10
$\begingroup$

Let $Z_1, Z_2, \dots$ be a Poisson point process on $[0, 1]$ with intensity function $1/z$. What is the distribution of the sum $Z = \sum_{i=1}^\infty Z_i$?

One can construct $Z_1, Z_2, \dots$ by taking a standard Poisson point process $X_1, X_2, \dots$ on $[0, \infty)$ with intensity $1$ (so the number of points in any interval $I$ is Poisson-distributed with mean $|I|$) and setting $Z_i = \exp(-X_i)$.

The sum $Z$ seems to have mean $1$ and variance $1/2$. Empirical CDF and distribution function below. It looks like $Z$ is uniformly distributed on $[0,1]$ and exponentially distributed above $1$. Is something like that true?

CDF distribution function

$\endgroup$
5
  • 1
    $\begingroup$ "$Z$ seems to have mean $1$ and variance $1/2$": note $Z = \sum_{j=1}^\infty \prod_{i=1}^j e^{-W_i}$, where the $W_i$'s are i.i.d., each with PDF $e^{-x}$. $\endgroup$ Commented Nov 4, 2022 at 15:42
  • 1
    $\begingroup$ By "seems to" I really meant something like "does" $\endgroup$ Commented Nov 4, 2022 at 15:51
  • 1
    $\begingroup$ Down-voter: any constructive feedback? I believe the question to be mathematically precise and I was mainly wondering if some expert in probability might be familiar with this already. Happy to comment further if something is unclear. $\endgroup$ Commented Nov 4, 2022 at 17:24
  • 1
    $\begingroup$ Also, I thought it was surprising that the distribution appears to be uniform between $0$ and $1$. $\endgroup$ Commented Nov 4, 2022 at 17:26
  • 1
    $\begingroup$ On $(0, 1)$, the density indeed seems to be equal to $e^{-\gamma}$, where $\gamma$ is the Euler gamma constant. But it is not exactly exponential on $(1, \infty)$, I think. DRJ's answer gives a way to prove this rigorously: the Laplace transform of $Z$ is $\exp(-\gamma - \Gamma(0, z) - \log z)$, which has the leading term equal to $e^{-\gamma} (1 - e^{-z}) / z$, and the remainder which decays faster than $e^{-z}$ in the right complex half-plane. Thus, the remainder is the Laplace transform of a function supported in $(1, \infty)$. $\endgroup$ Commented Nov 4, 2022 at 21:33

2 Answers 2

10
$\begingroup$

The density function of $Z$ is the Dickman $\rho$ function, normalized (that is, divided by its mass, $e^{\gamma}$).

This function, $\rho\colon [0,\infty)\to (0,\infty)$, is defined via $\rho(t)=1$ for $t \in [0,1]$ and $\rho'(t)=-\rho(t-1)/t$ for $t \ge 1$. Equivalently, $u\rho(u) = \int_{u-1}^{u} \rho(t)dt$. This already shows $\rho(t) \le 1/\Gamma(t+1)$, so it decays superexponentially.

Its Laplace transform is given by $$\hat{\rho}(s):=\int_{0}^{\infty} e^{-st}\rho(t)dt = e^{\gamma+I(-s)}$$ where $I(s) = \int_{0}^{s} \frac{e^t-1}{t}dt$ (see Theorem III.5.10 in Tenenbaum's book, "Introduction to Analytic and Probabilistic Number Theory"; the same chapter contains a wealth of information on $\rho$). Plugging $s=0$ we get that $\int_{0}^{\infty} \rho(t)dt = e^{\gamma}$, so $\rho(t) e^{-\gamma}$ is indeed a density function.

From DRJ's answer, we know that if $f$ is your density function then $$\int_{0}^{\infty} e^{-st} f(t)dt = e^{-\int_{0}^{s} \frac{1-e^{-u}}{u}du}=e^{I(-s)}$$ by change of variables $u=-v$. Standard uniqueness properties imply $f \equiv \rho e^{-\gamma}$.


This function features prominently in number theory and probability. In number theory $\rho(u)$ arises as the probability that a number $x$ is $x^{1/u}$-smooth (or friable). Equivalently, the CDF of the random variable $\log P(n)/\log n$ ($n$ random from $\mathbb{Z}\cap[1,x]$, $P(n)$ the largest prime factor of $n$) tends to $\rho(1/\cdot)$ as $x \to \infty$. Relatedly, in probability, $\rho(1/\cdot)$ arises as the CDF of the first coordinate of a Poisson-Dirichlet process.

