Define a *partition* of to be a finite or infinite multiset of real numbers in the interval (that is, an unordered set of real numbers in , possibly with multiplicity) whose total sum is : . For instance, is a partition of . Such partitions arise naturally when trying to decompose a large object into smaller ones, for instance:

- (Prime factorisation) Given a natural number , one can decompose it into prime factors (counting multiplicity), and then the multiset
is a partition of .

- (Cycle decomposition) Given a permutation on labels , one can decompose into cycles , and then the multiset
is a partition of .

- (Normalisation) Given a multiset of positive real numbers whose sum is finite and non-zero, the multiset
is a partition of .

In the spirit of the universality phenomenon, one can ask what is the natural distribution for what a “typical” partition should look like; thus one seeks a natural probability distribution on the space of all partitions, analogous to (say) the gaussian distributions on the real line, or GUE distributions on point processes on the line, and so forth. It turns out that there is one natural such distribution which is related to all three examples above, known as the *Poisson-Dirichlet distribution*. To describe this distribution, we first have to deal with the problem that it is not immediately obvious how to cleanly parameterise the space of partitions, given that the cardinality of the partition can be finite or infinite, that multiplicity is allowed, and that we would like to identify two partitions that are permutations of each other

One way to proceed is to random partition as a type of point process on the interval , with the constraint that , in which case one can study statistics such as the counting functions

(where the cardinality here counts multiplicity). This can certainly be done, although in the case of the Poisson-Dirichlet process, the formulae for the joint distribution of such counting functions is moderately complicated. Another way to proceed is to order the elements of in decreasing order

with the convention that one pads the sequence by an infinite number of zeroes if is finite; this identifies the space of partitions with an infinite dimensional simplex

However, it turns out that the process of ordering the elements is not “smooth” (basically because functions such as and are not smooth) and the formulae for the joint distribution in the case of the Poisson-Dirichlet process is again complicated.

It turns out that there is a better (or at least “smoother”) way to enumerate the elements of a partition than the ordered method, although it is random rather than deterministic. This procedure (which I learned from this paper of Donnelly and Grimmett) works as follows.

- Given a partition , let be an element of chosen at random, with each element having a probability of being chosen as (so if occurs with multiplicity , the net probability that is chosen as is actually ). Note that this is well-defined since the elements of sum to .
- Now suppose is chosen. If is empty, we set all equal to zero and stop. Otherwise, let be an element of chosen at random, with each element having a probability of being chosen as . (For instance, if occurred with some multiplicity in , then can equal with probability .)
- Now suppose are both chosen. If is empty, we set all equal to zero and stop. Otherwise, let be an element of , with ech element having a probability of being chosen as .
- We continue this process indefinitely to create elements .

We denote the random sequence formed from a partition in the above manner as the *random normalised enumeration* of ; this is a random variable in the infinite unit cube , and can be defined recursively by the formula

with drawn randomly from , with each element chosen with probability , except when in which case we instead have

Note that one can recover from any of its random normalised enumerations by the formula

with the convention that one discards any zero elements on the right-hand side. Thus can be viewed as a (stochastic) parameterisation of the space of partitions by the unit cube , which is a simpler domain to work with than the infinite-dimensional simplex mentioned earlier.

Note that this random enumeration procedure can also be adapted to the three models described earlier:

- Given a natural number , one can randomly enumerate its prime factors by letting each prime factor of be equal to with probability , then once is chosen, let each remaining prime factor of be equal to with probability , and so forth.
- Given a permutation , one can randomly enumerate its cycles by letting each cycle in be equal to with probability , and once is chosen, let each remaining cycle be equalto with probability , and so forth. Alternatively, one traverse the elements of in random order, then let be the first cycle one encounters when performing this traversal, let be the next cycle (not equal to one encounters when performing this traversal, and so forth.
- Given a multiset of positive real numbers whose sum is finite, we can randomly enumerate the elements of this sequence by letting each have a probability of being set equal to , and then once is chosen, let each remaining have a probability of being set equal to , and so forth.

We then have the following result:

Proposition 1 (Existence of the Poisson-Dirichlet process)There exists a random partition whose random enumeration has the uniform distribution on , thus are independently and identically distributed copies of the uniform distribution on .

A random partition with this property will be called the *Poisson-Dirichlet process*. This process, first introduced by Kingman, can be described explicitly using (1) as

where are iid copies of the uniform distribution of , although it is not immediately obvious from this definition that is indeed uniformly distributed on . We prove this proposition below the fold.

