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.



Ratio of expected diameter and height of a conditioned Galton-Watson tree
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).


