# Sum of a Poisson point process

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?

• "$Z$ 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 $e^{-x}$. Commented Nov 4, 2022 at 15:42
• By "seems to" I really meant something like "does" Commented Nov 4, 2022 at 15:51
• 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. Commented Nov 4, 2022 at 17:24
• Also, I thought it was surprising that the distribution appears to be uniform between $0$ and $1$. Commented Nov 4, 2022 at 17:26
• 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 $(1, \infty)$. Commented Nov 4, 2022 at 21:33

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.)

• I guess you meant $\rho(t) = 1$ for $t \in (0, 1)$, not $\rho(t) = 0$, right? Commented Nov 5, 2022 at 8:41
• Thanks Ofir! On reflection I should have thought of the Dickman function Commented Nov 5, 2022 at 10:15
• BTW, your last paragraph is exactly why I asked Commented Nov 5, 2022 at 10:16
• @MateuszKwaśnicki Thanks, corrected. Commented Nov 5, 2022 at 12:05
• @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. Commented 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...

• 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? Commented Nov 4, 2022 at 20:52
• 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.
– DRJ
Commented Nov 5, 2022 at 0:11