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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,

$$\begin{align} \Pr \{A_1 \leq a,A_2 \leq a\} &= E \left[ \mathbf{1}_{A_1 \leq a,A_2 \leq a}\right]\\ &=E \left[ \mathbf{1}_{A_1 \leq a} \mathbf{1}_{A_2 \leq a}\right]\\ &=E \left[ \mathbf{1}_{A_1 \leq a} \right] E \left[ \mathbf{1}_{A_2 \leq a}\right]+\operatorname{Cov}\left(\mathbf{1}_{A_1 \leq a},\mathbf{1}_{A_2 \leq a} \right)\end{align}$$

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    $\begingroup$ Are the $A_i$ events or random variables? Because if they are events, then I don't understand what $A_i \leq a$ means. $\endgroup$
    – Pedro M.
    Commented Jan 27, 2015 at 13:03
  • $\begingroup$ Thanks for pointing out mistake..Yes $A_i$ are random variables and $A_i \leq a$ are events. I have edited it. $\endgroup$
    – Oliver
    Commented Jan 28, 2015 at 0:25
  • $\begingroup$ If $N\geqslant3$, one can expect no formula in terms of expectations and covariances only, which are unable to describe higher order dependencies. $\endgroup$
    – Did
    Commented Jun 15, 2016 at 23:13
  • $\begingroup$ Maybe a covariance matrix would be useful? Or how about $E(XYZ)=E(X)E(YZ)+Cov(X,YZ)$ and work your way up from there? $\endgroup$
    – jdods
    Commented Jun 16, 2016 at 2:46

1 Answer 1

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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.

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  • $\begingroup$ Thanks for your observation. However, I want the expression for $N$ events and not for two events. If you have answer please let me know. $\endgroup$
    – Oliver
    Commented Jan 28, 2015 at 1:15

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