In the last examples it arises as a CDF but in your situation it is a density function, so let me give an example where it arises as the latter. It is the density function for the limit of $\sum_{k=1}^{n} k Z_k/n$, where $Z_k$ are independent Poisson with parameters $1/k$. This is intuitive if we work with Laplace transforms: $$\mathbb{E} e^{-s \sum_{k=1}^{n} k Z_k/n} = \prod_{i=1}^{n} \mathbb{E} e^{-s kZ_k/n} = \prod_{k=1}^{n} e^{\frac{1}{k}\left( e^{-sk/n}-1\right)} = e^{\sum_{k=1}^{n} \frac{1}{k}(e^{-sk/n}-1)}$$ and the exponent tends to $I(-s)$ with $n$. This is mentioned in "On strong and almost sure local limit theorems for a probabilistic model of the Dickman distribution" by La Bretèche and Tenenbaum, along with references.

A related nice fact: if you condition on $\sum_{k=1}^{n} k Z_k=n$, then the vector $(Z_1,\ldots,Z_k)$ becomes distributed like $(C_1(\pi_n),\ldots,C_n(\pi_n))$ where $\pi_n$ is a permutation chosen uniformly at random from $S_n$ and $C_i$ is its number of cycles of size $i$. This appears in the book "Logarithmic Combinatorial Structures" by Arartia, Barbour and Tavaré, specifically equation (1.15) and the discussion on page 26. (It could be that the book includes a discussion of the previous fact as well but I couldn't find it.)

$\endgroup$
6
  • $\begingroup$ I guess you meant $\rho(t) = 1$ for $t \in (0, 1)$, not $\rho(t) = 0$, right? $\endgroup$ Commented Nov 5, 2022 at 8:41
  • $\begingroup$ Thanks Ofir! On reflection I should have thought of the Dickman function $\endgroup$ Commented Nov 5, 2022 at 10:15
  • $\begingroup$ BTW, your last paragraph is exactly why I asked $\endgroup$ Commented Nov 5, 2022 at 10:16
  • $\begingroup$ @MateuszKwaśnicki Thanks, corrected. $\endgroup$ Commented Nov 5, 2022 at 12:05
  • $\begingroup$ @SeanEberhard Cool! Glad to help. This model with $Z_k$ is not that well known. I've added another fact that's good to know. $\endgroup$ Commented Nov 5, 2022 at 12:06
5
$\begingroup$

There is a formula for the Laplace transform of any additive functional of a Poisson process with intensity measure $\lambda$. Specifically, for any non-negative measurable function $f$,

$$E\left[e^{-\sum_{i=1}^\infty f(Z_i)}\right]=e^{-\int_0^\infty(1-e^{-f(u)})\lambda(du)}.$$

In your case, fixing $\theta>0$ and choosing $f(u)=\theta u$ yields

$$E\left[e^{-\theta Z}\right]=e^{-\int_0^\theta\left(\frac{1-e^{- u}}{u}\right) du}.$$

You can then easily compare this with the Laplace transform of any guess you might have for the distribution of $Z$. For example, if $Z$ had a density of the form $$f_Z(z) \propto 1_{(0,1)}(z)+e^{-\beta(z-1)}1_{(1,\infty)}(z),$$ for some constant $\beta>0$ as you suggest, then one would have $$E\left[e^{-\theta Z}\right]\propto \frac{1-e^{-\theta}}{\theta}+\frac{e^{-\theta}}{\theta+\beta}.$$ Those two expressions do not seem to match, so the answer to your question should be no...

$\endgroup$
2
  • 1
    $\begingroup$ The density seems to be $\gamma \cdot 1_{(0,1)}(z) + e^{-\beta(z-1)} 1_{(1,\infty}(z)$ with some $\beta, \gamma > 0$. May it be correct now? $\endgroup$ Commented Nov 4, 2022 at 20:52
  • 2
    $\begingroup$ I took $\gamma=1$ because the second plot of Sean indicates that the density is continuous at 1. Anyway, your more general version will still not work: the above computation shows that the Laplace transform $L$ of $Z$ solves the differential equation $L'(\theta)=L(\theta)\frac{1-e^{-\theta}}{\theta}$ and it is easy to check that the Laplace transform of your new candidate does not solve the same equation. $\endgroup$
    – DRJ
    Commented Nov 5, 2022 at 0:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.