An equivalent definition of a Poisson-Dirichlet process is a random partition with the property that

where is a random element of with each having a probability of being equal to , is a uniform variable on that is independent of , and denotes equality of distribution. This can be viewed as a sort of stochastic self-similarity property of : if one randomly removes one element from and rescales, one gets a new copy of .

It turns out that each of the three ways to generate partitions listed above can lead to the Poisson-Dirichlet process, either directly or in a suitable limit. We begin with the third way, namely by normalising a Poisson process to have sum :

Proposition 2 (Poisson-Dirichlet processes via Poisson processes)Let , and let be a Poisson process on with intensity function . Then the sum is almost surely finite, and the normalisation is a Poisson-Dirichlet process.

Again, we prove this proposition below the fold. Now we turn to the second way (a topic, incidentally, that was briefly touched upon in this previous blog post):

Proposition 3 (Large cycles of a typical permutation)For each natural number , let be a permutation drawn uniformly at random from . Then the random partition converges in the limit to a Poisson-Dirichlet process in the following sense: given any fixed sequence of intervals (independent of ), the joint discrete random variable converges in distribution to .

Finally, we turn to the first way:

Proposition 4 (Large prime factors of a typical number)Let , and let be a random natural number chosen according to one of the following three rules:

- (Uniform distribution) is drawn uniformly at random from the natural numbers in .
- (Shifted uniform distribution) is drawn uniformly at random from the natural numbers in .
- (Zeta distribution) Each natural number has a probability of being equal to , where and .
Then converges as to a Poisson-Dirichlet process in the same fashion as in Proposition 3.

The process was first studied by Billingsley (and also later by Knuth-Trabb Pardo and by Vershik, but the formulae were initially rather complicated; the proposition above is due to of Donnelly and Grimmett, although the third case of the proposition is substantially easier and appears in the earlier work of Lloyd. We prove the proposition below the fold.

The previous two propositions suggests an interesting analogy between large random integers and large random permutations; see this ICM article of Vershik and this non-technical article of Granville (which, incidentally, was once adapted into a play) for further discussion.

As a sample application, consider the problem of estimating the number of integers up to which are not divisible by any prime larger than (i.e. they are –smooth), where is a fixed real number. This is essentially (modulo some inessential technicalities concerning the distinction between the intervals and ) the probability that avoids , which by the above theorem converges to the probability that avoids . Below the fold we will show that this function is given by the Dickman function, defined by setting for and for , thus recovering the classical result of Dickman that .

I thank Andrew Granville and Anatoly Vershik for showing me the nice link between prime factors and the Poisson-Dirichlet process. The material here is standard, and (like many of the other notes on this blog) was primarily written for my own benefit, but it may be of interest to some readers. In preparing this article I found this exposition by Kingman to be helpful.

Note: this article will emphasise the computations rather than rigour, and in particular will rely on informal use of infinitesimals to avoid dealing with stochastic calculus or other technicalities. We adopt the convention that we will neglect higher order terms in infinitesimal calculations, e.g. if is infinitesimal then we will abbreviate simply as .

** — 1. Constructing the Poisson-Dirichlet process — **

We now prove Proposition 2, which of course also implies Proposition 1.

Let , and let be as in Proposition 2. We note that , so if one wished, one could normalise . However, for minor technical reasons it will be convenient to allow to to be arbitrary.

We first understand the law of the sum of the Poisson process .

Lemma 5Let be as in Proposition 2. Then is an exponential variable with probability distribution . In particular, is almost surely finite.

*Proof:* Taking Laplace transforms, it suffices to show that

for all . We come the left-hand side by an informal argument using infinitesimals, leaving the rigorous version of the argument as an exercise to the reader. Partitioning into infinitesimal intervals , we expect each interval to contain exactly one element of with probability about , and to contain two or more elements with negligible probability. We can then write

(ignoring higher order terms) and thus

and thus on sending to zero

The quantity vanishes when and has an -derivative of , so on integrating we have

and (3) follows.

The same analysis can be performed for Poisson processes with more general intensity functions, leading to Campbell’s theorem. However, the exponential nature of leads to an important decoupling property:

Proposition 6Let be as in Proposition 2. Then the random variable and the normalisation are independent.

*Proof:* Let be disjoint infinitesimal intervals in , and let be another infinitesimal interval in , with infinitesimal compared with . We compute the probability of of the (infinitesimal) event that lies in and each of is occupied by an element of ; our goal is to show that this probability decouples into the product of a function of and a function of the . This event is (up to higher order errors) empty if , and is otherwise equal to the event that each of the is occupied by an element of , and that the sum of the remaining elements lies in . By definition of Poisson process, the former event occurs with probability

Once we condition on this event, is again a Poisson process of intensity , so by Lemma 5 is an exponential variable of parameter , so it lies in with probability

Since is infinitesimally close to , we conclude that the probability of is

giving the claim.

