Total number of nodes in critical Galton-Watson process Consider the following critical Galton-Watson process: initially there is a population of $Z(0) = z_0$. The distribution of children for each node follows a binomial law, with maximum value $d$; i.e. the probability that a node has $k$ children is
$$
p_k = \binom{d}{k} q^k (1-q)^k
$$
where $q = 1/d$.
In this case, the expected number of children is $1$, so this is a critical process.
Now, my question is the following: define the random variable $S = \sum_{i=0}^{\infty} Z(i)$, the total number of nodes in the full process. With probability one we have $S < \infty$. What can we say about the distribution of $S$? That is, can one estimate $P(S = i)$?  
I have some preliminary calculations that suggest that for $z_0 = 1$ we have $P(S = i) \approx i^{-3/2}$. I don't have a proof of this and I am not sure what happens when $z_0$ becomes large.
Thanks for help or pointers
 A: Yes, this distribution can be described (although not always very explicitly).  More generally, let $S$ be the total number of individuals of a Galton--Watson process started from $z_0$ individuals and with offspring distribution $\mu$.
Let $(W_n, n \geq 0)$ be a random walk on $\mathbb{Z}$ with initial value $z_0$ and
jump distribution $\nu(k) = \mu(k + 1)$ for every $k \geq −1$. Set
$T = \inf\{n \geq 1 : W_n = −1\}$. Then  $S$ and $T$ have the same distribution (see e.g. Corollary 1.6 http://www.math.u-psud.fr/~jflegall/Cornell.pdf), and by Kemperman's formula (see e.g. Eq.(278) in http://www.stat.berkeley.edu/~pitman/621.pdf):
$$\mathbb{P}(S=n)= \frac{z_0}{n} \mathbb{P}(W_n=-z_0).$$
Then, depending on $\mu$, you can get estimates of $\mathbb{P}(W_n=-z_0)$. For example, if $\mu$ is critical, then the random walk $(W_n)$ is centered. If $\mu$ belongs to the domain of attraction of a stable law, you may use local limit theorems (in particular, if $\mu$ is critical and has finite variance $\sigma^2$, you have $$ \mathbb{P}(S=n) \quad \sim \quad \frac{1}{\sqrt{2\pi \sigma^2}} \cdot \frac{1}{n^{3/2}}.$$ ) If $\mu$ belongs to the subexponential class, you can get estimates as well.
