# Bayesian Network misunderstanding

I am learning about Bayesian networks (and PGMs in general) and I am stuck into this. Basically I'm trying to find a mistake in my reasoning.

Consider this example (copied from https://youtu.be/Nis3O4CVJAU?t=632):

I can obtain the given correct result by using the chain rule if I traverse the nodes from top to bottom (i.e. from $$N$$ to $$1$$):

\begin{align} p(x_1, \dots, x_N) &= \prod_{k=1}^N p(x_k | x_{k-1}, \dots, x_1) \\ &= p(x_N | x_{N-1} \dots, x_1) \cdots p(x_2 | x_1) p(x_1) \end{align}

In this case $$N=7$$ and, from the graph: \begin{align} p(x_7 | x_6 \dots, x_1) &= p(x_7 | x_4, x_5) \\ p(x_6 | x_5 \dots, x_1) &= p(x_6 | x_4) \\ &\cdots \\ p(x_3 | x_2, x_1) &= p(x_3) \\ p(x_2 | x_1) &= p(x_2) \\ p(x_1) &= p(x_1) \\ \end{align}

The problem appears when I try to traverse the nodes from $$1$$ to $$N$$:

\begin{align} p(x_1, \dots, x_N) &= p(x_1 | x_{2}, \dots, x_7) p(x_2 | x_{3}, \dots, x_7) \cdots p(x_7) \\ \end{align}

For the root nodes, this seems reasonable to me: \begin{align} p(x_1 | x_{2}, \dots, x_7) &= p(x_1)\\ p(x_2 | x_{3}, \dots, x_7) &= p(x_2)\\ p(x_3 | x_{4}, \dots, x_7) &= p(x_3)\\ \end{align}

But then I have:

\begin{align} p(x_4 | x_{5}, \dots, x_7) &= p(x_4) \\ p(x_5 | x_{6}, x_7) &= p(x_5) \\ p(x_6 | x_7) &= p(x_6) \\ p(x_7) &= p(x_7) \\ \end{align}

Which, on the one hand, has to be wrong, because: $$p(x_1, \dots, x_N) =^{??} p(x_1)\cdots p(x_7)$$

And therefore the variables would be independent (!?).

But, on the other hand, we have to apply the property defined in the graph (is this right??): $$p(x_3 | x_1, x_2, x_4, x_5, x_6, x_7) = p(x_3)$$

So, I'm guessing the mistake is that it doesn't follow from that property that $$p(x_3 | x_6, x_7) = p(x_3)$$

But if that's the case, why is it the following correct? $$p(x_3 | x_1, x_2, x_4, \dots, x_7) = p(x_3) \implies p(x_3 | x_1, x_2) = p(x_3)$$

I suspect I may have the definition of the property wrong, but I'm not sure.

• You have to be careful, $x_1, x_4$ and $x_5$ are dependent, in particular $p(x_1 | x_2, ..., x_7) = p(x_1 | x_4, x_5)$. – mbartczak Sep 13 at 22:44

Your independence assumptions are in fact not reasonable at all.

I'm not sure what gave you the idea that, for example,

$$p(x_1 | x_2,\ldots,x_7)=p(x_1).$$

No variable in a Bayesian network can be assumed to be independent of its immediate neighbors.

There should be some discussion in your text about d-separation, Markov blankets, colliders, etc. but essentially for A-junctions, as seen in the relationship between $$x_1,x_4,x_5$$. The variables $$x_4$$ and $$x_5$$ are independent precisely when one conditions on $$x_1$$, while for V-junctions, such as seen in the relationship between $$x_1,x_2,x_4$$, the variables $$x_1$$ and $$x_2$$ are independent precisely when one doesn't condition on $$x_4$$.