Note that the above calculation in fact computes the -point intensity function of as

We can now show the Poisson-Dirichlet property (2):

Proposition 7Let , and let be a random element, with each having a probability of equalling . Then is distributed uniformly in , and after one conditions on the value of , has the same distribution as .

*Proof:* We first establish that is distributed uniformly in . Consider the event that lies in an infinitesimal interval for some . This requires the interval to contain an element of , which occurs with probability , and then there is a probability that this element will be selected as . Thus has probability , giving the claim of uniform distribution.

Now consider the probability of the event that lies in , and that meets each of the disjoint infinitesimal , where we choose to be infinitesimal compared with . This event is empty if , and is otherwise equal (up to higher order errors) as the event that and that meets each of the . By (4) for applied at , we conclude that occurs with probability

which simplifies to

If we condition out by which has probability , we recover the same -point correlation function (4) as , and the claim follows.

Remark 8As a crude zeroth approximation to the process , one can consider the Poisson process on with intensity function . By (4), this process has the same -point intensity function as the true Poisson-Dirichlet process , but has the wrong higher point correlation functions (the cutoff is replaced by ). Related to this, is not a partition in general as there is no reason why should equal . Nevertheless, this zeroth approximation can be used as a crude heuristic in some cases. In the number theoretic setting, it corresponds to a Cramer-type random model in which each prime less than a “typical” number has an independent probability of of dividing .

Remark 9If one starts with a Poisson process of intensity for some parameter and normaliises it as above, one arrives at a more general Poisson-Dirichlet process whose enumeration variables are distributed according to a beta distribution rather than the uniform distribution.

** — 2. Cycle decomposition of random permutations — **

The proof of Proposition 3 is essentially immediate from the following claim:

Proposition 10Let and be as in Proposition 3. Let be a random element of , with each occuring with probability . Then is distributed uniformly among the multiples of , and after conditioning on a given value of , has the distribution of .

Heuristically at least, in the limit , this proposition asserts that the limiting distribution of the obeys the defining relation (2) of the Poisson-Dirichlet process. It is not difficult to make this argument rigorous, but we leave this as an exercise to the reader.

*Proof:* We first compute the probability that is equal to . Using the cycle interpretation of the enumeration of , this is equal to the probability that a randomly chosen element of lies in a cycle of of length . By permutation symmetry, it suffices to compute the probability that lies in a cycle of length . To count the number of permutations with this property, observe from direct counting that there are ways to select the cycle of length containing , and once this cycle is chosen there are ways to select the rest of the permutation , leading to ways in all. Dividing by the total number of permutations, we see that the cycle length occurs with probability , giving the required uniform distribution.

The same analysis shows that once one fixes the cycle of length that contains , the partition associated to the remaining cycles has the distribution of . Averaging over all the cycles of length containing , we obtain the second claim of the proposition.

** — 3. Asymptotics of prime divisors — **

We now prove Proposition 4. We first give a simple argument of Lloyd (as described by Kingman), which handles the much smoother (and hence much easier) third case of the theorem. In this case one can take advantage of Euler product factorisation. Indeed, if we define to be the point process of primes defined by setting each prime to occur with multiplicity in with probability , independently in , then we see from the fundamental theorem of arithmetic and the Euler product that the random variable has the same distribution as , where the product is with multiplicity. Setting and , we thus see that the point process associated to the prime divisors of this random number is the same process as . On the other hand, it is easy to see (from the same Mertens’ theorem calculation in the introduction) that in the limit as (or ), the point process converges (in the sense of Theorem 3) to the Poisson process with intensity , and the claim follows in this case from Proposition 2.

Now we prove the theorem in the first and second cases. Here, the Euler product factorisation is not available (unless one allows to be complex and uses something like Perron’s formula, but this turns the probabilities involved into complex numbers, and it is difficult to see how this can be made at all rigorous). Instead, we use (a slightly informal version of) the argument of Donnelly and Grimmett. The key proposition is the following analogue of Proposition 11, stated slightly informally:

Proposition 11Let be as in the first or second case of Proposition 4. Let be a random prime factor of , with each prime factor of having a probability of of being chosen as . Then the distribution of converges in the vague topology to the uniform distribution as . Also, if one conditions on a specific value of , then has the same distribution as (up to negligible errors).

Again, upon taking logarithms and sending we conclude, heuristically at least, that the limiting distribution of converges to a process that obeys (2) and is thus a Poisson-Dirichlet process. Again, one can make this argument rigorous; see the paper of Donnelly and Grimmett for details.

