There's a problem on indicator variables that involves calculating intersections for an indicator variable. I am stumped, I thought I might reach out for help on the theory.
The problem is
"Let $A,B,C$ be three events, and let $X = I_A, Y = I_B,$ and $Z = I_C$, be the associated indicator random variables. Give an algebraic expression, in the spirit of $X*Y$ is the indicator random variable of the event $A \cap C$, involving $X,Y$ and $Z$ for the indicator random variable of the following events."
The event $A^{c} \cap B \cap C$
This seems pretty clear, it is the probability of not $A$ and $B$ and $C$ and I thought it might be $(1 - X) * Y * Z$
The part that really stumps is the last part. It says at most two of the events $A,B,C$ occurred. My thinking is that in set notation this would be $(A \cup B) \cap (A \cup C) \cap (B \cup C)$ but then when I think about indicator random variables, unions stop making sense.
How do you describe the union of variables through algebra?
Use $$I_{A\cup B}=I_A+I_B-I_AI_B=1-(1-I_A)(1-I_B).$$
Then $$I_{A\cup B\cup C}=I_{A\cup B}+I_C-I_{A\cup B}I_C =I_A+I_B+I_C-I_AI_B-I_AI_C-I_BI_C+I_AI_BI_C.$$ Note the resemblance to inclusion/exclusion. Alternatively, $$I_{A\cup B\cup C}=1-(1-I_{A\cup B})(1-I_C) =1-(1-I_A)(1-I_B)(1-I_C).$$
But the event "at most two of $A$, $B$, $C$ occur" is $(A\cap B\cap C)^c$ with indicator function $1-I_{A\cap B\cap C}$,