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Question: Suppose that $ M_t $ and $ N_t $ are Poisson processes with intensities $ \lambda_1 > 0 $ and $ \lambda_2 > 0$ respectively. Let $ S_k $ and $ T_k $ be the arrival times for $ M_t $ and $ N_t $ respectively. Assume that $S_{k_1}$ and $T_{k_2}$ are independent for all $k_1,k_2 \in \mathbb{N}$. What is the probability distribution of $ N_{S_k} $?

Attempt: I found that $ P(N_{S_1} = n) = \left(\frac{\lambda_2}{\lambda_1 + \lambda_2}\right)^n\frac{\lambda_1}{\lambda_1 + \lambda_2}$, which is a geometric distribution. My approach was first determining $ P(N_{S_1} \geq n) = P(T_n \leq S_1)$ and then $$ P(N_{S_1} = n) = P(N_{S_1} \geq n) - P(N_{S_1} \geq n+1).$$ However, when I did it for general $k \in \mathbb{N}$, I couldn't evaluate the double integral.

Is there a better approach? Any help would be greatly appreciated.

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    $\begingroup$ Why not apply the formula $$P(N_{S_k}=n)=\int_0^\infty P(N_t=n)\,f_{S_k}(t)\,dt\ ?$$ $\endgroup$ – Did Apr 19 '18 at 17:12
  • $\begingroup$ I have never seen that formula, could you please tell me how do you get the formula? $\endgroup$ – FuroCharu Apr 19 '18 at 17:22
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    $\begingroup$ Direct consequence of the conditional probability $$P(N_{S_k}=n\mid S_k=t)=P(N_t=n\mid S_k=t)=P(N_t=n)$$ valid by independence, and of the general decomposition $$P(X=n)=\int_0^\infty P(X=n\mid S_k=t)\,f_{S_k}(t)\,dt$$ applied to $$X=N_{S_k}$$ $\endgroup$ – Did Apr 19 '18 at 17:27
  • $\begingroup$ That's clear, thanks. $\endgroup$ – FuroCharu Apr 19 '18 at 17:46

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