The indicator function for a probability event $A \subset \Omega$ is given by
$ \mathbf{1}_A(x) =\begin{cases} 1 & \text{if }x \in A \\ 0 & \text{if }x \notin A. \end{cases}$
Consider $N$ dependent events $A_1 \leq a,A_2 \leq a,\cdots,A_N \leq a \subset \Omega.$ Now, we want to evaluate the probability $ \Pr \{A_1 \leq a,A_2 \leq a,\cdots,A_N \leq a\}$, which can be written in terms of the indicator functions as
$\Pr \{A_1 \leq a,A_2 \leq a,\cdots,A_N \leq a\} = E \left[ \mathbf{1}_{A_1 \leq a,A_2 \leq a,\cdots,A_N \leq a}\right].$ How to arrive at the next step in terms of covariance of the indicator random variables?
Example: Consider the 2 events case. Then,
$$\begin{align} \Pr \{A_1 \leq a,A_2 \leq a\} &= E \left[ \mathbf{1}_{A_1 \leq a,A_2 \leq a}\right]\\
&=E \left[ \mathbf{1}_{A_1 \leq a} \mathbf{1}_{A_2 \leq a}\right]\\ &=E \left[ \mathbf{1}_{A_1 \leq a} \right] E \left[ \mathbf{1}_{A_2 \leq a}\right]+\operatorname{Cov}\left(\mathbf{1}_{A_1 \leq a},\mathbf{1}_{A_2 \leq a} \right)\end{align}$$
Best Answer
By the definition of covariance, $$\text{Cov}(X,Y)=E[XY]-E[X]E[Y]$$
In this case, we take $X=1_{A_1\leq a}$ and $Y=1_{A_2\leq a}$. Then $XY=1_{A_1\leq a,A_2\leq a}$ and your result follows.