Branching process: The expected value of number of cells

Suppose that there are two kinds of cells, type $$A$$ and type $$B$$.

Let $$A_t$$ and $$B_t$$ be the number of cells (i.e. $$\mathbb{N}_{0}$$-valued) of type $$A$$ and type $$B$$, respectively, at time $$t\in\mathbb{R}_{+}$$.

Each cell of type $$A$$, independent of all other cells is, after a time that is exponential distribution with parameter $$\lambda>0$$, dived into

• two cells of type $$A$$ with probability $$p_1> 0$$,
• two cells of type $$B$$ with probability $$p_2> 0$$,and
• one cell of type $$A$$ and one cell of type $$B$$ with probability $$p_3> 0$$,

where $$p_1+p_2+p_3 = 1$$.

Each cell of type $$B$$, independent of all other cells, after a time that is exponential distribution with parameter $$\gamma>0$$, dies. i.e. the number of cells of type $$B$$ decreases by one.

Let $$a\in\mathbb{N}$$ and $$b\in\mathbb{N}_{0}$$. What is \begin{align} \mathbb{E}[A_t\mid A_0=a] \end{align} and \begin{align} \mathbb{E}[B_t\mid A_0=a,B_0=b] \end{align} for each time $$t\in\mathbb{R}_+$$?

My attempt based on the suggestion by mjqxxxx:

I'd argue that something like \begin{align} \partial_t\mathbb{E}[A_t\mid A_0=a] = \lambda(p_1-p_2) \mathbb{E}[A_t\mid A_0=a] \end{align} should hold and therefore the solution is \begin{align} \mathbb{E}[A_t\mid A_0=a] = a \exp(\lambda(p_1-p_2)t) \end{align} for each time $$t\in\mathbb{R}_+$$. Moreover, \begin{align} \partial_t\mathbb{E}[B_t\mid A_0=a,B_0=b] = -\gamma \mathbb{E}[B_t\mid A_0=a,B_0=b] + \lambda(2p_2 + p_3)\mathbb{E}[A_t\mid A_0=a] \end{align} and therefore \begin{align} \mathbb{E}[B_t\mid A_0=a,B_0=b] = \begin{cases} b \exp(-\gamma t) + \frac{\lambda(2p_2+p_3)a\exp(-\gamma t)}{\gamma +\lambda(p_1 - p_2)}\left(\exp((\gamma +\lambda(p_1 - p_2))t)-1\right) & \text{if } \gamma +\lambda(p_1 - p_2) \neq 0 \\ (b+\lambda(2p_2+p_3)a t)\exp(-\gamma t) & \text{otherwise} \end{cases} \end{align} for each time $$t\in\mathbb{R}_+$$.

Is this true? How do you formalize this?

• You should be able to write a first-order linear differential equation expressing the rate of change of expected populations of $A$ and $B$ in terms of their current values. Feb 10, 2022 at 7:28

Let $$\mathcal{N}_{\mu}\left(\cdot\right)$$ be a Poisson point measure: $$\mathcal{N}_{\mu}\left(A\right)$$ counts the number of points generated by a Poisson point process over some Borel set $$A$$ -- it is a random measure. Then, from the model you described (and if I am not missing anything), we can write the pathwise dynamics as

$$A_t=A_0+\int_{0}^t\sum_{k=1}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}\mathcal{N}_{p_1 k \lambda}\left(ds\right)-\int_{0}^t\sum_{k=1}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}\mathcal{N}_{p_2 k \lambda}\left(ds\right).$$

Remark that $$A_t\geq 0$$: Let $$\tau$$ be the hitting time whereby the process $$\left(A_t\right)_{t\geq 0}$$ hits zero, then $$\mathbf{1}_{\left\{A_{s-}=k\right\}}=0$$ for all $$k\geq 1$$ and $$s> \tau$$. Therefore, $$A_t=0$$ for all $$t\geq \tau$$, as the integration above is zero over $$\left(\tau,t\right)$$ for any $$t\geq \tau$$.

Further, by compensating the Poisson point processes, we recover an integral equation only in terms of $$\left(A_t\right)_{t\geq 0}$$ up to a zero-mean martingale term:

$$A_t=A_0+\int_{0}^t\sum_{k=0}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}\left(\mathcal{N}_{p_1 k \lambda}\left(ds\right)-p_1k\lambda ds+p_1k\lambda ds\right)\\-\int_{0}^t\sum_{k=0}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}\left(\mathcal{N}_{p_2 k \lambda}\left(ds\right)-p_2 k \lambda ds+p_2 k \lambda ds\right),$$

which yields

$$A_t=A_0+\underbrace{\int_{0}^t\sum_{k=0}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}\left(\mathcal{N}_{p_1 k \lambda}\left(ds\right)-p_1k\lambda ds\right)-\int_{0}^t\sum_{k=0}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}\left(\mathcal{N}_{p_2 k \lambda}\left(ds\right)-p_2 k \lambda ds\right)}_{:=M_{A}(t)} \\ +\int_{0}^t\sum_{k=0}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}p_1k\lambda ds-\int_{0}^t\sum_{k=0}^{\infty} \mathbf{1}_{\left\{A_{s-}=k\right\}}p_2 k \lambda ds,$$

where $$M_{A}(t)$$ is a zero-mean martingale (e.g., [1]). Therefore,

$$A_t = A_0 + M_{A}(t)+(p_1-p_2)\lambda\int_{0}^t A_{s-}ds.$$

In the same vain, we can write the (pathwise) dynamics for $$B_t$$

$$B_t = B_0 + M_{B}(t)+\left(\lambda p_3+2\lambda p_2\right)\int_{0}^t A_{s-} ds -\gamma\int_{0}^t B_{s-} ds.$$

Now, you can take $$E\left[\cdot\left|A_0,\,B_0\right.\right]$$ over the first equation (dynamics of $$A_t$$) and over the second one (dynamics of $$B_t$$) to obtain the ODEs you conjectured.

[1] Rogers and Williams, Diffusions, Markov Processes and Martingales: Foundations, 2nd edition, vol. 2, Cambridge University Press, September 2000.