We define the continuous-time, multi-type branching process $(X(t))_{t\ge0}$ as follows: $(X(0))=\alpha\in\mathbb{R}^d$, where $\alpha$ is the urn initial composition, meaning that, at time $0$, there are $\alpha_i$ particles of type $i$ alive in the system. Each particle reproduces (or “splits”) independently from the rest at rate $1$, and at a reproduction event triggered by a particle of type $i$, we add to the system $R_{i,j}$ particles of type $j$, for all $1\le i,j\le d$.
For all $1\le i\le d$, we let $X^{(i)}$ be the urn process of initial composition $e_i$ (the canonical basis of $\mathbb{R}^d$, where the $i$-th position is equal to $1$ and $0$ in all the other coordinates) and replacement matrix $R$. We want to calculate $\mathbb{E}[X^{(i)}(t)]$, for all $t\ge 0$. We look at the time when the ball in the urn at time zero splits (with probability $e^{-t}$, the initial ball hasn’t split yet at time $t$): for all $t\ge 0$, the computation is given by $$\mathbb{E}[X^{(i)}(t)]=e_ie^{-t}+\int^t_0e^{-s}\text{d}s\mathbb{E}\bigg[\sum^d_{j=1}\sum^{R_{i,j}+\delta_{i,j}}_{k=1}X^{(j,k)}(t-s)\bigg]$$where, for all $1\le j\le d$, $(X^{(j,k)})_k\ge1$ is a sequence of i.i.d. copies of $X^{(j)}$, and the double-indexed sequence $(X^{(j,k)})_{i,j\ge1}$ is a sequence of independent processes.
I don't understand how to get this formula: I would split the expectation in two events, the first "the ball hasn't split yet at time $t$" and "the ball splits", so the first term $e_ie^{-t}$ concerns the first event, while, for the second event I think I should get the integral written above. The term $e^{-s}$ is the probability, which follows an exponential distribution with parameter 1, but I dont' know how to get $\mathbb{E}\big[\sum^d_{j=1}\sum^{R_{i,j}+\delta_{i,j}}_{k=1}X^{(j,k)}(t-s)\big]$
The mechanism I know is, that if at time $s<t$, a ball "$i$" splits, it is replaced by $R_{i,j}+\delta_{i,j}$ balls "$j$". Any help on how I should get that formula?
