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

  • 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 $e^{-x}$. $\endgroup$ Nov 4, 2022 at 15:42
  • 1
    $\begingroup$ By "seems to" I really meant something like "does" $\endgroup$ 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. $\endgroup$ Nov 4, 2022 at 17:24
  • 1
    $\begingroup$ Also, I thought it was surprising that the distribution appears to be uniform between $0$ and $1$. $\endgroup$ 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 $(1, \infty)$. $\endgroup$ Nov 4, 2022 at 21:33

2 Answers 2


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.)

  • $\begingroup$ I guess you meant $\rho(t) = 1$ for $t \in (0, 1)$, not $\rho(t) = 0$, right? $\endgroup$ Nov 5, 2022 at 8:41
  • $\begingroup$ Thanks Ofir! On reflection I should have thought of the Dickman function $\endgroup$ Nov 5, 2022 at 10:15
  • $\begingroup$ BTW, your last paragraph is exactly why I asked $\endgroup$ Nov 5, 2022 at 10:16
  • $\begingroup$ @MateuszKwaśnicki Thanks, corrected. $\endgroup$ 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. $\endgroup$ Nov 5, 2022 at 12:06

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...

  • 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? $\endgroup$ 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
    Nov 5, 2022 at 0:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .