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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?

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1 Answer 1

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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 $d=1$.

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