How can I show that these two variables in a Markov network are marginally independent? I am reading "Bayesian Reasoning And Machine Learning" and I'm doing exercise 4.2 on page 79.
This is the exercise:
Consider the Markov network
$$p(a,b,c)=\phi(a,b)\phi(b,c)$$
Nominally, by summing over $b$, the variables $a$ and $c$ are dependent. For binary $b$, explain a situation in which this is not the case, so that marginally, $a$ and $c$ are independent.
My attempt:
We have that $$p(a,b,c)=p(a,c)=\sum_bp(a,b,c)=\sum_b\phi(a,b)\phi(b,c) \\ p(a)=\sum_{a,b}\phi(a,b)\phi(b,c)\\p(c)=\sum_{a,b}\phi(a,b)\phi(b,c)\\ \Rightarrow p(a,c)\neq p(a)p(c)$$
We let $b\in\{-1,1\}$,
$$\phi(a,b=1)=0\\ \phi(a,b=-1)=1 \\ \phi(c,b=1)=0 \\ \phi(c,b=-1)=1$$
Now, $$p(a,c)=\sum_b\phi(a,b)\phi(b,c)=0(0)+1(1)=1\\ p(a)p(c)=\sum_{a,b}\phi(a,b)\phi(b,c)\sum_{a,b}\phi(a,b)\phi(b,c)\\=(0+1)(0+1)=1$$
So in this situation $p(a,c)=p(a)p(c)$ and hence $a$ and $c$ are marginally independent.
Is this correct?

The conditional independency induced by this Markov network is
$$
a \perp c \mid b
$$
Perhaps a simpler way of answering this question is to first write
\begin{align}
p(a,c) &= \sum_b p(a,c \mid b) \cdot p(b) \\
&= p(a,c \mid b = 0) \cdot p(b = 0) + p(a,c \mid b = 1) \cdot p(b = 1)
\end{align}
We could then ask: how do we choose $p(b = 0)$ (or $p(b = 1)$) such that
$$
p(a,c) = p(a) \cdot p(c)
$$
A good choise would be to let $p(b = 0) = 1$. In other words, $b$ is a constant equal to $0$. This means that the equation above simplifies to
$$
p(a,c) = p(a,c \mid b = 0)
$$
We then use the conditional independency derived above to write
$$
p(a,c) = p(a \mid b = 0) \cdot p(c \mid b = 0)
$$
Since we know that $b$ is a constant, then observing it does not add any more information about $a$ or $c$ (see this question for details), and so $a$ is independent of $b$ and $c$ is independent of $b$, such that
$$
p(a,c) = p(a) \cdot p(c)
$$
