Excursion Lengths in a Uniform Eulerian Circuit of the Complete Graph: Aldous's Conjecture Holds if and only if i = o(n^(3/2))

Let N_i be the number of excursions of length i from a fixed vertex in a uniformly random Eulerian circuit of the complete graph K_n in which every edge is replaced by two opposite arcs. Aldous and Yu (Open Problems in Mathematics, 2014) stated as the natural conjecture that E N_i is asymptotic to e^(−i/n). We prove that for every fixed K, uniformly in 2 ≤ i ≤ K n^(3/2), E N_i = exp(−i/n − i²/(2n³))(1 + O(n^(−1/2) log n)), and that E N_i e^(i/n) → 0 when i/n^(3/2) → ∞. Consequently E N_i ∼ e^(−i/n) holds if and only if i = o(n^(3/2)). This range contains fixed i, the scale i ∼ xn and essentially all excursions, and the length of a uniformly chosen excursion, divided by n, converges in distribution to the standard exponential law. At the scale i ∼ y n^(3/2) the ratio E N_i/e^(−i/n) tends to e^(−y²/2), so the conjecture read literally for all i is false. The proof is elementary. It combines the construction of a uniform Eulerian circuit from a uniform spanning tree (the BEST theorem, in the form used by Kandel, Matias, Unger and Winkler) with a hazard representation of the first excursion and a convexity bound. Exact computations for n ≤ 6 confirm the hazard representation and the closed forms for E N_2 and E N_3, and Monte Carlo simulations up to n = 6400 are consistent with the asymptotic results. This is an unrefereed note.