2
$\begingroup$

In the paper by Airoldi, Blei, Fienberg, Xing, 2008, sometimes I see a distinction between $\phi_{p\rightarrow q}$ and $\phi_{q\leftarrow p}$ and sometimes not (the non-distinction in the generative process) when considering the asymmetric directed graphs.

Does this distinction exist or not?

In other words, in a directed network of $N$ individuals, do we have $N-1$ of $\phi_{p\rightarrow .}$'s and $N-1$ of $\phi_{.\leftarrow p}$'s for each individual $p$?

$\endgroup$
2
$\begingroup$

They are certainly distinct: Each $\phi_{p \to q}$ and $\phi_{p \leftarrow q}$ is a variational parameter corresponding to $z_{p \to q}$ and $z_{p \leftarrow q}$. These refer, respectively, to the (unobserved, latent) group membership of $p,q$ when $p$ connects to $q$. From the NIPS paper on the model:

The indicator vector $z_{p \to q}$ denotes the specific block membership of node $p$ when it connects to node $q$, while $z_{p\leftarrow q}$ denotes the specific block membership of node $q$ when it is connected from node $p$.

In other words, the ordering of the nodes in the subscript denotes the direction of the connection: $p$ before $q$ implies $p$ is connecting to $q$. The tail of the arrow faces the node whose membership is referenced.

There exist $2N^2$ of them, which you can see intuitively in Fig. 1:

enter image description here

$\endgroup$
  • $\begingroup$ In the generative process for a link between $p$ and $q$, meaning $p\rightarrow q$, the community indicators drawn are $z_{p\rightarrow q}$ and $z_{q\rightarrow p}$. In the picture however it is $z_{p\rightarrow q}$ and $z_{p\leftarrow q}$. My question is whether $z_{q\rightarrow p}$ is any different from $z_{p\leftarrow q}$ $\endgroup$ – A.Yazdiha Nov 15 '16 at 9:59
  • $\begingroup$ @A.Yazdiha Why don't you e-mail one of the four authors to ask? $\endgroup$ – Chill2Macht Nov 18 '16 at 14:28
  • $\begingroup$ @A.Yazdiha Ah, I see I erred in typing my arrows, see my edit. Though, the description in the monks example of the fuller is somewhat at odds with how I interpret the description quoted above, so you may wish to follow William's advice and contact one of the authors. $\endgroup$ – Sean Easter Nov 22 '16 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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