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Let's say we are given a $n$-dimensional Hawkes point process. To be more precise, let $N = (N_1,\dots,N_n)$ be a point process and $\mathcal{H}_t=\sigma\{N_1(s),\dots,N_n(s):0\leq s\leq t\}$ be the history of the process up to time $t$. Let $\lambda_i(t) = \nu_i+\sum_{k=1}^n \int_0^t g_{ik}(t-s) \, dN^k_s$ be the $\mathcal{H}_t$-conditional intensity of the $i$-th process, in a sense that

$$E(N^i_{t+h}-N^i_t \mid \mathcal{H}_t)=E\left(\int_t^{t+h} \lambda_i (s) \, ds \mid \mathcal{H}_t\right),$$

where $\{g_{ij}(t)\}_{i,j\in\{1,\dots,n\}}$ is the matrix of interaction kernels, which are assumed to be bounded, integrable functions.

Now, say that we have, for every function in the interaction matrix - $g_{ij}(t)=I_{[d,+\infty)}(t)f_{ij}(t)$ for some bounded, integrable function $f_{ij}(t)$, where $d$ is a positive number.

Then, I can show that $\lambda_i(t+d)$ is $\mathcal{H}_t$-measurable, $\forall t\geq 0$. Furthermore, $\int_t^{t+d}\lambda_i(s) \, ds$ is $\mathcal{H}_t$-measurable.

What I would like to be true is that the increments $N^i_{t+d}-N^i_t$ and $N^j_{t+d}-N^j_t$ are conditionally independent given $\mathcal{H}_t$ for every $i$ and $j$.

Can one prove this from these assumptions? Is there a counter-example?

Help is much appreciated.

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Yes, because, as you point out, under these assumptions $\lambda_i(t)$ only depends on the history of the process up to time $t-d$. So, conditioned on ${\cal H}_t$, each $N_u^i$ behaves as an inhomogeneous Poisson process for $u$ between $t$ and $ t+d$, with rate dependent only on the history up to time $t$.

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Correct me if I'm wrong, but what you're talking about only works if $\lambda_{i}(t)$ is deterministic (or $\mathcal{H}_{0}$-measurable). Then, by Watanabe's characterization theorem, it's a doubly stochastic Poisson process. There is no such assumption here. –  Stojan Jovanović Feb 6 '13 at 13:44
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