When programmers hear the phrase “random tree,” they most likely think of a random binary search tree, i.e., a binary search tree built from the insertion of a uniformly random permutation of $$n$$ keys—denote such a tree by $$\mathrm{BST}_n$$. A mathematician might instead think that a random tree’’ is more likely to be a uniformly random tree taken from some class, like the set of all ordered binary trees with $$n$$ nodes—denote such a tree by $$\mathrm{UNIF}_n$$. Of course, neither would be wrong. It should be clear, though, that these two distributions on the space of binary trees are quite different. In particular, most undergraduate students of computer science learn, through the study of comparison-based sorting algorithms, that $\mathbf{E}\{\mathrm{height}(\mathrm{BST}_n)\} = \varTheta(\log n) ,$ and some will learn that $\mathbf{E}\{\mathrm{height}(\mathrm{UNIF}_n)\} = \varTheta(\sqrt{n}) .$ Though random binary search trees might seem more immediately relevant to programmers, uniformly random binary trees are part of a bigger picture which is comparatively more versatile in the probabilistic analysis of algorithms. To this end, we introduce the concept of Galton-Watson trees.

Galton-Watson trees were originally used to model the spread and ultimate extinction of aristocratic family names, as part of the eugenics craze of the late 19-th and early 20-th century. It is a mathematical model of asexual reproduction. Starting with one individual, each individual has an independently random number of children, which each have their own equidistributed independently random number of children, and so on. The distribution of the underlying ordered family tree is known as a Galton-Watson tree, which is determined by its offspring distribution, i.e., the distribution of the number of children which each individual births. We write $$\xi$$ for a random variable distributed as such. For example, we might have the following offspring distribution: $\mathbf{P}\{\xi = 0\} = \mathbf{P}\{\xi = 2\} = 1/4, \quad \mathbf{P}\{\xi = 1\} = 1/2 ;$ in words, each individual has $$0$$ or $$2$$ children with probability $$1/4$$, and $$1$$ child with probability $$1/2$$. Importantly, in this case, there is a positive probability that individuals can have no progeny, so there is a positive probability that the Galton-Watson tree determined by $$\xi$$ dies off. If each individual represents a person with a given family name, this would indicate the extinction of the family name.

What follows is a fascinating and classical result about Galton-Watson tree extinction:

Theorem 1.
Let $$q$$ denote the probability that a $$\xi$$-Galton-Watson tree goes extinct. Then, $$q = 1$$ if and only if $$\mathbf{E} \xi \le 1$$.

Accordingly, there are three regimes:

• the subcritical regime, when $$\mathbf{E} \xi < 1$$;

• the critical regime, when $$\mathbf{E} \xi = 1$$, and;

• the supercritical regime, when $$\mathbf{E} \xi > 1$$.

The critical regime is of paramount importance. Though it may not be obvious to a first-time observer, uniformly random binary trees are actually critical Galton-Watson trees, and indeed, critical Galton-Watson trees model many kinds of “uniform” random trees.

To see this fact about uniformly random binary trees, we must first control the number of nodes in a Galton-Watson tree. We define a conditioned Galton-Watson tree to be a Galton-Watson tree conditioned to have $$n$$ nodes, and denote it by $$T_n$$. So, for the $$\xi$$-Galton-Watson tree $$T$$ and specific tree $$t$$ on $$n$$ nodes, $\mathbf{P}\{T_n = t\} = \mathbf{P}\{T = t \mid |T| = n\} = \frac{\mathbf{P}\{T = t\}}{\mathbf{P}\{|T| = n\}} ,$ where $$|T|$$ denotes the number of nodes in $$T$$.

For certain offspring distributions, we may be conditioning on an event which occurs with zero probability. For example, if $\mathbf{P}\{\xi = 0\} = \mathbf{P}\{\xi = 2\} = 1/2 ,$ then every $$\xi$$-Galton-Watson tree has an odd number of nodes, so $$n$$ must be odd. We will ignore this technicality.

Sticking with this choice of $$\xi$$, observe that if $$t$$ is a binary tree with $$n = 2k + 1$$ nodes, then $$n$$ has $$k$$ internal nodes and $$k + 1$$ leaves, so by independence of the number of offspring of each node in the Galton-Watson process, $\mathbf{P}\{T = t\} = ( \mathbf{P} \{\xi = 0\} )^{k + 1} ( \mathbf{P}\{\xi = 2 \} )^k = 2^{-n} .$ Since $$t$$ was an arbitrary binary tree on $$n$$ nodes, this means that $$T_n$$ is, after all, a uniformly random binary tree.

This is just one way of viewing a uniformly random tree as a critical Galton-Watson tree. Some other important offspring distributions include:

• the $$\mathrm{Geometric}(1/2)$$ offspring distribution, $\mathbf{P}\{\xi = i\} = 2^{- (i + 1) } ,$ corresponding to uniformly random ordered trees (with any valence);

• the $$\mathrm{Poisson}(1)$$ offspring distribution, $\mathbf{P}\{\xi = i\} = \frac{e^{-1}}{i!} ,$ corresponding to uniformly random unordered labelled trees, and;

• the $$\mathrm{Binomial}(d, 1/d)$$ offspring distribution, $\mathbf{P}\{\xi = i\} = \binom{d}{i} \frac{1}{d^i} \left(1 - \frac{1}{d}\right)^{d - i} ,$ corresponding to uniformly random (up to-) $$d$$-ary trees.

