In a Bayesian graphical model, suppose that we have a random variable $B$ whose parent is the random variable $A$. So there is an arrow from $A$ to $B$, and this means that the joint distribution is mechanistically factored as: $\Pr(A,B) = \Pr(A) \Pr(B|A)$. In particular, the probability of any realization $B=b$ is specified by single realizations of $A$, e.g. $\Pr(B=b | A=a)$.

I'd like to know how to graphically represent a situation where $B$ depends not just on a realization of $A$ but also on the expectation value of $A$. So I want something like $\Pr(B=b | A=a, \langle A\rangle)$. An example might be that $B$ is normally distributed given $A=a$ with the mean $\mu = \langle A\rangle - a$. As far as I can tell, there isn't a simple way to graphically represent this sort of dependence.

One route would be to introduce another random variable $C$ which is a distribution over distributions of $A$. But since I my interest is alwys with a single joint distribution $\Pr(A,B)$ it seems silly to introduce some peaked distribution over distributions. Even then, I don't know how I'd then show the dependence on a particular realization of $A$.

So what is a clear and useful way to represent a distributional dependency in a directed graphical model?


If you take a Bayesian perspective (in which parameters are treated as random variables), you can introduce a new variable C that represents the mean of A, and have both A and B depend on it, and B still depends on A as well.

  • $\begingroup$ This works for the simple situation I described. Any thoughts on how it could be extended to handle a case such as: $A \to B \to C \to D \to E$. I would want, $\Pr(E | D=d)$ to also depend on the distribution $\Pr(D)$, not just on the realization $d$. In this situation, the only input that relates to your suggestion goes into $A$ and then this induces a distribution $\Pr(D)$. $\endgroup$
    – Tom
    May 18 '15 at 21:44
  • 1
    $\begingroup$ Can you follow the same pattern as the first situation? Have a new node related to the distribution of D, which connects to both D and E? $\endgroup$
    – DavidR
    May 19 '15 at 2:03

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