What is the variance of the difference of two random-variable indicators with a chance of intersection between events? Let $X$ be an event whose probability P($X$) = $p$ and let $Y$  be an event whose probability is P($Y$) = $q$. The probabilit$Y$  of $X$ intersection with $Y$, $P(X \cap Y)$ = $r$.
$I_X$ is the indicator of the event $X$, i.e., when event $X$ happens, $I_X$ is equal to 1, otherwise, it's 0. Similarly, $I_Y$  is the indicator of $Y$  and it behaves in the same way $I_{X}$ does.

What is the variance of $(I_{X}- I_{Y})$?

My attempt so far is to take $(I_X - I_{Y})$ as $R$ and proceed with $var(R) = E(R^2) - (E(R))^2$ , where $E(.)$ is the expected value operator.
First, I'd calculate $E(R)$ as
$E(R) = E(I_X - I_{y} ) = E(I_{x} ) - E(I_{y} ) = (p - q)$. 
Thus, $(E(R))^2$ is $(p - q)^2$. 
Should I somehow account for a possible intersection here? I know the union of events M and N is calculated as:
$P(M \cup N) = P(M) + P(N) - P(M \cap N)$
But I don't know how to account for it in the difference of the indicator variables.
Second, 
$E(R^2) = E(I_{X}^2 - 2.I_{x} .I_Y  + I_{Y}^2) = E(I_{X}^2) - 2.E(I_{X} .I_{Y}) + E(I_{Y}^2)$. 
$I_{X}^2$ happens over $\{0,1\}^2 = \{0,1\}$, hence, the probability ratio between the possible outcomes is preserved. Thus, $E(I_{X}^2) = E(I_{X}) = p$. Is this assumption correct? If it is, consequently, $E(I_{Y}^2) = E(I_{Y}) = q$.
I could then use these results and plug them into $var(R) = E(R^2) - (E(R))^2$ but how would will I deal with the possible intersection?
I guess that's the part I feel unsure of: how does one account for a possible intersection?
In the way the problem is stated, I'd say they're independent events, ergo, 
$E(I_{X} .I_{Y} ) = E(I_{X} ) . E(I_{Y} ) = r$.
But I don't know if such an assumption would be correct.
 A: You don't need fancy notation at all. $I_X-I_Y=R$ is a random variable that takes on value $1$ if event $X$ occurs but event $Y$ does not (probability $p-r$), value $-1$ if event $Y$ occurs but $X$ does not(probability $q-r$, and value $0$ otherwise (when either both events occur or both do not) whose probability I could calculate but won't since the value is irrelevant. 
So,
$$E[I_X-I_Y] = E[R] = 1\cdot (p-r) + (-1)\cdot (q-r) + 0\cdot (\text{irrelevant})$$ and
$$E[(I_X-I_Y)^2] = E[R^2] = (1)^2\cdot (p-r) + (-1)^2\cdot (q-r) + 0\cdot (\text{irrelevant}).$$
Can you take it from here?
A: $E[I_X^2]=E[I_X]$ is correct because $I_X\in \{0,1\}$. This is same for $I_Y$. For the interaction term, $E[I_XI_Y]$, we have
$$\begin{align}E[I_XI_Y]&=P(I_X=1\cap I_Y=1)\times 1\\&+P(I_X=1\cap I_Y=0)\times 0\\&+P(I_X=0\cap I_Y=1)\times 0\\&+P(I_X=0\cap I_Y=0)\times 0\\&=P(I_X=1\cap I_Y=1)\\&=P(X\cap Y)=r\end{align}$$
The rest is substitution. You don't need to assume independence. In independence case, you'd have $pq=r$.
