I was reading the paper on arixv.

I was confused the equation of nth moment of Poisson distribution.

The detail and partial paper as follow:


For large N, this connection probability takes the form [28] $$\Pi_T (a,a^\prime) = 1 − exp[−λ(a + a^\prime)]$$, (1) where $\lambda = T/N$ and we have set $m = 1$ to simplify calculations. The topological properties of the integrated network are encoded in the propagator $g_T (k|a)$, defined as the probability that a node with activity $a$ has integrated degree $k$ at time $T$ , and whose generating function $\hat g_T (z|a) = \sum_k g_T (k|a)z^k$ satisfies the general equation [27,28] $$ln \hat g_T (z|a) = N\sum_{a^\prime} F(a^\prime) ln[1 − (1 − z)\Pi_T(a,a^\prime)]$$. (2) From the propagator, the degree distribution of the integrated network at time $T$ is trivially given by $$P_T(k) =\sum_a F(a)g_T (k|a)$$. (3) In the limit of small $\lambda$ or $N\gg T$ with constant $T$, which we assume from now on, the connection probability Eq. (1) can be approximated as $$\Pi_T(a,a^\prime)\simeq \lambda(a + a^\prime)$$. (4) In this same limit $\lambda\rightarrow 0$, Eq. (2) can be solved, leading to a propagator with the form of a Poisson distribution with mean $T (a + \langle a\rangle)$. From it, we obtain an asymptotic degree distribution [28] $$P_T(k)\simeq \frac{1}{T}F(\frac{k}{T}-\langle a \rangle)$$. (5)


Here says Poisson distribution with mean $T (a + \langle a\rangle)$.


The position of the percolation threshold can be simply obtained by considering that $u = 1$ is always a solution of Eq. (8), corresponding to the lack of giant component. A physical solution $u < 1$, corresponding to a macroscopic giant component, can only take place whenever $G_1^\prime(1) > 1$, which leads to the Molloy-Reed criterion [32]: $$\frac{\langle k^2\rangle_T}{\langle k\rangle_T}> 2$$, (9) where $\langle k^n\rangle_T=\sum_k k^n P_T(k)$ is the $n$th moment of the degree distribution at time $T$ . In the case of the activity-driven model, these moments can be computed noticing, from Eq. (3), that $\langle k^n\rangle_T = \sum_a F(a)\sum_k k^n g_T(k|a)$. Since the propagator has the form of a Poisson distribution, the moments of the degree distribution simply read as $$\langle k^n\rangle_T =\sum_{m=1}^n\genfrac{\{}{\}}{0pt}{}{n}{m}T^m\kappa_m$$, (10) where $\genfrac{\{}{\}}{0pt}{}{n}{m}$ are the Stirling numbers of the second kind [33] and $$\kappa_m =\sum_a F(a)(a+\langle a\rangle)^m=\sum_{i=0}^m \binom mi \langle a^i\rangle \langle a\rangle^{m-i}$$. (11)

Well, I find that

Moments of the Poisson distribution

If X is a random variable with a Poisson distribution with expected value $\lambda$, then its $n$th moment is


So, Does $T\kappa$ is the expected value of Poisson distribution in the paper?

Why $\kappa$ is not $(\sum_a F(a)(a+\langle a\rangle))^m$ or $(a + \langle a\rangle)^m$? I am confused with $\kappa_m$.

Should I supply more information to my question? Did the title suite for my question?


1 Answer 1


It is the conditional distribution $g(k|a)$ that has a Poisson distribution, with mean $\lambda(T,a)=T(a+\langle a\rangle)$. So the unconditional average $$\langle k^n\rangle = \sum_a F(a)\langle k^n|a\rangle,$$ with $$\langle k^n|a\rangle\equiv\sum_k k^n g(k|a)=\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}\lambda^m$$ is given by $$\langle k^n\rangle= \sum_a F(a)\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}\bigl[T(a+\langle a\rangle)\bigr]^m.$$ This is the equation from the paper, with $$\kappa_m=\sum_a F(a)(a+\langle a\rangle)^m=T^{-m}\sum_a F(a)\lambda^m.$$

This is the relation between the expected value $\lambda$ of the Poisson distribution and $\kappa_m$ requested in the OP.

  • $\begingroup$ $\langle k^n \rangle = \sum_a F(a)\langle k^n|a \rangle=\sum_a F(a) \sum_k k^n g(k|a) =\sum_a F(a) \sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}\lambda^m =\sum_a F(a)\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}[T(a+\langle a\rangle)]^m =\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}T^m\sum_a F(a)(a+\langle a\rangle)^m =\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}T^m\kappa_m$ $\endgroup$
    – Nick Dong
    Dec 14, 2017 at 16:04

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