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The Poisson process $\left(N_t\right)_{t\in\left[0,\infty\right)}$ is supposed to be a Markov process, but a Markov process $\left(X_t\right)_{t\in I}$ should be coupled with a family of distributions $\left(\mathrm{P}_e\right)_{e\in E}$ ($E$ being the process's state set) that satisfy the condition $$\forall e\in E, \mathrm{P}_e\left(X_0=e\right)=1$$ However, the Poisson process satisfies by definition $$N_0=0$$ It doesn't add up. I'd like to revise my textbook's definition of a Poisson process to make it a proper Markov process.

According to my textbook (Section 5.5, "The Poisson Process"), with $$\mathcal{I} := \left\{(a,b]:\space a,b\in[0,\infty), a\leq b\right\}$$ and with $\left(N_I,\space I\in\mathcal{I}\right)$ being a family of random variables with values in $\mathbb{N}_0$, $N_I$ to be interpreted as the number of occurrences during the time interval $I$, $\left(N_I,\space I\in\mathcal{I}\right)$ is a Poisson process iff it satisfies the following five axioms:

(A1) $N_{I∪J} = N_I + N_J$ if $I\cap J = \emptyset$ and $I\cup J\in\mathcal{I}$.

(A2) The distribution of $N_I$ depends only on the length of $I$: $P_{N_I} = P_{N_J}$ for all $I,J\in\mathcal{I}$ with $\ell\left(I\right) = \ell\left(J\right)$.

(A3) If $\mathcal{J}\subseteq\mathcal{I}$ with $I\cap J=\emptyset$ forall $I,J\in\mathcal{J}$ with $I\neq J$, then $\left(N_J, J\in\mathcal{J}\right)$ is an independent family.

(A4) For any $I\in\mathcal{I}$, we have $\mathrm{E}\left[N_I\right]<\infty$.

(A5) $\lim_{\varepsilon\downarrow0}\varepsilon^{−1}\mathrm{P}\left[N_{(0,\varepsilon]}\geq2\right] = 0$.

How can i revise these axioms to suit my purpose?

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In the case of Poisson processes, it is not difficult to write down $P_i$ for every $i\geqslant0$. So, if the question is about a possible inconsistency in an exposition of the subject that you are reading right now, I am afraid you will simply have to live with it. –  Did Sep 11 '12 at 5:37
    
@did: ....Done. –  Evan Aad Sep 11 '12 at 15:49
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It doesn't add up. I'd like to revise my textbook's definition of a Poisson process to make it a proper Markov process.

To expand on my comment, it does add up and there is nothing to revise.

Let us recall that the process $(N_I)_{I\in\mathcal I}$ is indexed by intervals, not real numbers, hence the notation $(N_t)_{t\geqslant0}$ is of your own making, in the context of the definitions you recall, and in particular $N_0$ simply does not exist. To recover a Markov process $(X_t)_{t\geqslant0}$, one should ask that, for every $t\gt0$, $$ X_t=X_0+N_{(0,t]}, $$ and it would probably be healthier to add the hypothesis that $X_0$ and $(N_I)_{I\in\mathcal I}$ are independent.

With this in mind, it is true that one often chooses $X_0=0$, in which case $X_t=N_{(0,t]}$ for every $t\geqslant0$, the case $t=0$ included.

To sum up, the process $(N_I)_{I\in\mathcal I}$ enumerates some events which occur in intervals $I$ of $(0,+\infty)$ hence it describes only the increments of a Markov process $(X_t)_{t\geqslant0}$. Either one considers that zero event happens at time $0$, then the number of events up to time $t$ is $X_t=N_{(0,t]}$ for every $t\geqslant0$. Or, $X_0$ events occur at time $0$, then the number of events up to time $t$ is $X_t=X_0+N_{(0,t]}$ for every $t\geqslant0$.

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I see. It makes sense. Thanks. –  Evan Aad Sep 11 '12 at 15:50
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Call $\left(N_I\right)_{I\in\mathcal{I}}$ the set of Poisson increments and define a proto-Poisson process to be any $\mathbb{N}_0$-valued stochastic process $X=\left(X_t\right)_{t\in[0,\infty)}$ with Poisson increments, i.e. let $X$ be such that for every $a,b\in[0,\infty)\space\space a<b$, $$\mathrm{P}_{X_b-X_a}=\mathrm{P}_{N_{(a,b]}}$$

Define a Poisson process to be any $\mathbb{N}_0$-valued stochastic process on $\left(\mathbb{N}_0^{[0,\infty)}, \mathbb{P}\left(\mathbb{N}_0\right)^{\otimes{[0,\infty)}}\right)$ with the standard projections acting as its random variables and with distributions $\left(\mathrm{P}_{X^{(n)}}\right)_{n\in\mathbb{N}_0}$ for some set of proto-Poisson processes $\left(X^{(n)}\right)_{n\in\mathbb{N}_0}$ that satisfy $\mathrm{P}\left(X_0^{(n)}=n\right)=1.$

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