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Let $X, X_{n,k}$ for $k,n\in\mathbb{N}$ denote independent random variables with values in $\mathbb{N}_0$. $X$ is identically distributed as all $X_{n,k}$. Define $N_0:=1$ and for $n\in\mathbb{N}$ set $$ N_n:=\begin{cases}0, & \text{ if }N_{n-1}=0\\X_{n,1}+\cdots+X_{n,N_{n-1}}, & \text{ if }N_{n-1}>0.\end{cases} $$ Find conditions on the distribution of $X$ for which the probability $$ q:=P(\exists n\in\mathbb{N}: N_n=0) $$ satisfies $q=1$.

Defining the hitting time of $0$ as $H(0)=\inf\left\{n\geq 0: N_n=0\right\}$, it is $$ q=P(H(0)<\infty). $$

Now the only idea I have is to start like this: $$ P(H(0)<\infty)=\sum_{n\in\mathbb{N}_0}P(H(0)=n)\\ =\sum_n P(N_n=0|N_{n-1}>0,...,N_0=1) $$ Markov property: $$ =\sum_n P(N_n=0|N_{n-1}>0)\\ =\sum_{n}\frac{P(N_n=0,N_{n-1}>0)}{P(N_{n-1}>0)}\\ =\sum_n \frac{P(X_{n,1}+\cdots+X_{n,N_{n-1}}=0)}{P(X_{n-1,1}+\cdots+N_{n-1,N_{n-2}}>0)} $$

X identically distributed as all $X_{n,k}$:

$$ =\sum_{n}\frac{P(N_{n-1}\cdot X=0)}{P(N_{n-2}\cdot X>0)} $$

In case that this makes any sense: How do I have to continue? Give me some help, please.

Salamo
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    The necessary and sufficient condition is that $E(X)\leqslant1$ (and that $P(X=1)\ne1$ if $E(X)=1$). This is basic theory of Galton-Watson processes. – Did Nov 07 '14 at 22:02
  • I really wonder why this is on worksheet, because we did not have that. -- So my idea is absolute useless? – Salamo Nov 07 '14 at 22:09
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    Are you able to use generating functions? – Did Nov 07 '14 at 22:11
  • If you mean if we had that in the lecture: yes. We defined what a generating function is and had some Basic properties; for a random variable $X$ with values in $\mathbb{N}0\Cup\left{\infty\right}$ we had $g_X(z)=\sum{n\in\mathbb{N}_0}P(X=n)z^n, z\in\mathbb{C}$ as the generating function. Moreover we had that $g_X(1)=P(X<\infty)$. – Salamo Nov 07 '14 at 22:14
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    More importantly, $P(N_n=0)=g_{N_n}(0)$. What can you say about $g_{N_n}$? – Did Nov 07 '14 at 22:18
  • I had a look in my lecture. I list all things that I found that I can say about $g_{N_n}$, namely, (1.) it is non-negative, monotonically increasing and continuous (2.) it is convex (3.) $g_{N_n}(z)=E(z^{N_n}|N_n<\infty)P(N_n<\infty)$. – Salamo Nov 07 '14 at 22:29
  • Does any of this help me to continue? (3.) says that $g_{N_n}(z)=E(z^{N_n}|N_n<\infty)=\sum_{m\in\mathbb{N}0}E(z^{N_n}|N_n=m)\cdot 1{\left{N_n=m\right}}$. – Salamo Nov 07 '14 at 22:59

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