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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.

harisf
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2 Answers2

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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 $t$.

gelas
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  • Thank you! I was too quick to set $\sigma(N(s)) = 2^{{0,1}}$ without consideration of preimages. – harisf Jan 25 '22 at 07:55
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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 $t\,.$

Kurt G.
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