A cell has been placed in a biological solution at time $t=0$. After an exponential time of rate $\mu$, it is divided, producing $k$ cells with probability $p_{k}, k=0,1, \ldots$, with the mean value $\rho=\sum_{k=1}^{\infty} k p_{k}(k=0$ means that the cell dies $)$. The same mechanism is applied to each of the living cells, independently.

(a) Let $M_{t}$ be the number of living cells in the solution by time $t>0$. Prove that $\mathbb{E} M_{t}=\exp [t \mu(\rho-1)]$. [You may use without proof, if you wish, the fact that, if a positive function $a(t)$ satisfies $a(t+s)=a(t) a(s)$ for $t, s \geqslant 0$ and is differentiable at zero, then $a(t)=e^{\alpha t}, t \geqslant 0$, for some $\left.\alpha .\right]$

Let $\phi_{t}(s)=\mathbb{E} s^{M_{t}}$ be the probability generating function of $M_{t}$. Prove that it satisfies the following differential equation

$$\frac{\mathrm{d}}{\mathrm{d} t} \phi_{t}(s)=\mu\left(-\phi_{t}(s)+\sum_{k=0}^{\infty} p_{k}\left[\phi_{t}(s)\right]^{k}\right), \quad \text { with } \quad \phi_{0}(s)=s$$

(b) Now consider the case where each cell is divided in two cells $\left(p_{2}=1\right)$. Let $N_{t}=M_{t}-1$ be the number of cells produced in the solution by time $t$.

Calculate the distribution of $N_{t}$. Is $\left(N_{t}\right)$ an inhomogeneous Poisson process? If so, what is its rate $\lambda(t)$ ? Justify your answer.