Solution
= Solution
The graph factorizes as
$$
p(x)=p(x_2)p(x_1\mid x_2)p(x_4\mid x_2)
p(x_3\mid x_1,x_4)p(x_5\mid x_2)
p(x_6\mid x_4,x_5).
$$
Conditioning on all variables except $X_1$, terms not involving $x_1$ cancel, leaving
$$
p(x_1\mid x_2,x_3,x_4,x_5,x_6)
\propto p(x_1\mid x_2)p(x_3\mid x_1,x_4).
$$
This depends only on $(x_2,x_3,x_4)$, so
$$
X_1\perp(X_5,X_6)\mid(X_2,X_3,X_4).
$$
Thus $(X_2,X_3,X_4)$ is a <Markov blanket> of $X_1$.