Critical Galton-Watson trees have many marvelous qualities which are too numerous to even summarize here. In particular, writing $$\mathrm{Var}\{\xi\} = \sigma^2$$, if $$0 < \sigma^2 < \infty$$, then $\mathbf{E}\{\mathrm{height}(T_n)\} = \varTheta(\sqrt{n}) ,$ and this height is even highly concentrated (Addario-Berry et al., 2013).

Even more fascinating is the local limit to the size-biased tree, also known as Kesten’s tree. To describe this structure, first we introduce the size-biased random variable $$\hat{\xi}$$, where $\mathbf{P}\{\hat{\xi} = i\} = i \mathbf{P}\{\xi = i\} .$

Since $$\xi$$ is critical, then the size-biased distribution is indeed a probability distribution, i.e., $\sum_{i \ge 0} \mathbf{P}\{\hat{\xi} = i\} = \sum_{i \ge 0} i \mathbf{P}\{\xi = i\} = \mathbf{E} \xi = 1 .$

With this, we construct the infinite Kesten’s tree, denoted by $$T_\infty$$. This random tree has a set of mortal and immortal nodes, starting with one immortal node at the root. Each immortal node has an independently random number of children distributed as $$\hat{\xi}$$, where one among these is chosen as its immortal successor, and all others are made mortal. Each mortal node has an independently random number of children distributed as $$\xi$$.

Observe that $$\mathbf{P}\{\hat{\xi} = 0\} = 0$$, so each immortal has at least one child, and on average $$\mathbf{E} \hat{\xi} = 1 + \sigma^2$$ children. Thus, Kesten’s tree has an infinite “spine” along the sequence of immortal nodes, with several independent $$\xi$$-Galton-Watson trees hanging off this spine on either side.

For a tree $$t$$, let $$[t]_r$$ denote the tree $$t$$ truncated to include only the nodes of depth at most $$r$$. The following classical result describes the local limit to Kesten’s tree mentioned earlier:

Theorem 2
Suppose that $$\mathbf{E} \xi = 1$$. Then, for each $$r \ge 1$$ and each tree $$t$$, $\lim_{n \to \infty} \mathbf{P}\{ [T_n]_r = t \} = \mathbf{P}\{ [T_\infty]_r = t\} .$

So, in the limit, conditioned Galton-Watson trees look like Kesten’s tree.

It may seem that $$T_n$$ “grows into” $$T_\infty$$ in some sense, if only because they are the same in the limit and because $$T_\infty$$ is “larger” than $$T_n$$, but this is not always true. To be more precise, the question is whether or not there exists a coupling of $$T_n$$ and $$T_\infty$$ for which $$T_n \subset T_\infty$$, where “$$\subset$$” denotes subtree inclusion at the root along some subsequence of children. More generously, some have asked whether or not conditioned Galton-Watson trees can be grown incrementally, i.e., whether or not there exists a coupling between $$T_n$$ and $$T_{n + 1}$$ for which $$T_n \subset T_{n + 1}$$. Equivalently, this is asking whether or not $$T_{n + 1}$$ can be grown from $$T_n$$ by the insertion of a leaf. Recall that inserting leaves uniformly at random in a binary tree creates a random binary search tree, so certainly if we have any hope of creating uniform random binary trees, the insertion profile must be radically nonuniform.

Note that if $$T_n \subset T_{n + 1}$$ for each $$n$$, then $$T_n \subset T_\infty$$, so incremental growth implies growth into Kesten’s tree. Observe also that this kind of incremental growth can be extremely useful to study monotone properties of $$T_n$$, since it allows us to use the coupling between $$T_n$$ and $$T_\infty$$ to transfer the study of a property of $$T_n$$ onto the nearly-Galton-Watson Kesten’s tree $$T_\infty$$, which carries a lot of independence. If finiteness of the tree is required, we can even truncate $$T_\infty$$ at some level around $$\sqrt{n}$$, since the height of $$T_n$$ is highly concentrated.

In (Luczak and Winkler, 2004), it is shown that for any $$d \ge 2$$ and $$\xi \sim \mathrm{Binomial}(d, 1/d)$$, $$T_n$$ can indeed be grown incrementally. There are a few more results about growing conditioned Galton-Watson trees—see (Lyons et al., 2008), (Broman, 2014), (Broman, 2016)—but as far as I am aware, these are virtually the only papers including positive results about which offspring distributions admit incremental growth or growth into Kesten’s tree.

In the reverse direction, (Janson, 2006) shows that for some critical offspring distributions and for some $$n$$, there is no coupling for which $$T_n \subset T_{n + 1}$$, or $$T_n \subset T_\infty$$. We summarize the argument here, since it is quite simple.

