# Increments of Random Partitions

The following is problem 2.1.4. from Combinatorial Stochastic Processes:

Let $X_i$ be the indicator of the event that i the least element of some block of an exchangeable random partition $\Pi_n$ of [n]. Show that the joint law of the $(X_i,1\leq i\leq n)$ determines the law of $\Pi_n$.

As Pitman says, this is a result by Serban Nacu, the paper for which can be found here. In this post I’m going to explain what an exchangeable random partition is, how to prove the result, and a couple of consequences.

The starting point is the question ‘what is an exchangeable random partition?’ The most confusing aspect is that there are multiple definitions depending on whether the blocks of the partition are sets or just integers corresponding to a size. Eg, {1,2,4} u {3} is a partition of [4], corresponding to the partition 3+1 of 4. Obviously one induces the other, and in an exchangeable setting the laws of one may determine the laws of the other.

In the second case, we assume 3+1 is the same partition as 1+3. If order does matter then we call it a composition instead. This gets a bit annoying for set partitions, as we don’t want these to be ordered either. But if we want actually to talk about the sets in question we have to give them labels, which becomes an ordering, so we need some canonical way to assign these labels. Typically we will say $\Pi_n=\{A_1,\ldots,A_k\}$, where the curly brackets indicate that we don’t care about order, and we choose the labels by order of appearance, so by increasing order of least elements.

We say that a random partition $\Pi_n$ of [n] is exchangeable if its distribution is invariant the action on partitions induced by the symmetric group. That is, relabelling doesn’t change probabilities. We can express this functionally by saying

$\mathbb{P}(\Pi_n=\{A_1,\ldots,A_k\})=p(|A_1|,\ldots,|A_k|),$

for p a symmetric function. This function is then called the exchangeable partition probability function (EPPF) by Pitman.

Consider a partition of 4 into sets of sizes 3 and 1. There is a danger that this definition looks like it might be saying that the probability that A_1 is the set of size 3 is the same as the probability that A_1 is the set of size 1. This would be a problem because we expect to see some size-biasing to the labelling. Larger sets are more likely to contain small elements, merely because they contain more elements. Fortunately the definition is not broken after all. The statement above makes no reference to the probabilities of seeing various sizes for A_1 etc. For that, we would have to sum over all partitions with that property. It merely says that the partitions:

$\{1,2,3\}\cup\{4\},\quad \{1,2,4\}\cup\{3\},\quad\{1,3,4\}\cup\{2\},\quad \{2,3,4\}\cup\{1\}$

have respective probabilities:

$p(3,1),\quad p(3,1),\quad p(3,1),\quad p(1,3),$

and furthermore these are equal.

Anyway, now let’s turn to the problem. The key idea is that we want to be looking at strings of 0s and 1s that can only arise in one way. For example, the string 10…01 can only arise corresponding to the partitions {1,2,…,n-1} u {n} and {1,2,…,n-2,n} u {n-1}. So now we know p(n-1,1) and so also p(1,n-1). Furthermore, note that 10…0 and 11…1 give the probabilities of 1 block of size n and n blocks of size 1 respectively at once.

So then the string 10…010 can only arise from partitions {1,2,…,n-2,n} u {n-1} or {1,2,…,n-2} u {n-1,n}. We can calculate the probability that it came from the former using the previously found value of p(n-1,1) and a combinatorial weighting, so the remaining probability is given by p(2,n-2). Keep going. It is clear what ‘keep going’ means in the case of p(a,b) but for partitions with more than two blocks it seems a bit more complicated.

Let’s fix k the number of blocks in partitions under consideration, and start talking about compositions, that is $a_1+\ldots+a_k=n$. The problem we might face in trying to generalise the previous argument is that potentially lots of compositions might generate the same sequence of 0s and 1s, so the ‘first time’ we consider a composition might be the same for more than one composition. Trying it out in the case k=3 makes it clear that this is not going to happen, but we need some partial ordering structure to explain why this is the case.

Recall that a composition with k blocks is a sequence $a=(a_1,\ldots,a_k)$ which sums to n. Let’s say a majorizes b if all its partial sums are at least as large. That is $a_1+\ldots+a_l\geq b_1+\ldots+b_l$ for all $1\leq l \leq k$. We say this is strict if at least one of the inequalities is strict. It is not hard to see that if a majorizes b then this is strict unless a = b.

Since we don’t care about ordering, we assume for now that all compositions are arranged in non-increasing order. So we find a partition corresponding to some such composition $a_1,\ldots,a_k$. The partition is:

$\{1,\ldots,a_1\}\cup\{a_1+1,\ldots,a_1+a_2\}\cup\{a_1+a_2+1,\ldots,a_1+a_2+a_3\}\cup\ldots\cup\{n-a_k,\ldots,n\}.$

This generates a sequence of 0s and 1s as describe above, with $a_i-1$ 0s between the i’th 1 and the (i+1)th 1. The claim is that given some composition which admits a partition with this same corresponding sequence, that composition must majorize a. Proof by induction on l. So in fact we can prove Nacu’s result inductively down the partial ordering described. We know the probability of the sequence of 0s and 1s corresponding to the partition of [n] described by assumption. We know the probability of any partition corresponding to a composition which majorizes a by induction, and we know how many partitions with this sequence each such composition generates. Combining all of this, we can find the probability corresponding to a.

Actually I’m not going to say much about consequences of this except to paraphrase very briefly what Nacu says in the paper. One of the neat consequences of this result is that it allows us to prove in a fairly straightforward way that the only infinite family of exchangeable random partitions with independent increments is the so-called Chinese Restaurant process.

Instead of attempting to prove this, I will explain what all the bits mean. First, the Chinese Restaurant process is the main topic of the next chapter of the book, so I won’t say any more about it right now, except that its definition is almost exact what is required to make this particular result true.

We can’t extend the definition of exchangeable to infinite partitions immediately, because considering invariance under the symmetric group on the integers is not very nice, in particular because there’s a danger all the probabilities will end up being zero. Instead, we consider restrictions of the partition to $[n]\subset\mathbb{N}$, and demand that these nest appropriately, and are exchangeable.

Independent increments is a meaningful thing to consider since one way to construct a partition, infinite or otherwise, is to consider elements one at a time in the standard ordering, either adding the new element to an already present block, or starting block. Since 0 or 1 in the increment sequence corresponds precisely to these events, it is meaningful to talk about independent increments.