# Conditioned Galton Watson Process

Let $$S_n$$ be a Galton Watson Process with offspring distribution $$p_k$$. We assume that $$p_0$$ > 0 and that $$\sum_{i=0}^{\infty} k p_k > 0$$. Also $$S_0$$ = 1.

We define $$T_0$$ := inf {n > 0 : $$S_n$$ = 0). Let d := P($$T_0$$ < $$\infty$$). Also

G(z) := $$\sum_{i=0}^{\infty} z^k P(S_1 = k)$$

is the generating function of $$S_1$$.

We have to show that

a) P($$T_0$$ < $$\infty$$ | $$S_1$$ = k) = $$d^k$$

b) Let $$Z_n$$ be a Galton Watson Process with offspring distribution $$r_k$$ = $$d^{k-1}$$ P($$S_1$$ = k), then $$Z_n$$ is going to die out almost surely.

Do you have any ideas how to solve this?

Conditioned on $$S_1=k$$, extinction is equivalent to the extinction of $$k$$ independent branching processes with the same offspring distribution, hence $$\mathbb P(T_0<\infty\mid S_1=k) = d^k$$.
For b), it seems that you've defined $$\mathbb P(S_1=k) = d^{k-1}\mathbb P(S_1=k)$$ for all $$k$$. In which case, assuming $$\mathbb P(S_1=k)>0$$, dividing yields $$d^{k-1}=1$$ and hence $$d=1$$.