How can I derive the joint distribution for this Markov network?

Question

I am reading Bayesian Reasoning And Machine Learning and I'm not sure how to do exercise 4.6 on p.80.

The undirected graph:

represents a Markov network with nodes $x1, x2, x3, x4, x5$, counting clockwise around the pentagon with potentials $\phi(x_i,x_j)$. Show that the joint distribution can be written as $$p(x_1,x_2,x_3,x_4,x_5)=\frac{p(x_1,x_2,x_5)p(x_2,x_4,x_5)p(x_2,x_3,x_4)}{p(x_2,x_5)p(x_2,x_4)}$$

What I have tried so far:

We have that \begin{align} \frac{p(x_1,x_2,x_5)p(x_2,x_4,x_5)p(x_2,x_3,x_4)}{p(x_2,x_5)p(x_2,x_4)} &= \frac{p(x_1|x_2,x_5)p(x_2,x_5))p(x_5|x_2,x_4)p(x_2,x_4)p(x_2|x_3,x_4)p(x_3|x_4)p(x_4)}{p(x_2,x_5)p(x_2,x_4)} \\ &=p(x_1|x_2,x_5)p(x_5|x_4)p(x_2|x_4)p(x_3|x_4)p(x_4) \\ &=p(x_1|x_2,x_5)\phi(x_4,x_5)\phi(x_2,x_3)\phi(x_3,x_4)p(x_4) \end{align}

I'm not sure how to factorize it any further. Could anyone please help?

Answer 1

The conditional independencies induced by this Markov network are $$ x_1 \perp x_3,x_4 \mid x_2,x_5 \\ x_2 \perp x_4,x_5 \mid x_1,x_3 \\ x_3 \perp x_5,x_1 \mid x_2,x_4 \\ x_4 \perp x_1,x_2 \mid x_3,x_5 \\ x_5 \perp x_2,x_3 \mid x_1,x_4 $$ Using the chain rule, the joint distribution can be written as $$ p(x_1,x_2,x_3,x_4,x_5) = p(x_1 \mid x_2,x_3,x_4,x_5) \cdot p(x_3 \mid x_2,x_4,x_5) \cdot p(x_2,x_4,x_5) $$ Apply the conditional independencies derived above such that $$ p(x_1 \mid x_2,x_3,x_4,x_5) = p(x_1 \mid x_2,x_5) \\ p(x_3 \mid x_2,x_4,x_5) = p(x_3 \mid x_2,x_4) $$ So, $$ p(x_1,x_2,x_3,x_4,x_5) = p(x_1 \mid x_2,x_5) \cdot p(x_3 \mid x_2,x_4) \cdot p(x_2,x_4,x_5) $$ Using Bayes' rule, \begin{align} p(x_1,x_2,x_3,x_4,x_5) &= \frac{p(x_1,x_2,x_5)}{p(x_2,x_5)} \cdot \frac{p(x_3,x_2,x_4)}{p(x_2,x_4)} \cdot p(x_2,x_4,x_5) \\ &= \frac{p(x_1,x_2,x_5) \cdot p(x_3,x_2,x_4) \cdot p(x_2,x_4,x_5)}{p(x_2,x_5) \cdot p(x_2,x_4)} \end{align} which is the desired result.