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A Galton-Watson tree is the family tree of a Galton-Watson process. Let $T_n$ denote a Galton-Watson tree conditioned on total population size $n$. The time of extinction is its height $H(T_n)$ and its diameter $D(T_n)$ is the length of a longest path in the tree. It is known that for certain offspring distributions the ratio $\mathbb{E}[D(T_n)] / \mathbb{E}[H(T_n)]$ converges to $4/3$. The same property is expected to hold in a more general context, as argued heuristically by Aldous in the paper "The continuum random tree II: an overview" (1991).

I would like to know if a formal proof of this is known in the case of an arbitrary offspring distribution with expected value 1 and finite nonzero variance.

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Yes, it is known. By results of Haas & Miermont [http://arxiv.org/pdf/1003.3632v3.pdf - Lemma 33] (see also Addario--Berry, Devroy & Janson [http://arxiv.org/pdf/1011.4121v1.pdf - Theorem 1.2 - for stronger bounds], $ H(T_n)/\sqrt{n}$ is bounded in $L^2$, and so is $D(T_n)/\sqrt{n}$ because $D(T_n) \leq 2 H(T_n)$. Since $H(T_n)/\sqrt{n}$ and $D(T_n)/\sqrt{n}$ converge in distribution respectively to the height and diameter of the Brownian CRT times the same constant, it follows that $\mathbb{E}[H(T_n)]/\mathbb{E}[D(T_n)]$ converges towards $\mathbb{E}[H]/\mathbb{E}[D]$, where $H$ is the height of the Brownian CRT and $D$ its diameter.

Since it is known that $\mathbb{E}[H(T_n)]/\mathbb{E}[D(T_n)]$ converges towards $4/3$ for some offspring distributions, we get that $\mathbb{E}[H]/\mathbb{E}[D]=4/3$, implying that $\mathbb{E}[H(T_n)]/\mathbb{E}[D(T_n)]$ converges actually towards $4/3$ for all offspring distributions (critical with finite positive variance).

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Thanks for the helpful answer. –  Benedikt Stufler Jul 20 at 18:21

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