The second largest outdegree of conditioned Galton–Watson trees with infinite variance

Let $\xi$ be a random variable with values in $\{0,1,2,\dots\}$ and $\mathbb E\xi=1$, let $Y_{(1)}\ge Y_{(2)}\ge\dots$ be the outdegrees, in decreasing order, of a Galton–Watson tree with offspring distribution $\xi$ conditioned to have $n$ vertices, and let $\xi_{(1)}\ge\xi_{(2)}\ge\dots$ be the order statistics of $n$ independent copies of $\xi$. Janson proved that if $\xi$ has finite variance then, for every fixed $j$, $Y_{(j)}$ and $\xi_{(j)}$ have asymptotically the same distribution. He asked whether two consequences of this remain true when the variance is infinite: that $Y_{(1)}<h(n)$ with high probability when $n\mathbb P(\xi\ge h(n))\to0$, and that $Y_{(j)}\ge h(n)$ with high probability for every fixed $j$ when $n\mathbb P(\xi\ge h(n))\to\infty$. Kortchemski and Richier showed that the first can fail. We show that the second can fail as well. We construct an offspring distribution with mean $1$ and infinite variance, and integers $n_r\to\infty$ and $L_r$ with $n_r\mathbb P(\xi\ge L_r)\ge r$, such that with probability at least $1-1/r$ the conditioned tree with $n_r$ vertices has exactly one vertex of outdegree at least $L_r$, of outdegree $(r+\frac12)L_r$, while all other vertices have outdegree at most $L_r/(64(\log n_r)^2)$. Along $n_r$ the largest outdegree is of larger order than $\xi_{(1)}$ and the second largest is of smaller order than $\xi_{(2)}$, and each of the six conclusions of Janson's theorem fails. The same holds for the random allocation of $n-1$ balls into $n$ boxes.