1
$\begingroup$

Let $\Gamma_0,\Gamma_1,...$ be regeneration epochs.

If $(X_n)_{n \in \mathbb{N}}$ is a $\lambda$ biased random walk on a Galton-Watson tree, than the regeneration epochs are defined as:

$\Gamma_0:=\inf\{\iota \ | X_i\neq X_{\iota} \ \forall i\leq \iota \ \text{and} \ X_{j}\neq (X_{\iota})_* \ \forall j\geq \iota \}$,

$\Gamma_k:=\inf\{\iota \ | \iota>\Gamma_{k-1}: X_i\neq X_{\iota} \ \forall i\leq \iota \ \text{and} \ X_{j}\neq (X_{\iota})_* \ \forall j\geq \iota \}$

I want to show that

$\{\Gamma_0 < \infty \}=\mathcal{S}$, where $\mathcal{S}$ is the survival set.

The statement makes sense, of course. But i do not know how to start the proof.

Maybe someone has experience in the subject area and could help me.

Greetings : Fynn

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

You need to assume that the bias satisfies $\lambda<m$ where $m$ is the mean offspring. Then you can find the proof in Lemma 3.3 page 253 of [1].

[1] Lyons, Russell, Robin Pemantle, and Yuval Peres. "Biased random walks on Galton–Watson trees." Probability theory and related fields 106, no. 2 (1996): 249-264.

https://link.springer.com/content/pdf/10.1007/s004400050064.pdf

Improve this answer
$\endgroup$
2
  • $\begingroup$ Can we therefore say: For a bias $\lambda<m$ holds $\mathcal{S}\subseteq \{ \Gamma_0<\infty\}$ because of infinitely many regeneration epochs and $ \{ \Gamma_0<\infty\}\subseteq \mathcal{S}$ because in the case $\Gamma_0<\infty$ the tree must be infinite? $\endgroup$
    – Fynn13
    Commented Oct 3, 2021 at 12:32
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Yuval Peres
    Commented Oct 3, 2021 at 18:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .