There is a paper in JSTOR, The continuum random tree, I, by David Aldous, that is similar to your question. That is the first of a three part series of papers. However, it doesn't seem like Aldous' tree (or tree ensemble) can be the same as your tree. Your tree would have infinite valence everywhere, because the harmonic series diverges.

To expand a little on this alternate answer that was studied by David Aldous: Suppose that, instead of the process that Ian suggested, you take a random spanning tree of the complete graph on $n$ vertices. Or more properly, you take the ensemble of spanning trees of the complete graph, and then attempt a limit of those ensembles as $n \to \infty$. You can obtain a good ensemble in that limit by looking, for each $k$, at the $k$-neighborhood (in the tree) of a marked vertex, and then sending $k$ to infinity after $n$ is sent to infinity. Then you get a well-defined ensemble of infinite rooted trees, which is hopefuless the same as what Aldous studies. I think that it is a valid model of what is sometimes called "the" random tree.

For instance, what is the probability that the root is a leaf? Using the formula that there are $n^{n-2}$ labelled trees on $n$ vertices, you can compute that it is $1/e$. If I set up the tree properly, then it would mean that $1/e$ of the vertices are leaves, almost surely.

"The" random tree in this sense is statistically the same everywhere, to every finite size of neighborhood. However, I think that it can't be the same tree out to infinity every time, because it contains an infinite amount of data to distinguish samples. For instance the spacing between leaves is an infinite amount of data. I'm not expert enough to arrive at this rigorously, but I believe that a "translation invariant" random process that really gives you the same tree every time would have to give you a boring tree with the same valence everywhere. (This is what I meant by my joke that it would be barking up the wrong tree.) On the other hand, if you do make a "translation invariant" ensemble of trees, then it could be "the" tree in the sense that you might recover all of the local statistics of the entire ensemble just by average over one typical sample.

An analogue that I understand rather better is the Penrose tiling. It is "the" Penrose tiling in the sense that any one Penrose tiling has the local statistics of all Penrose tilings. On the other hand, there are uncountably many different Penrose tilings. (In a natural sense, the Penrose tilings form an ergodic family; they come from a foliation of an $A_4$-shaped torus by parallel planes.)

`i`

, take of the edges between`i`

and some smaller vertex at random; then repeat. The tree will have possess the same two properties, so it's all the same again. $\endgroup$ – Ilya Nikokoshev Nov 18 '09 at 20:24