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What type of Markov process relates to an inhomogeneous Poisson process?

A homogeneous Poisson process-- one where the rate, $\lambda$, is constant-- is a pure birth continuous time Markov chain (with a constant birth rate). That is, if $q_{ij}$ is the transition rate then $q_{ij} = 0 $ unless $j=i+1$ i which case $q_i,i+1 = \lambda$

What happens when $\lambda = \lambda \left(t\right)$? Does, $q \rightarrow q\left(t\right)$? And, how is this represented in a transition matrix?

For a homogeneous Poisson process, the matrix of transition probabilities for the corresponding Markov process is one that is zero every where except the band immediately above the diagonal, all entries of which are $\lambda$.

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up vote 1 down vote accepted

Well, let's try this... Let $\{N_{t}\}_{t\geq0}$ be an inhomogeneous Poisson process with rate function $\{\lambda(t)\}_{t\geq0}$. Let's try to compute the transition probabilites.

By definition of an inhomogeneous Poisson process, we have

$$P\{N(t+h)-N(t)=1\}=\lambda(t)h+o(h)$$ $$P\{N(t+h)-N(t)>1\}=o(h)$$

Since $\{N_{t}\}_{t\geq0}$ has independent increments by definition, we have

$$P\{N(t+h)-N(t)=1|N(t)=k\}=\lambda(t)h+o(h)$$ $$P\{N(t+h)-N(t)>1|N(t)=k\}=o(h)$$

for any $t\geq0$, any $k\in\mathbb{N}_{0}$ and "small" $h$.

Dividing previous equations by $h$, and letting $h\rightarrow 0$, we get

$$\lim_{h\rightarrow0}\frac{1}{h}P\{N(t+h)-N(t)=i+1|N(t)=i\}=\lambda(t)$$ $$\lim_{h\rightarrow0}\frac{1}{h}P\{N(t+h)-N(t)=j|N(t)=i\}=0$$ for $j>i+1$ or $j<i$.

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I see, so it is the case that the probability of transition still tracks $\lambda$. Thanks. – mac389 Jun 15 '13 at 14:29
What about the case where $\lambda\left(t\right)$ is periodic and the period $\geq 10h$? – mac389 Jun 15 '13 at 14:30
Well, the reasoning was sort of independent of the form of $\lambda(t)$, so the answer would be the same, I guess... – Stojan Jovanović Jun 15 '13 at 20:39

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