How to get this formula for expectation of continuous-time urn process

Question

We define the continuous-time, multi-type branching process $(X(t))_{t\ge0}$ as follows: $(X(0))=\alpha\in\mathbb{R}^d$, where $\alpha=(\alpha_1,\dots,\alpha_d)$ 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 $1\le i,j\le d$, where $R$ is a replacement matrix of the urn process $X(t)$.

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 $R$. E.g. $X^(1)$ will have as initiali composition the vector $e_1=(1,0,\dots,0)$.

We want to calculate $\mathbb{E}[X^{(i)}(t)]$, for all $t\ge 0$. This 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.

This apparently comes from the following reasoning: "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 $t$)"

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?

Reference: PÒLYA URNS AND OTHER REINFORCEMENT PROCESSES

Answer 1

As a toy example, consider a Yule process $(Y_t, t\geq 0)$: It's law can be described as follows:

• $Y_0=1$, a.s.
• If $Y_t=n$, then the transition to $n+1$ happens at rate $n$.

This is very similar to your urn process. You can think of this as a population evolving in continuous time where every individual branches independently into two at rate $1$ and their offspring evolves independently.

Let's compute $\mathbb{E}[Y(t)]$ by considering the time of the first branching event, denote this time by $\tau$. $$m(t) := \mathbb{E}[Y(t)] = \mathbb{P}(\tau > t) + \int_0^t \mathbb{E}[Y(t) \vert \tau = s] \mathbb{P}({\tau \in ds}),$$ where the second equation follows from conditioning on the value of $\tau$. Next, we observe that $\tau$ is distributed like an exponential random variable with rate $1$. Further note that the two offspring that are created evolve independently. For an independent copy $Y'$ of $Y$ and by the Markov property $$\mathbb{E}[Y(t) \vert \tau = s] = \mathbb{E}[Y(t-s)+Y'(t-s)] = 2 \mathbb{E}[Y(t-s)]= 2 m(t-s).$$ If we combine this with the distribution of $\tau$ we obtain the following integral equation (the renewal equation) for $m$: $$m(t) = e^{-t}+\int_0^t 2m(t-s)e^{-s}ds$$ Combine this with $m(0)=1$ to obtain $m(t)=e^t$ (hint: differentiate).

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The same exact approach also works for multi-type branching processes. Here the difference is what you will get for the offspring process: in the Yule process you get two independent copies $Y,Y'$ of the same process where as in your process you get $$1+ \sum_{j=1}^d R_{i,j}$$ new process -- one for the initial particle (that will have type $1$) and $R_{i,j}$ new particles of type $j$. The processes emanating from these particles are denoted by $X^{(j,k)}$. They will all evolve independently starting from time $s$, so until time $t$ there is $t-s$ time left after applying the Markov property. That's why in the end you sum over $X^{(j,k)}(t-s)$.

I hope this helps!