Determining direct cause with some "realization" of `V \ {X, Y}` I'm bouncing around "Causality" by Judea Pearl.
On page 222 it offers this definition of a direct cause:

"$X$ is a direct cause of $Y$" if there exist two values $x$ and $x'$ of $X$ and a value $u$ of $U$ such that $Y_{xr}(u) \not= Y_{x'r}(u)$ where $r$ is some realization of $V  \setminus \{X, Y\}$.

My questions are:

*

*What is a "realization", is it the same as Wikipedia's Realization (probability) definition?

*What does the $\setminus$ symbol mean in the context of the two functions $X$ and $Y$? Can you give me an example?

*Finally, how do I use this definition in practice?

Let's say I have two structural causal models:

*

*$X \rightarrow Y$

*$X \rightarrow Q \rightarrow Y$
How does this definition of direct cause allow me to discover that $X$ is a direct cause in the first case, and $X$ is not a direct cause in the second case?
 A: Here Pearl is defining under what condition $X$ is a direct cause of $Y$.
In his notation $U$ refers to exogenous variables (variables coming from outside the model) and $V$ refers to endogenous variables (variables inside the model). See wikipedia for a bit more detail.
Q2. The \ symbol is typically used in reference to sets to mean "exclusion". So $V\backslash \{X,Y\}$ means the set of endogonous variables (not including $X$ and $Y$).
Q1. By realization, he is refering to the statistical interpretation to which you linked. He means, let's imagine that the variables in $V$ (not including $X$ and $Y$) take some values which we call $r$.
Q3. This question is a little more vague to answer, but the definition is (effectively) saying that "all other things being equal", changes in the value of $X$ will change the observed value of $Y$.
Q4. The definition let's you distinguish these two use cases since in the second case $V\backslash \{X,Y\}=Q$ and so a different value of $X$ will create a different realized value of $r$. I'm not 100% certain on this point though so perhaps someone else can explain this better.
A: I think Adam Kells has it mostly right, but let me write up a few more comments on your last question (which Adam calls Q4):
The notation $Y_x(u)$ is referring to the value of the effect $Y$ when the exogenous variables are equal to $u$, and the cause $X$ has been forced, via the do operator, to have the value $x.$ It follows that the expression
$$(\exists\,x,x')[x\not=x'\implies Y_{xr}(u)\not=Y_{x'r}(u)]$$
means that there exists $x,x'$ such that when we are force $V\setminus\{X,Y\}$ to have the realization $r$ for both cases of $x$ and $x',$ as well as forcing $X=x$ and $X=x'$ in two different scenarios, that the effect $Y$ takes on different values.
In my comment to Adam's answer, I have convinced myself that if the values of $Q$ in your $X\to Q\to Y$ causal graph are the same, $Y$ must be the same. Causal information can only flow all the way from $X$ to $Y$ if $X$ can "wiggle" $Q,$ which then "wiggles" $Y.$ By "wiggle", I mean "force a change in".
So in your scenario, if you have $x, x'$ for $X,$ but only one $Q=q=r,$ then $Y$ can only take on one value. Hence, $X$ cannot be a direct cause of $Y.$
