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.