Let $\{X_t\}, \{Y_t\}$ be two independent Poisson processes, of rates $\lambda,~ \mu$ respectively. Find the probability of the event: "between two incidents of the process $\{X_t\}$, there are $4$ incidents of the process $\{Y_t\}$".
Attempt. Let $W_1$ be the time between two incidents of $\{X_t\}$, so $W_1\sim Exp(\lambda)$, so we are looking for the probability of the event $Y_t=4$, while $t$ belongs on an interval defined by $W_1$. How can this be managed?
Thank you in advance!
Let $T=\inf\{t>0: X_t=0\}$ be the first arrival of the process $\{X_t\}$. Then $T$ is exponentially distributed with rate $\lambda$, and so conditioning on $T$ we compute \begin{align} \mathbb P(Y_{T}=4) &= \int_0^\infty \mathbb P(Y_{T}=4\mid T=t)f_T(t)\,\mathsf dt\\ &= \int_0^\infty \frac{(\mu t)^4}{4!}e^{-\mu t} \lambda e^{-\lambda t}\,\mathsf dt\\ &= \frac{\lambda\mu^4}{4!}\int_0^\infty t^4e^{-(\lambda+\mu)t}\,\mathsf dt\\ &= \frac{\lambda\mu^4}{(\lambda+\mu)4!}\int_0^\infty t^4(\lambda+\mu)e^{-(\lambda+\mu)t}\,\mathsf dt\\ &= \frac{\lambda\mu^4}{(\lambda+\mu)^5}. \end{align}