Determining unconditional independence in Markov Networks I would like to know whether $E \perp\kern-5pt\perp A $ in the following Markov Network and would like to know if my reasoning is correct:

So, since this is a Pairwise Markov Network, it factorizes into potentials. We have that $P(A,B,C,D,E) = \phi (A,B) \phi (A,C) \phi(B,D)\phi (C,D)\phi(D,E)\frac{1}{Z}$
where $Z$ is a normalization constant.
And so we have that the marginal for $E$ and $A$ is $P(E,A) \propto \sum_{B,C,D}\phi (A,B) \phi (A,C) \phi(B,D)\phi (C,D)\phi(D,E)$
Since $\sum_{x_1} \phi(x_1,x_2) \phi(x_1,x_3) = \phi(x_2,x_3)$
We have that $ \sum_{B,C,D}\phi (A,B) \phi (A,C) \phi(B,D)\phi (C,D)\phi(D,E) = \sum_{C,A} \phi(A,B)\phi(A,C)\phi(B,E,C) = \phi(B,E,C,A)$. Since $E$ and $A$ are inside one potential, this makes them dependent.
Would this be a correct way to determine unconditional independence in Markov Networks? Is there a faster, non-formula way to solve this?
 A: $\newcommand{\pperp}{\perp\kern-5pt\perp}
\newcommand{\npperp}{\not\perp\kern-8pt\perp}
$
I am not sure whether I understand your argument. In general, to show that $A\npperp E$, I don't think it is sufficient to show that $p(A, E) = \phi(A, E)$ for some potential $\phi$, but one has to show that this $\phi(A,E)$ cannot be factored as $\phi_A(A)\phi_B(B)$.
So my approach would be as follows:
To be extra precise, I think what you want to show is the following: The condition, that a probability density function (pdf) has as I-map your Markov Network (MN), is not sufficient to deduce $A\pperp E$. I.e., we have to show that there exists a pdf that has the MN as I-map and that has $A\npperp E$.
Thus, let's presume that all variables are binary. Then, such a pdf could be:
$$
\begin{align}
A &\sim U(\{0, 1\})\quad\mbox{(uniform binary random variable)}\\
B &= A\\
C &= A\\
D &= I(B+C-2)\\
E &= D,
\end{align}
$$
where $I()$ is the function that is zero everywhere except at zero where it is equal to one. Then:
$$
\begin{align}
p(E=0|A=0) &= 1 \\
p(E=0|A=1) &= 0,
\end{align}
$$
which means $A\npperp E$.
(If it vexes you a bit that some conditional probabilities are undefined in this MN, just jitter the Markov transition matrices only a tiny bit, so that all probabilities become positive. This would not change the dependence of $A$ and $E$.)

Your question: "Is there a faster, non-formula way to solve this?":
Yes, there is the easier, graph-based, way. For MNs, if you can presume that the graph is a P-map for the pdf, then $A \pperp E$ if and only if there is no path in the MN connecting $A$ and $E$.
