# Thread: Age dependent branching process

1. ## Age dependent branching process

Let $\displaystyle Z_n$ be the size of the nth generation in an age-dependent branching process $\displaystyle Z(t)$, the lifetime distribution of which is exponential with parameter $\displaystyle \lambda$. If $\displaystyle Z(0) = 1$, show that the probability generating function $\displaystyle G_t (s)$ of $\displaystyle Z(t)$ satisfies

$\displaystyle \frac{\partial}{\partial t} G_t(s) = \lambda {G(G_t(s)) - G_t(s)}$

i get:

from a theorem:

$\displaystyle G_t(s) = \int^t_0 G(G_{t-u})f_T(u)\ du + \int^\infty_t sf_T(u) \ du$

differentiating with respect to t

$\displaystyle \frac{\partial}{\partial t} G_t(s) = G(G_0(s))f_T(t) - sf_T(t)$

however i am not sure this right as the Generating functions are confusing and i am uncertain as where to go next.

2. ## Re: Age dependent branching process

Hello,

When you're differentiating the first integral, the constant is when u=t, and the part to be differentiated is when u=0.

This would rather give $\displaystyle \frac{\partial}{\partial t} G_t(s)=G(G_t(s))f_T(0)+sf_T(t)$

and since $\displaystyle f_T$ is the pdf of an exponential, $\displaystyle f_T(0)=\lambda$. But as to get $\displaystyle sf_T(t)=G_t(s)$, I don't know, sorry...