The jumping times $\tau_k$ of a $\mathcal F$-poisson process $N$ are $\mathcal F$-stopping times and $N_t=\sum_k1_{\left\{\:\tau_k\:\le\:t\:\right\}}$

Question

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$ and $(N_t)_{t\ge0}$ be a $\mathcal F$-Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\lambda>0$, i.e.

1. $N$ is an $\mathbb N_0$-valued $\mathcal F$-adapted process on $(\Omega,\mathcal A,\operatorname P)$;
2. $N_0=0$;
3. $N_t-N_s$ and $\mathcal F_s$ are independent for all $t\ge s\ge0$;
4. $N_t-N_s$ is Poisson distributed with parameter $\lambda(t-s)$ for all $t\ge s\ge0$.

Assuming $N$ is almost surely right-continuous, I was able to show that

5. $N$ is almost surely nondecreasing

and assuming that $N$ is almost surely càdlàg, I was able to show that$^1$

6. $\operatorname P\left[\forall t\ge0:\Delta N_t\in\{0,1\}\right]=1$.

Assume $N$ is (surely) càdlàg. Let $\tau_0:=0$ and $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t\ne0\right\}$$ for $k\in\mathbb N$.

Are we able to show that

7. $\operatorname P\left[\forall n\in\mathbb N_0:\exists t\ge0:N_t=n\right]=1$;
8. $\tau_k$ is a $\mathcal F$-stopping time for all $k\in\mathbb N$;
9. $\tau_k$ is almost surely finite for all $k\in\mathbb N$.
10. $N_t=\sum_{k\in\mathbb N_0}1_{\left\{\:\tau_k\:\le\:t\:\right\}}$ for all $t\ge0$ almost surely.

All these claims are intuitively trivial, but I really worry about some of the technical details we need to prove them rigorously. For example, does (8.) really hold or do we need to replace $\mathcal F$ with the right-continuous filtration generated by it?

For (7.), I've tried to consider $$\operatorname P\left[N_t<n\right]=\sum_{k=0}^{n-1}\operatorname P\left[N_t=k\right]=e^{-\lambda t}\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!}\tag{11}$$ for all $t\ge0$ and $n\in\mathbb N$. $(11)$ would tend to $0$ as $t\to\infty$ if $t^ke^{-\lambda t}\xrightarrow{t\to\infty}0$, but does this really hold? We clearly got $t^ke^{-\lambda t}=e^{k\ln t-\lambda t}$, but now the only useful inequality I'm aware of is $\ln t\le t-1$ for all $t>0$, which is not enough to conclude $k\ln t-\lambda t\xrightarrow{t\to\infty}-\infty$.

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$^1$ As usual, if $x:[0,\infty)\to\mathbb R$ is càdlàg, then $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$.

Answer 1

8. $\{\tau_1\le t\} = \{N_t\ge 1\}\in\mathcal F_t$. Likewise for subsequent $\tau_k$.

9. ${\rm P}[\tau_1<\infty] = \lim_{t\to\infty}{\rm P}[\tau_1\le t]=\lim_{t\to\infty}{\rm P}[N_t\ge 1]=\lim_{t\to\infty}1-e^{-\lambda t}=1$. Similarly for subsequent $\tau_k$.

10. This should read $N_t = \sum_{k\ge 1}1_{\{\tau_k\le t\}}$. Both sides of the equality-to-be are non-decreasing, cadlag, with jumps of size 1, and initial values $0$. To see that they agree you can check that they have the same jumps at each time $t$. But this is clear because $\Delta N_t=1$ if and only if $t=\tau_k$ for some $k\ge 1$.

11. This follows from 6. and 8.

Answer 2

From Thm 3.13 in Chapter $1$ of Karatzas-Shreve, a submartingale $X_t$ for a filtration satisfying the usual conditions has an RCLL (cadlag) modification that remains a submartingale under the same filtration , if and only if the function $t \to EX_t$ is continuous. One easily checks that this is the case for a Poisson process, and therefore the Poisson process, under the filtration being right-continuous and complete, has a cadlag modification that is still a submartingale. WLOG we will now assume that we are dealing with the modification, and this also answers the question about whether the filtration can be anything : it cannot.

Let $N'_t$ be the modification of $N_t$. Then $P[N'_t = N_t] = 1$ for all $t \geq 0$, is the definition of being a modification. We must show that $N'_t$ is a Poisson process if $N_t$ is, otherwise we can't work with it.

