I am reading supplementary information of the paper Activity driven modeling of dynamic networks. It analogys the number of out degree of a activity node by Polya urns problem:
it will equal to the number of different balls extracted from a urn with $N$ balls, performing $Tma_i$ extractions. The probability of extracting $d$ balls will be given by
$$P(d)= \begin{pmatrix} N \\ d \end{pmatrix} p^d(1-p)^{(N-d)}$$
where
$$p=1-(1-\frac{1}{N})^{Tma_i}$$
is the probability of extracting at least one ball in the urn.
I just even dont know why to treated this as Binormial Distribution? How to get the probability of extracting at least one ball in the urn? What's the solving detail? Thanks for your consideration. Am I clear about my question? Do you need more information for answering my question?
It should looks like:
If there are more than one edges between two nodes, only one edge should be chosen. e6 e1 e8 treated as same ball. For the figure above there should be 9 balls in a urn.
Here is an account of what that maths does model.
Consider an urn with $M$ balls and one golden ball. Consider $E:= Tma_i$ selections, $X_1,X_2,\dots,X_E$ from the urn with replacement. $\mathbb{P}[X_i=\text{gold}]=\frac{1}{N}$ and the $X_i$ are pairwise independent.
What is the probability that the golden ball is selected at some point?
$$\begin{align} \mathbb{P}[\text{not selected}]&=\mathbb{P}[(X_1\neq \text{gold})\cap(X_2\neq\text{gold})\cap\cdots\cap(X_E\neq\text{gold})] \\&\prod_{i=1}^E\mathbb{P}[X_i\neq\text{gold}] \\&=\left(1-\frac{1}{N}\right)^E \\\Rightarrow \mathbb{P}[\text{selected}]&=1-\left(1-\frac{1}{N}\right)^E:=p. \end{align}$$
Now consider $N$ urns, each with $M$ balls, and each with a golden ball (suppose in addition the golden ball in urn $i$ has a little stamp "$i$" on it).
Now if we are going to make $E$ selections from the $N$ urns, then the probability that we select the golden ball at least once is $p$ in each case. Furthermore these probabilities are independent. Therefore if $Y$ is the number of urns for which the $E$ selections yields a golden ball, $Y\sim \operatorname{Bin}[N,p]$ and so $$\mathbb{P}[Y=d]={N\choose d}p^d(1-p)^{N-d}.$$