Let $Z_t$ denote the population of a Galton Watson tree at time $t$, starting at $Z_0 = 1$. Let $X$ with $\mathbb{P}(X=k) = p(1-p)^k$ be the number of children of each individual. Let $D$ be the last moment with positive population, i.e. $D = \min\{t \in \mathbb{N}:Z_t > 0\}$.
I'm trying to find $\mathbb{P}(D = t)$ in terms of $p$ and $t$ explicitly. Using the fact that $\mathbb{P}(D\leq t) = G(\mathbb{P}(D \leq t-1))$ were $G$ is the probability generating function, I found it as the difference of a ($t$ times) repeating fraction.
I wonder whether there can be a nicer expression (without the repeating fraction), by for example calculating $\mathbb{P}(D=t) = \mathbb{P}(Z_{t+1} = 0 \cap Z_t > 0) = \sum_{k=1}^\infty \mathbb{P}(Z_{t+1} = 0 | Z_t =k)\mathbb{P}(Z_t=k) = \sum_{k=1}^\infty p^k \mathbb{P}(Z_t=k)$.
At this point I'm stuck. What would be a good approach to find $\mathbb{P}(Z_t=k)$?
$\sum\limits_{k=1}^{\infty}p^{k}P(Z_t=k)=Ep^{Z_t}$. If $f(s)=Es^{Z_1}$ then $Es^{Z_t}$ is the composition of $f$ with itself $t$ times, usually denoted by $f_t(s)$. Thus $P(D=t)=f_t(p)$.