Define a Poisson process as a Levy process where the increments have a Poisson distribution with parameter $\lambda$*"length of increment".

I want to prove these properties:

It has almost surely jumps of value 1.

It is almost surely increasing.

When it changes, the change it is almost surely integer-valued.

It is almost surely positive. I think this will follow from the above.

Earlier I tried to prove that the jumps is almost surely of value 1 here: Proving that the Poisson process has a.s. jumps of value 1.

But looking back on the proof I think it is wrong. Because that proof is built on that the if for instance $N((k+1)*T/n)-N(kT/n)$ in our entire partition is either 0 or 1, then this event is contained in the event that all our jumps are of value either 0 or 1. However it could theoretically move up and down more in each interval.

So do you see how to prove the properties I wrote in the start? My only idea is to some way look at partitions that approaces 0 in length, and use that then the probability that $N((k+1)*T/n)-N(kT/n)$ is either 0 or 1 approaches 1. But the problem is that even if $N((k+1)*T/n)-N(kT/n)=1$ the process may have had two jumps where it went down 2.4 and then up 3.4.

Another problem is the increasing part. For any interval, there is a positive probability that the end value of the process at the interval minus the start value is positive, is 1.(Since the distribution here is Poisson) However, this does not say that the process is increasing throughout the interval?

How do we show that it behaves the way we know it does, a.s.?

Or can we maybe only say something about the process if we fix the interval first? If we fix a given interval, we can say with probability 1 that it increases and that the increasing value is of integer value? But we can't prove that the process has a.s. increasing sample paths? So that in essence we can only prove things about finitely many points, but not the sample path as a whole? But then, what happens with jumps of value 1 a.s.? Maybe we can not even prove this?

UPDATE: Cleaner version of the quesiton Maybe it is not so easy to see what I am wondering about in the quesiton above. I'll try to explain it a little more precise.

Assume that you only have the interval [0,T]. Lets say that you have a sequence of partitions that converges to 0. Then you only look at the respective values of the process in the partition points. You can then prove that these values are increasing with probability 1. But if you looked at the entire sample path, could you then prove that the path is increasing with probability 1?

In each partition you can prove with probability 1 that the function only changes with integer values at the partition points. Can you prove this property for the entire sample-path a.s.?

You can prove that as the parition intervals goes to zero, the probability of the jumps at two points next to each other are bigger than 1, converges to 0.(I have only prove this myself for partitions of equal length but I assume it holds in the general case). Can this property be extended to the entire sample path?, so when the function changes value, it does so by increasing its value to 1. (as)

So in summary: If you look at the finite dimensional distribution of a Poisson process, where you look at points close to each other, you have that with probability 1 it is increasing, and the jumps are integer -valued. And you have with a probability close to 1(depending on the partition) that the jumps are of value 1. Can these properties in some way be extended to the entire sample-path?.

  • $\begingroup$ "where the increments have a Poisson distribution with parameter $\lambda$" needs some correcting. 1. Break up the interval into $dt$ pieces and use the fact that probability of more than $1$ jump in the small piece is $o(dt)$. Use union bound over $\frac T {dt}$ pieces. 2. Increasing = non-decreasing. Since increments are non-negative, the process is non-decreasing. 3. The changes are only integer-valued by definition. $\endgroup$ – A.S. Feb 23 '16 at 18:57
  • $\begingroup$ @A.S. You say that the probability of more than 1 jump is $o(dt)$. But that does not follow from the definiton. If you have an interval of length $\delta t$, you have that $N(a+\delta t)-N(a)$ is Poisson distributed with parameter $\lambda \delta t$. This tells us that indeed the probability of $N(a+ \delta t)-N(a)$ having more than 1 in value is $o(\delta t)$, however, it doesn't say anything about the number of jumps in the interval. It only says something about the end of the interval - the start of the interval. cont.. $\endgroup$ – user119615 Feb 23 '16 at 19:11
  • $\begingroup$ That was what I was trying to say in the post, even if we can show that $N(a+ \delta t)-N(a)$ having values bigger than 1 is $o(\delta t)$ we haven't said anything about for instance excluding two jumps in the interval where one is 2, and one is -1. $\endgroup$ – user119615 Feb 23 '16 at 19:11
  • $\begingroup$ @A.S. And yes I agree that we can prove that increments are non-decreasing with integer values. But this only work if you fix the interval, what I am asking about is about the sample path as a whole. Maybe it is trivial, but I don't see how: any increment is increasing a.s. $\rightarrow$ the sample path is increasing a.s. $\endgroup$ – user119615 Feb 23 '16 at 19:14
  • $\begingroup$ And to the last point you say that it is integer valued by definition. Again, it is the increments that are integer valued. But I am wondering if this in some way can be shown to hold for the entire sample-path aswell, that every sample path is a step function that increases with the value 1, when it changes, and this must hold a.s. $\endgroup$ – user119615 Feb 23 '16 at 19:16

Let $(X_t)_{t \geq 0}$ be a Poisson process with intensity $\lambda$.

Step 1: $(X_t)_{t \geq 0}$ has almost surely increasing sample paths.

Proof: Fix $s \leq t$. Since $(X_t)_{t \geq 0}$ has stationary increments and $X_{t-s}$ is Poisson distributed, we have

$$\mathbb{P}(X_s>X_t) = \mathbb{P}(X_t-X_s < 0) = \mathbb{P}(X_{t-s}<0)=0,$$

i.e. $X_s \leq X_t$ almost surely. As $\mathbb{Q}_+$ is countable, this implies

$$\mathbb{P}(\forall q \leq r, q,r \in \mathbb{Q}_+: X_q \leq X_r)=1.$$

Since $(X_t)_{t \geq 0}$ has càdlàg sample paths, this already implies

$$\mathbb{P}(\forall s \leq t: X_s \leq X_t)= 1.$$

Step 2: $(X_t)_{t \geq 0}$ takes almost surely only integer values.

