Let $\{A_n\}_{n=1}^\infty$ be a sequence of events.
If we let $I_n$ be the indicator r.v. such that $I_n=1$ when $A_n$ occurs and $I_n=0$ if $A_n$ does not occur.
I know that $P(A_n) = E(I_n)$
But why is it that $P(A_j \cap A_k) = E(I_j I_k)$?
Further: If we define $\eta_n = \sum_{k=1}^n I_k$, then
$\sum_{j=1}^n\sum_{k=1}^n P(A_j \cap A_k) = E(\eta_n^n)$
I don't understand this statement either, can someone please help
Best Answer
The distribution of the product $I_j I_k$ is $$I_j I_k = \begin{cases} 1 & \text{if } I_j = 1 \thinspace \land\thinspace I_k = 1 \iff A_j \cap A_k ,\\ 0 & \text{otherwise.} \end{cases}$$ So $\mathbb{E}(I_j I_k) = 1 \cdot \mathbb{P}(A_j \cap A_k) + 0 = \mathbb{P}(A_j \cap A_k)$.
For the second part: $$\sum_{j = 1}^n \sum_{k = 1}^n \mathbb{P}(A_j \cap A_k) = \sum_{j = 1}^n \sum_{k = 1}^n \mathbb{E}(I_j I_k) = \mathbb{E}\left(\sum_{j = 1}^n \sum_{k = 1}^n I_j I_k\right) = \mathbb{E}\left(\sum_{j = 1}^n I_j \sum_{k = 1}^n I_k\right) = \mathbb{E}(\eta_n^2).$$