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In a uniform linearized chord diagram, $\mathrm{Pr}(\sqrt{n}{R}_1\ge x)=(1-x^2/n{)}^n\to e^{-x^2}$. Conditional on the right endpoints, $d_1(n)=2+{\sum }_{i=2}^n{1}_{\{L_i\le R_1\}}$, with independent success probabilities $R_1/R_i$. A Chernoff bound gives $R_i=(1+o(1))\sqrt{i/n}$ uniformly for $i\ge n^{1/10}$. The early indicators are $o(\sqrt{n})$, and the remaining conditional mean is $(2+o(1))n{R}_1$. Their conditional variance is $O_{\mathbb{P}}(\sqrt{n})$. Hence $d_1(n)/\sqrt{n}-2\sqrt{n}{R}_1\to 0$ in probability, proving the displayed tail limit. A self-loop's contribution is included in the constant two.

Draw $n$ independent pairs of independent uniform random variables on $[0,1]$. In each pair call the minimum $L$ and maximum $R$, then relabel whole pairs so $R_1<\cdots <R_n$. Merge all endpoints in $(R_{i-1},R_i]$ into vertex $i$, with $R_0=0$, and join the blocks containing the two endpoints of each pair, orienting from the right endpoint to the left. Self-loops are retained. This gives the uniform linearized chord diagram representation.

Before ordering, a pair has joint density $2$ on $0<L<R<1$. Consequently its maximum has distribution $\mathrm{Pr}(R\le r)=r^2$ and density $2r$; conditional on $R=r$, its minimum is uniform on $[0,r]$. Independence and relabelling give the equivalent precise descriptionEvery one of the $(2n)!$ relative orderings of the independently drawn labelled endpoints has the same probability. After forgetting pair labels and within-pair labels, each matching of $2n$ ordered points is produced by $2^{n}n!$ orderings. Thus its matching is uniform among the $(2n-1)!!$ pairings.

To prove equivalence with the recursive graph law, remove the last endpoint and its mate from a uniform pairing. The remaining ordered points have a uniform $(n-1)$-pairing, and the mate's old rank is uniform among the $2n-1$ possible ranks, independently of that smaller pairing. Conversely insert the mate into a uniformly chosen one of its $2n-1$ gaps, and pair it with the new final endpoint. For each old vertex block, the number of gaps assigned to it is exactly the number of endpoints already in that block, hence its old total vertex degree. These are the gaps immediately before each endpoint of that block. The last gap, after the last old endpoint, makes both new endpoints belong to the new final block and produces a self-loop. Therefore the attachment probabilities are $d_i/(2n-1)$ for old vertices and $1/(2n-1)$ for the new one. Induction proves the continuous construction and the recursive LCD model have the same distribution. Inserting a left endpoint into an old block raises exactly that vertex's vertex degree without splitting any old block.