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Let $N^a$, $N^b$ be two jump process with stochastic intensity process $(\lambda^a_t)_{t\in\mathbb{R}}$, $(\lambda^b_t)_{t\in\mathbb{R}}$ (the lambdas are $\mathcal{F}_t$ -adapted). Let $N$ defined by : $N_t := N^a_t + N^b_t$. Now define $\Delta N_t = N_t - N_{t-} \in \{0, 1\} a.s.$ if $N$ jumps at time $t$, what is the probability that this jump is made by $N^a$ ? More precisly, what is:

$P(\Delta N_{T_n}^a=1 \lvert \mathcal{F}_{T_n}), \; \forall n \in \mathbb{N}$

Where $T_1 < T_2 < T_3 < ...$ are the jump times of the jump process $N$

I have an intuition that $P(\Delta N_{T_n}^a=1 \lvert \mathcal{F}_{T_n}) = \frac{\lambda^a_{T_n}}{\lambda^a_{T_n} + \lambda^b_{T_n}}$ but I don't know where to start the proof. Can anyone please help me prove this please?

Sabrebar
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    Your intuition is correct if $N^a$ and $N^b$ are independent Poisson processes. However, why would you expect the probability to have such a simple form if the two processes aren't independent? or if at least one of the two processes doesn't have independent increments? – AOS Apr 22 '24 at 13:11
  • @AOS Yes you might be right. In the first place, how would you prove this result if $N^a$ and $N^b$ are independent? – Sabrebar Apr 24 '24 at 15:15
  • Otherwise, if they are not independent, I believe that this information would be encoded in the intensities and thus the result would still be true. For instance if $N^a$ and $N^b$ jump one after the other, then at $T_{n-}$, $\lambda^b_{T_n-}$ is zero if the $n-1$ th jump is from $N^b$ and $\frac{\lambda^a_{T_n}}{\lambda^a_{T_n} + \lambda^b_{T_n}} = 1$ – Sabrebar Apr 24 '24 at 15:27

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