$$\int_{\theta =0}^T \left\{ \sum\limits_{u=0}^{S_1-1} \sum\limits_{m=0}^{S_1-u-1} p( m,\lambda_{0,1}(T-\theta)) p(u,\lambda_1\theta) \right\}\lambda _0 p(S_0-1,\lambda_0\theta ) \,d\theta $$
I am trying to work out this integral, I will appreciate if someone help. Thanks!
There are two independent Poisson process with rate of $\lambda_1$ and $\lambda_0$.
Rearrange the order of sum and integral, you should get
$$\frac{\lambda_0^{S_0} e^{-\lambda_{01} T}}{(S_0-1)!} \sum_{u=0}^{S_0-1} \frac{\lambda_1^u}{u!} \sum_{m=0}^{S_1-u-1}\frac{\lambda_{01}^m}{m!} \int_0^T d\theta \: (T-\theta)^m \theta^{S_0+u-1} e^{-(\lambda_0 + \lambda_1 - \lambda_{01}) \theta} $$
The integral may be expressed in terms of a confluent hypergeometric function $M$:
$$\frac{(\lambda_0 T)^{S_0} e^{-\lambda_{01} T}}{(S_0-1)!} \sum_{u=0}^{S_0-1} \frac{(\lambda_1 T)^u}{u!} \sum_{m=0}^{S_1-u-1}\frac{(\lambda_{01} T)^m}{m!} \frac{ m! (S_0+u-1)!}{(m+S_0+u-1)!} M[m,m+S_0+u-1,-(\lambda_0 + \lambda_1 - \lambda_{01})T] $$
From here, I do not see any path to simplification.