We have $\mathbb{P}(X_q \in \mathbb{N}_0)=1$ for all $q \in \mathbb{Q}_+$. Hence, $\mathbb{P}(\forall q \in \mathbb{Q}_+: X_q \in \mathbb{N}_0)=1$. As $(X_t)_{t \geq 0}$ has càdlàg sample paths, we get

$$\mathbb{P}(\forall t \geq 0: X_t \in \mathbb{N}_0)=1.$$

(Note that $\Omega \backslash \{\forall t \geq 0: X_t \in \mathbb{N}_0\} \subseteq \{\exists q \in \mathbb{Q}_+: X_q \notin \mathbb{N}_0$ and that the latter is a $\mathbb{P}$-null set.)

Step 3: $(X_t)_{t \geq 0}$ has almost surely integer-valued jump heights.

We already know from step 1 and 2 that there exists a null set $N$ such that $X_t(\omega) \in \mathbb{N}_0$ and $t \mapsto X_t(\omega)$ is non-decreasing for all $\omega \in \Omega \backslash N$. Consequently, we have

$$X_s(\omega) - X_t(\omega) \in \mathbb{N}_0$$

for any $s \geq t$ and $\omega \in \Omega \backslash N$. On the other hand, we know that the limit

$$\lim_{t \uparrow s} (X_s(\omega)-X_t(\omega)) = \Delta X_s(\omega) $$

exists. Combing both considerations yields $\Delta X_s(\omega) \in \mathbb{N}_0$. (Check that the following statement is true: If $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N}_0$ and the limit $a:=\lim_n a_n$ exists, then $a \in \mathbb{N}_0$.) Since this holds for any $s \geq 0$, we get

$$\Delta X_s(\omega) \in \mathbb{N}_0 \qquad \text{for all $\omega \in \Omega \backslash N$, $s \geq 0$},$$

i.e. $$\mathbb{P}(\forall s \geq 0: \Delta X_s \in \mathbb{N}_0)=1.$$

Step 4: $(X_t)_{t \geq 0}$ has almost surely jumps of height $1$.

By step 3, it suffices to show that $$\mathbb{P}(\exists t \geq 0: \Delta X_t \geq 2)=0.$$

Since the countable union of null sets is a null set, it suffices to show

$$p(T) := \mathbb{P}(\exists t \in [0,T]: \Delta X_t \geq 2)=0$$

for all $T>0$. To this end, we first note that

\begin{align*} \Omega_0 \cap \{\exists t \in [0,T]: \Delta X_t \geq 2\} &\subseteq \Omega_0 \cap \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\} \\ &\subseteq \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\} \end{align*}

for all $k \in \mathbb{N}$ where $$\Omega_0 := \{\omega; s \mapsto X_s \, \, \text{is non-decreasing}\}.$$ Using that $\mathbb{P}(\Omega_0)=1$ (by Step 1) and the fact that the increments $X_{\frac{j}{k}}-X_{\frac{j-1}{k}}$ are independent Poisson distributed random variables with parameter $\lambda/k$, we get

$$\begin{align*} p(T) &=\mathbb{P}(\Omega_0 \cap \{\exists t \in [0,T]: \Delta X_t \geq 2\}) \\ &\leq \sum_{j=1}^{kT} \mathbb{P}(X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2) \\ &= kT \mathbb{P}(X_{\frac{1}{k}} \geq 2) = kT \left(1-e^{-\lambda/k} \left[1+\frac{\lambda}{k} \right]\right) \\ &= \lambda T \frac{1-e^{-\lambda/k} \left(1+\frac{\lambda}{k} \right)}{\frac{\lambda}{k}}. \end{align*}$$

Letting $k \to \infty$, we find

$$p(T) \leq \lambda T \frac{d}{dx} (-e^{-x}(1+x)) \bigg|_{x=0} = 0.$$

  • $\begingroup$ Thank you very much! But just something on point 3, if you assume cadlag, and the way you define the jumps, wont those jumps be zero in the limit? $\endgroup$ – user119615 Mar 3 '16 at 19:37
  • $\begingroup$ @user119615 Yeah, thanks; it should read $t \uparrow s$ instead of $t \downarrow s$ $\endgroup$ – saz Mar 3 '16 at 19:50
  • $\begingroup$ It is a very good and big answer, thank you very much. I just have two more questions, I hope it is ok, it is in regards to point 4. In the start in point 4 you say that it is enough by point 3 to prove that $\mathbb{P}(\exists t \geq 0: \Delta X_t \geq 2)=0.$, can you please say why this follow from point 3? And also you have this set containment equation: $$\{\exists t \in [0,T]: \Delta X_t \geq 2) \subseteq \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\}$$, but I do not see how this holds? For a trajectory, may there not for instance be a jump of +2 and -2 in the $\endgroup$ – user119615 Mar 3 '16 at 20:17
  • $\begingroup$ in the little intervall, and hence it is contained in the complement instead? $\endgroup$ – user119615 Mar 3 '16 at 20:17
  • $\begingroup$ @user119615 Well, from step 3 we know that $\mathbb{P}(\forall t: \Delta X_t \in \mathbb{N}_0)=1$. Therefore, in order to prove that $X_t$ has only jumps of height $0$ or $1$ it suffices to show that $$\mathbb{P}(\exists t: \Delta X_t \notin \{0,1\}) = \mathbb{P}(\exists t: \Delta X_t \in \mathbb{N}_0 \backslash \{0,1\}) \leq \mathbb{P}(\exists t: \Delta X_t \geq 2)=0.$$ $\endgroup$ – saz Mar 4 '16 at 6:38

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