To do this, first put $t=0$ so $N'_0 = 0$ a.s. for starters. Throw out the null set where this doesn't happen. Now, for any rational $q$, we have $P(N'_q = N_q) = 1$, and so in particular $P(N'_q \in \mathbb N_0) = 1$. Throw out all the countably many null sets (so in total probability zero) for each positive rational, to get that $P(N'_q \in \mathbb N_0 \forall q \in \mathbb Q^+) = 0$. But we are already within the realm of right-continuity, since $N_q'$ is RCLL : this shows that $P(N_r' \in \mathbb N_0 \forall r \in \mathbb R) = 1$ since the limit of a sequence of integers, if it exists (it does by continuity) is an integer. Thus $N'_t$ is a.s. $\mathbb N_0$ valued.

Now, consider $t > s \geq 0$ and the filtration $\mathcal F_s$ along with $N'_t - N'_s$.We want to show independence. Let $A \in \mathcal F_s$. We know that $N'_t - N'_s = N_t - N_s$ almost surely, and therefore $1_A(N'_t - N'_s) = 1_A(N_t - N_s)$ almost surely. Taking expectations, $E[1_A(N'_t - N'_s)] = E[1_A(N_t - N_s)]=E[1_A]E[N_t - N_s] = E[1_A]E[N'_t - N'_s]$, so independence follows.

Finally, the distribution doesn't change under a.s. equivalence so the Poisson condition is trivial. This shows that $N'_t$ is also a Poisson process under this filtration which has cadlag paths. We let $N_t = N'_t$ from now on WLOG.

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For $5$, note that for each pair of positive rationals $q<q'$ we have $N_{q'} \geq N_q$ almost surely by the Poisson condition. For a right-continuous process, we know that if $N_r < N_s$ for $r >s$ then there are rationals $p,q$ such that $N_p > N_q$, simply by taking rationals close enough to the right of $r$ and $s$ respectively. It follows then that $N$ has a.s. non-decreasing paths.

For $6$, you say you have done it so I will assume it done.

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Finally onto the points you mention.

For $7$ , write for $n \in \mathbb N$, $N_n = (N_{n} - N_{n-1}) + (N_{n-1} - N_{n-2}) + ... + (N_2-N_1) + N_1$. So $N_n$ is a sum of $n$ iid $Poi(\lambda)$ random variables. By $SLLN$(verify the condition, it is easy) we get $\frac{N_n}{n} \to E[Poi(\lambda)] > 0$ a.s. , and in particular $\lim_{t \to \infty} N_t = \infty$ a.s.. Finally, as $N_t$ has a.s. only jumps in the range $\{0,1\}$, we have that the range of $N_t$ is $\mathbb N_0$ a.s., which is the assertion.

For $8$ we must go by induction. $\tau_1$ is a stopping time since $\{\tau_1 \leq t\} = \{N_t \geq 1\} \in \mathcal F_t$. Now, if $\tau_k$ is a stopping time, then the two events equated below can be seen to differ symmetrically only by a null set (on which cadlag is violated), so by completeness that $\tau_{k+1}$ is a stopping time follows. $$ \{\tau_{k+1} > t\} = \{\tau_k > t\} \cup \left(\cup_{0<q \in \mathbb Q<t}\left(\{\tau_{k} \leq q\} \cap \left(\cap_{q<q' \in \mathbb Q<t} \{N_q = N_{q'}\}\right)\right)\right) $$

For $9$, the range of $N_t$ is $\mathbb N_0$ a.s., and for all such elements of the sample space, $\tau_k$ is finite, so this follows.

For $10$, we know that a.s., the set of jumps of $N_t$ is in $\{0,1\}$, hence fixing a $t$ we know that on this set we have $P(N_t \geq k)= P(\tau_k \leq t)$, since every jump is of size one. A similar identity $P(\sum_{k \in \mathbb N} 1_{\tau_k \leq t} \geq k) = P(\tau_k\geq t)$ is more obvious. So,$N_t$ and $\sum_{k \in \mathbb N} 1_{\tau_k \leq t}$ are modifications of each other, but both are cadlag (that the second is cadlag is left as an exercise)! It is a standard exercise(go via rationals) to then prove they are indistinguishable, as desired.