I asked this on math.stackexchange.com with no luck.. any thoughts?

I'm currently reading a text on Bayesian networks and the text is giving some very crude interpretations of what appear to be some of the most important foundations of the subject.

It states the following:

Theorem 1.2.7 (Parental Markov Condition): A necessary and sufficient condition for a probability distribution P to be Markov relative a [directional acyclic graph] G is that every variable be independent of all it's nondescendants (in G), conditional on it's parents.

I understand the premise of what is happening here. Given some point X in the graph, if you condition over all parent nodes, it should be independent of all non-descendants. I guess the confusion for me arrises around how this makes the process Markov relative. Maybe that is the weak part of my understanding. Could someone please provide a example of graph which fails this condition and hence fails being Markovian relative to this graph?

I've done some study on Markov Chains and I feel like maybe their definitions differ in some way that's been lost on me.


  • $\begingroup$ Crosspost: math.stackexchange.com/questions/909300. Posting a copy of the same question in both stats.stackexchange and math.stackexchange is not desired. Rather, you should have flagged the post at math.SE and asked a moderator to migrate it to stats.SE. $\endgroup$ Commented Aug 27, 2014 at 11:00
  • $\begingroup$ @JuhoKokkala Can easily do in the future. Would you recommend deleting this one and migrating the other? $\endgroup$ Commented Aug 28, 2014 at 0:29

1 Answer 1


Being Markov is a property of the distribution, not the graph (although it is only defined relative to a given graph). A graph can't be Markov or fail to be Markov, but a distribution can fail to be Markov relative to a given graph.

Here is an example in terms of causal networks. Say we know that $X_1$ influences $X_2$ and $X_3$, but $X_2$ and $X_3$ don't influence each other. Then we can represent the true causal relationships with the graph $G$:

G = X_2 <- X_1 -> X_3

The distribution $P(X_1, X_2, X_3)$ should represent that $X_2$ and $X_3$ become independent conditional on $X_1$, so the $P(X_1, X_2, X_3)$ is Markov relative to $G$.

Now let's say we did not observe $X_1$, but only $X_2$ and $X_3$. Remember that neither of $X_2$ or $X_3$ influences the other. So we could represent the direct causal relationships between $X_2$ and $X_3$ using $G'$, the subgraph over those nodes:

G' = X_2, X_3

In this case, the marginal distribution $P(X_2, X_3)$ is not Markov with respect to $G'$. $X_2$ is dependent on its non-descendent $X_3$, conditional on its parents in $G'$ (i.e. the empty set). This is because $G'$ omits the common cause $X_1$.

This example shows that the Markov condition is only reasonable when your graph includes all common causes of any pair of nodes in the graph (an assumption called "causal sufficiency"). Note that you can include a variable in the graph even if you did not observe it - $P(X_1, X_2, X_3)$ is Markov relative to $G$ even if I am missing data for $X_1$.

Markov chains are similar, and often used to represent time series. The Markov assumption represents that the next state of the time series depends only on the current state, and not on any previous states. Here's an example representation:

time series


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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