*Proof:* (Informal) For sake of concreteness let us work with the second case of Proposition 4, as the first case is similar (and also follows from the second case by a dyadic decomposition). We compute the probability that is equal to a given prime , and equal to a given multiple of . This requires lies in , and occurs with probability about

(This is slightly inaccurate in the case that is also divisible by , but this event is negligible in the limit ).

Summing in , we see that is equal to with probability about , and that once one conditions on the event , one essentially recovers the distribution of . Finally, the probability that lies in is essentially

which by Mertens’ theorems is asymptotic to , and the claim follows.

Remark 12The same argument also works when is drawn from the zeta distribution. Comparing this with Lloyd’s argument, this gives another proof of Proposition 2 (although ultimately it is just a discrete version of the previous, continuous, proof).

Remark 13A variant of Proposition 4 of relevance to the polymath8 project; if instead of selecting uniformly at random from , one instead performs a weighted distribution where each has a probability of being chosen as for some fixed , where is the number of prime factors of , then the same argument shows that is now asymptotic to a generalised Poisson-Dirichlet process in which the coordinates of the random enumeration remain independent, but now have the probability density function of rather than the uniform distribution. (This distribution is in fact of the form in Remark 9, with replaced by .) In particular, when is large, the elements of tend to cluster near the origin (most elements will have magnitude ), which helps explain the Motohashi-Pintz-Zhang phenomenon that when sieving for -tuples, the majority of moduli that one encounters in the sieve are quite smooth (or at least densely divisible).

** — 4. Dickman’s function — **

Let be the probability that the Poisson-Dirichlet process has no element larger than . Clearly for ; it is also not difficult to show that is continuous. It thus suffices to verify the delay-differential equation for .

Let . If we let be the process from Proposition 2, and , we see that is equal to the probability that for all . Taking (backwards) Newton quotients, we then conclude that is equal to the probability that for all , and that for the largest element of .

For any real number , let denote the event that , for all , and that , where is infinitesimal compared with . This is essentially the same as the event that contains an element in , that for all , and that . The first event occurs with probability . Conditioning on this event, is still a Poisson process with intensity with sum . By Lemma 5, the random variable is exponential with parameter and indepenent of the process . Thus the remaining two events defining are conditionally independent, with the former occuring with probability and the latter with probability . Thus the total probability that occurs is

which simplifies to ; integrating in , we conclude that is equal to , and the claim follows.

Remark 14Using (4) and the inclusion-exclusion formula, one also has the integral expressionfor , with the convention that the term is ; note that the summands vanish for and so only finitely many terms in this expression need to be evaluated. This gives another way to verify the delay-differential equation .

## 25 comments

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24 September, 2013 at 7:05 am

arch1In step 2 of the enumeration procedure, why isn’t the probability of equaling instead , since there are instances of remaining in the multiset at that point, each of which has probability of being chosen?

[Oops, that was a careless mistake; fixed now, thanks. -T.]24 September, 2013 at 5:53 pm

DThe steps 2–4 do not agree with the reconstruction in formula (1). When you pick u_1, do you divide remaining points by 1-u_1? In other words, u_2 does not come from the original list \Sigma, but from a rescaled list?

[Oops, you’re right; the way it was written was completely confusing. I hope I’ve fixed it now – T.]26 September, 2013 at 5:19 am

arch1In the recursive formula for Enum, should the multiplier therefore be ?

[Corrected, thanks – T.]12 February, 2020 at 9:52 pm

Peleg Michaeli پليج (@pelegmichaeli)I think this is still written wrongly, or at least unclear. Suppose , and suppose . Then is chosen from . Now suppose . You probably don’t want to have in the denominator in step 3.

[Corrected, thanks – T.]25 September, 2013 at 9:57 am

DavidThis Poisson-Dirichlet process is PD(1) in a more general family PD(t), see page 99 of Poisson processes by Kingman. The same construction, only the uniform distribution is replaced by Beta(1,t) distribution. The process PD(t) itself is PD(0,t) in more general family PD(a,t) in the paper of Pitman and Yor.

25 September, 2013 at 9:59 am

DIn Proposition 6, equality sign is missing in {\Sigma_N(\Gamma_a) = \frac{1}{S} \cdot \Gamma_a}.

[Corrected, thanks – T.]25 September, 2013 at 11:04 am

❧ ✲ ❂ ❂ ✲✳ ✴ ✶ ❂✯❀In Proposition 4, \Sigma_{PD} is \Sigma_{PF}.

