{N}_t$ for all $s, t$.
Let me explain.
For a fixed time $s$, we have $\sigma(\{N(s)\}) = 2^{\{0, 1\}}$, which is not time-dependent. So $\sigma(\{N(s): s \leq t)\} = 2^{\{0, 1\}}$.
It seems silly that the filtration is not growing in time. I hope you can point out where I went wrong.
 A: I assume that by $2^{\{0,1\}}$ you mean the $\sigma $-algebra generated by $\{0\}$ and $\{1\}$. But this is not the same as $\sigma (N(s))$.
By definition, for a random variable $X:\Omega\to \mathbb{R}$, $\sigma (X)$ is the $\sigma $-algebra generated by the preimages of measurable sets of $\mathbb{R}$ under $X$.
In our case, since $\sigma (N(s))$ may only take the values $0$ and $1$, $\sigma (N(s))$ is the $\sigma $-algebra generated by the preimages of $\{0\}$ and $\{1\}$ under $N(s)$. That's
$$\sigma (\{N(s)=0\},\{N(s)=1\})=\sigma (\{T>s\},\{T\leq s\})=\sigma (\{T\leq s\}).$$
So ${\cal N_t}=\sigma (\{N(s)\}_{s\leq t})=\sigma (\{T\leq s\}_{s\leq t})$.
This will generally vary (increase) with $t$.
A: The natural filtration of $N_t$ is a model of how much we know about the process at any given fixed time $t\,.$ To know if $N_t$ is zero or one is equivalent to know if the random time $T$ was before $t$ or will be strictly after $t\,.$ Therefore ${\cal N}_t$ must be the sigma algebra generated by all $\{T\le s\}$ with $s\le t\,.$ Note that if we know that $T$ was before $s\le t$ we know that it was before $t\,.$
