Proving independence of events Let $A$, $B$ and $C$ be mutually independent indicator variable (also called dummy variable, &gt;Boolean indicator, binary variable).
Prove that the events $(A = B)$, $(B = C)$ are independent if and only if:
a) $A$ is uniform (i.e. $P(A=1)=P(A=0)=0.5$)
OR
b) $C$ is uniform
OR
c) $B$ is constant(i.e. $P(B=1) = 0$ OR $1$).

I can't even figure out even where to start. Any help would be really appreciated.
Thanks a lot!
For convenience, let $a = P(A=1)$, $b=P(B=1)$, $c = P(C=1)$.  Then $P(A=B) = ab + (1-a)(1-b)$,
$P(B=C) = bc + (1-b)(1-c)$, and $P(A=B=C) = abc + (1-a)(1-b)(1-c)$.
You want to solve $P(A=B=C) = P(A=B) P(B=C)$.
Try expanding out $P(A=B+C) - P(A=B) P(B=C)$ and factoring.  If the problem is correct, $a - 1/2$, $c - 1/2$, $b - 1$ and $b$ should be factors.
I'm the OP. Could not comment above. So I'm writing here. 

What am I to solve exactly here? 

$$P(A=B=C)=P(A=B)P(B=C)$$ or
$$P(A=B+C)−P(A=B)P(B=C)$$

Also, what is $$P(A=B+C)$$ equal to?

One last thing: To show that the events [A=B] and [B=C], we need to show that $$P( A=B \cap B = C) = P(A=B) \cdot P(B = C)$$, right? I'm not sure how this relates to solving the above equations.

Many thanks!

-Nick!