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.


$E(R^2) = E(I_{X}^2 - 2.I_{x} .I_Y + I_{Y}^2) = E(I_{X}^2) - 2.E(I_{X} .I_{Y}) + E(I_{Y}^2)$.

$I_{X}^2$ 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.


2 Answers 2


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?

  • $\begingroup$ But what is the variance $\endgroup$ Jun 8, 2021 at 22:55

$E[I_X^2]=E[I_X]$ 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 $pq=r$.


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