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

Let ξ be a random variable with values in \0,1,2,\ and Eξ=1, let Y₍₁₎≥ Y₍₂₎≥ be the outdegrees, in decreasing order, of a Galton–Watson tree with offspring distribution ξ conditioned to have n vertices, and let ξ₍₁₎≥ξ₍₂₎≥ be the order statistics of n independent copies of ξ. Janson proved that if ξ has finite variance then, for every fixed j, Y₍ⱼ₎ and ξ₍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 P(ξ≥ h(n))→0, and that Y₍ⱼ₎≥ h(n) with high probability for every fixed j when n P(ξ≥ h(n))→∞. 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 nᵣ→∞ and Lᵣ with nᵣ P(ξ≥ Lᵣ)≥ r, such that with probability at least $1-1/r$ the conditioned tree with nᵣ vertices has exactly one vertex of outdegree at least Lᵣ, of outdegree (r+12)Lᵣ, while all other vertices have outdegree at most Lᵣ/(64(log nᵣ)²). Along nᵣ the largest outdegree is of larger order than ξ₍₁₎ and the second largest is of smaller order than ξ₍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.