Each vertex of the square has a value which is randomly chosen from a set.

Question

I don't need so much help computing this question (yet). However, I am still at the point of understanding what the question is really asking. If anyone understands it, and can explain it, that would be greatly appreciated!

I understand what pairwise and mutually independent are. I just don't understand how I would be able to show that. From what I understand, each vertex is randomly selected from {1, 2, ..., 6}. So say edge 1 is connected to a vertex that randomly selected 2, and another vertex that randomly selected 3 from the set, then the pdf would be X1 = 0 (because they are not the same value). Now say edge 2 connects two vertices that have randomly selected the same value, say 4, then its pdf would be X2 = 1 (since the vertices have the same value). I understand that they are pairwise independent (each set of pairs are independent), because they are randomly selected from the set {1, 2, ..., 6} but how can I show that? And how are X1, X2, X3, and X4 not mutually independent?

Answer 1

So say edge 1 is connected to a vertex that randomly selected 2, and another vertex that randomly selected 3 from the set, then the pdf would be $X_1 = 0$ (because they are not the same value).

Instead of fixing values for the vertices, look at the probability that $X_i=1$; let's call this $P(X_i)$. What is $P(X_1)$? What is $P(X_1 \cap X_2)$? How about $P(X_1 \cap X_2 \cap X_3)$ and $P(X_1 \cap X_2 \cap X_3 \cap X_4)$? Now use the definitions of pairwise independence and mutual independence.

Note also that for pairwise independence you should check the two kinds of pairs of edges: those that share a vertex, and those that are disjoint.

Answer 2

That $X_1$ and $X_3$ are independent is obvious.

Let $e_2=\{v_1,v_2\}$ and $e_3=\{v_2,v_3\}$. Whatever number vertex $v_2$ gets assigned we have $P(X_2=1)=P(X_3=1)={5\over6}$ and $P(X_2=1\ \wedge\ X_3=1)={25\over36}$; and similarly for the other cases.

On the other hand, if we, e.g., know that $X_1=X_2=X_3=0$ then we can be sure that $X_4=0$ as well. It follows that the $X_i$ are not mutually independent.