[Corrected, thanks – T.]25 September, 2013 at 6:44 pm

DIn the case of Zeta distribution, the comment after the proof of Theorem 1 in Kingman’s paper implies that (asymptotically in x) multiplicities of any finite number of largest prime divisors are equal to one. I am assuming this also holds in the case 1 of Uniform distribution, so is it some basic result in number theory that the density of numbers with this property is equal to one? How can the number of largest prime divisors grow with x so that one still has this property?

25 September, 2013 at 8:01 pm

Terence TaoYes, it is rare for a large number to be divisible by a large prime with multiplicity more than one. This is basically because the sum is convergent, so that the sum goes to zero as goes to infinity. Thus, the proportion of numbers which are divisible by the square of a prime greater than goes to zero as goes to infinity. Among other things, this shows that the set of numbers whose largest prime factors occur with multiplicity one, has density one for any fixed .

5 October, 2013 at 3:36 pm

vcpIn the first sentence following the box containing Proposition 4, “by first done” can be deleted.

[Corrected, thanks – T.]15 October, 2013 at 8:37 pm

thichchaytronIt very difficulty with me :(

17 October, 2013 at 12:28 pm

XI think that there are some typos in section 1: Resp. 2 and 3 equations below (3), the left-hand side should be E exp(-sS), not E exp(-tS)

Also, “Theorem 2/3” should be “as in Proposition 2/3”

[Corrected, thanks – T.]5 November, 2013 at 1:17 pm

AnitaIs it trivial that if they are non-zero and sums to 1, the multiset must be countable?

6 November, 2013 at 10:47 pm

AnitaOf course it is, ignore this.

13 November, 2013 at 3:30 am

Grigory FTypo in the third line after equation (1): should read instead.

[Corrected, thanks – T.]21 November, 2013 at 5:33 pm

Sary DSome signs should probably be swapped in section 4 : $u\rho'(u)=-\rho(u-1)$ and probability $-du\rho'(u)$, since rho is decreasing.

I love reading your blog !

[Corrected, thanks – T.]14 March, 2014 at 11:16 am

The Chinese Restaurant Process | Eventually Almost Everywhere[…] The Poisson-Dirichlet process, and large prime factors of a random number (terrytao.wordpress.com) […]

14 March, 2014 at 11:17 am

Urns and the Dirichlet Distribution | Eventually Almost Everywhere[…] The Poisson-Dirichlet process, and large prime factors of a random number (terrytao.wordpress.com) […]

15 July, 2015 at 2:44 pm

Cycles of a random permutation, and irreducible factors of a random polynomial | What's new[…] If is a randomly selected large integer of size , and is a randomly selected prime factor of (with each index being chosen with probability ), then is approximately uniformly distributed between and . (See Proposition 9 of this previous blog post.) […]

18 March, 2016 at 12:07 am

sarojcan any body give me the proof of ”(estimation of s^(2/B))<infinite''????

16 August, 2018 at 10:40 am

Karl KepplerDickman function arises in a problem proposed here:

https://math.stackexchange.com/questions/2130264/sum-of-random-decreasing-numbers-between-0-and-1-does-it-converge

31 October, 2018 at 8:53 am

MartinHi Terry!

Your blog post asked what’s new in prime numbers. I have something I think is new I would like to show you and the readers of your blog. I hope you don’t mind and are gracious enough to post this.

Kind regards,

Martin

Claim: “All prime numbers can be transformed into an integer sequence equivalent to one-half the sum of the first 2n + 1 primes.”

Proof:

1) Sum each consecutive pair of prime numbers (i.e. 2,3 then 3,5, etc.)

2) Subtract the difference between the two numbers.

3) Divide the result by 4.

4) Keep a continual sum from step 3.

5) At odd intervals the continual sum is equivalent to 1/2 half the sum of the first 2n + 1 primes.

1, 5, 14, 29, 50, 80, 119, 164, 220, 284, 356, 437, 530, 632, 740, 860, 994, 1138, 1292, 1457, 1633, 1819, 2014, 2219, 2444, 2675, 2915, 3169, 3435

30 December, 2019 at 8:53 am

Separating big sticks from little sticks, and applications – Random Permutations[…] Tao on the stick-breaking process, or the Poisson–Dirichlet process (which is essentially the same); […]

30 March, 2020 at 9:37 am

NoamI think there is a mistake in the calculation of the integral that appears in the proof that S is an exponential variable – I believe it should be log(a/(s+a)) and not – log(s+a).

[Corrected, thanks – T.]3 October, 2020 at 7:38 am

WillSome kind of malformed formatting tag in the proof of Proposition 7 has gone mad with power and underlined half this post.

[Corrected, thanks – T.]