Let $$Z_k(t)$$ denote the number of nodes at depth $$k$$ in the tree $$t$$. Observe that $\mathbf{E} Z_k(T_\infty) = \sigma^2 + \mathbf{E} Z_{k - 1}(T_\infty) = 1 + k \sigma^2 .$ If $$T_n \subset T_\infty$$, then $\mathbf{E} Z_k (T_n) \le \mathbf{E} Z_k (T_\infty) = 1 + k \sigma^2$ for each $$k$$, and if $$T_n \subset T_{n + 1}$$, then $\mathbf{E} Z_k (T_n) \le \mathbf{E} Z_k(T_{n + 1})$ for each $$k$$. Janson proposes the following offspring distribution: $\mathbf{P}\{\xi = 0\} = \mathbf{P}\{\xi = 2\} = \frac{1 - \varepsilon}{2}, \quad \mathbf{P}\{\xi = 1\} = \varepsilon ,$ for some fixed $$\varepsilon > 0$$, and then simply enumerates all small trees to compute that $\mathbf{E} Z_1 (T_3) > \mathbf{E} Z_1 (T_4)$ for a small enough choice of $$\varepsilon$$, where we emphasize that $$Z_1$$ is simply the degree of the root node.

So indeed, some small conditioned Galton-Watson trees do not grow incrementally. When I first saw this result, I thought that perhaps this is a small $$n$$ problem, and that maybe for large enough $$n$$, it may still be true that $$T_n \subset T_{n + 1}$$. After some more thinking, I realized that I was wrong.

Pick the following offspring distribution for some fixed $$d \ge 2$$: $\mathbf{P}\{\xi = 0\} = \left(1 - \frac{1}{d}\right)^2, \quad \mathbf{P}\{\xi = 1\} = \frac{1}{d}, \quad \mathbf{P}\{\xi = d\} = \frac{1}{d}\left(1 - \frac{1}{d}\right) .$ It can be verified that $$\mathbf{E} \xi = 1$$ and $$\sigma^2 = d - 2 + 1/d$$, so $\mathbf{E} Z_1 (T_\infty) = d - 1 + \frac{1}{d} .$ We also know the distribution of the root degree for conditioned Galton-Watson trees from (Janson, 2012), so \begin{aligned} \mathbf{P}\{Z_1(T_n) = i\} &= \frac{n}{n - 1} i \mathbf{P}\{\xi_1 = i \mid \xi_1 + \xi_2 + \dots + \xi_n = n - 1\} \\ &= \frac{n i \mathbf{P}\{S_{n - 1} = n - 1 - i\}}{(n - 1) \mathbf{P}\{S_n = n - 1\}} , \end{aligned} where $$\xi_1, \dots, \xi_n$$ are independent and identically distibuted as $$\xi$$, and $$S_n = \xi_1 + \dots + \xi_n$$. Using tedious computations through local limit theorems (Petrov, 1975), we can estimate the above probabilities in order, finally, to compute that $\mathbf{E} Z_1(T_n) = d - 1 + \frac{1}{d} + \frac{d^2 + o(d^2)}{2n} + o(1/n) ,$ where the function $$o(d^2)$$ does not depend on $$n$$.

What this implies is that we can pick $$d$$ large enough so that $$d^2$$ dominates $$o(d^2)$$ in the numerator, and then, for all sufficiently large $$n$$, $\mathbf{E} Z_1 (T_n) > \mathbf{E} Z_1 (T_\infty) .$ So, there are even arbitrarily large conditioned Galton-Watson trees which still cannot be grown within Kesten’s tree.

1. Broman, E.I., 2016. Stochastic ordering of infinite geometric Galton-Watson trees. J. Theoret. Probab. 29, 1069–1082. https://doi.org/10.1007/s10959-015-0608-x
2. Broman, E.I., 2014. Stochastic ordering of infinite binomial Galton-Watson trees. ALEA Lat. Am. J. Probab. Math. Stat. 11, 209–227.
3. Addario-Berry, L., Devroye, L., Janson, S., 2013. Sub-Gaussian tail bounds for the width and height of conditioned Galton-Watson trees. Ann. Probab. 41, 1072–1087. https://doi.org/10.1214/12-AOP758
4. Janson, S., 2012. Simply generated trees, conditioned Galton-Watson trees, random allocations and condensation. Probab. Surv. 9, 103–252. https://doi.org/10.1214/11-PS188
5. Lyons, R., Peled, R., Schramm, O., 2008. Growth of the number of spanning trees of the Erdos-Rényi giant component. Combin. Probab. Comput. 17, 711–726. https://doi.org/10.1017/S0963548308009188
6. Janson, S., 2006. Conditioned Galton-Watson trees do not grow, in: Proceedings, Fourth Colloquium on Mathematics and Computer Science Algorithms, Trees, Combinatorics and Probabilities (Nancy, 2006). pp. 331–334.
7. Luczak, M., Winkler, P., 2004. Building uniformly random subtrees. Random Structures Algorithms 24, 420–443. https://doi.org/10.1002/rsa.20011
8. Petrov, V.V., 1975. Sums of Independent Random Variables. Springer-Verlag Berlin Heidelberg New York.