Graphical dependence in the DAG X->Z<-Y In Barber's book pp. 40-41 he says that the belief network X->Z<-Y:

is "graphically dependent" since:
$$p(x,y|z) \propto p(z|x,y)p(x)p(y)$$
I don't understand why graphical dependence follows from the above equation? Is it because we don't see $p(x,y|z)\propto p(x|z)p(y|z)$?
Further, in a similar example he also says that the same graph represents $p(x,y)=p(x)p(y)$, i.e. that $x$ and $y$ are independent, which I assume follow from the definition: $p(x,y,z)=p(z|x,y)p(x)p(y)$.
Is he saying that $x$ and $y$ are independent, but dependent when conditioned on $z$?
Why is $x$ and $y$ graphically dependent in the above graph?
 A: The term "graphical dependence" seems to be specific to Barber's book. I have not seen it anywhere else. I think by graphical dependence, he means that there are arrows connecting variables together. So, the connected variables appear to be dependent. But actually they may or may not. They are just dependent (connected) graphically. 
In the graph, $x\rightarrow z \leftarrow y$, you have $x \not\perp y | z$ because you cannot get $p(x|z)p(y|z)$ from $p(x,y|z)$, as you said. In words, $x$ and $y$ are not conditionally dependent given $z$. In this graph, it is true that $x\perp y$. That is, $x$ is (marginally) independent of $y$ because
$p(x,y,z) = p(x)p(y) p(z|x,y)$
$p(x, y) = \sum_z p(x,y,z) = p(x)p(y) \overbrace{\sum_z p(z|x,y)}^1 = p(x)p(y).$
This graph kind of structure where there are two arrows pointing two one variable is called a V-structure. $z$ is also called a collider node.
A good concrete example to illustrate the concept is as follows. Imagine $x$ and $y$ are two independent coins with arbitrary probabilities for heads. Let $z \in \{0, 1\}$ where $z=1$ if $x=y$ and $z=0$ if $x\neq y$. By design, $x$ and $y$ are independent. So, $x \perp y$. Now assume $z=1$. Given $z=1$, it can be seen that knowing $x$ reveals the value of $y$. If $x=1$, you can deduce that $y=1$ because $z=1$ means $x$ and $y$ are the same. Here, given $z$ you  know more about $y$ by observing $x$ and vice versa. This effect is called "explaining away". Hence, $x \not\perp y | z$. In this example, we did not put any randomness in $z|x,y$. However, the principle is the same even with stochasticity.
