The method of indicators for probability This problem, which is in the book "Probability Theory"(fourth edition) written by M. Loeve, states as follows,
Rule: In order to compute $PB$, $B = f(A_{1}, A_{1}^{c}, \cdots, A_{m}, A_{m}^{c})$, take the following steps:

*

*Reduce the operations on events to complementations, intersections,
and sums;

*Replace each event with its indicator, expand, and take
the expectation.

In this way find
\begin{equation*}
P(\bigcup_{j=1}^{m} A_{j}) \text{ and } P(\bigcap_{j=1}^{k} A_{j} \bigcup_{j=k+1}^{m} A_{j}^{c}) \text { in terms of } P(\bigcap_{j=1}^{r} A_{j})\text{'s}.
\end{equation*}
I was wondering how can $\bigcup_{j=1}^{m} A_{j}$ can be represented only by a set of $\bigcap_{j=1}^{r} A_{j}$ since the sequence $\{\bigcap_{j=1}^{r} A_{j}\}$ is decreasing with the largest value is $A_{1}$.
Any hint for solving it? Thanks.
 A: For any event $B$, let $1_B$ and $B^c$ denote the indicator and the complement of $B$, respectively. Let $A:=\bigcup_{j=1}^{m} A_j$.
Let $\binom{[m]}r$ denote the set of all subsets of cardinality $r$ of the set $[m]:=\{1,\dots,m\}$.
Then
$$1_A=1-1_{A^c}=1-\prod_{j=1}^m 1_{A_j^c}
=1-\prod_{j=1}^m (1-1_{A_j})
=1-\sum_{r=0}^m(-1)^r \sum_{J\in\binom{[m]}r}\prod_{j\in J} 1_{A_j}
=\sum_{r=1}^m(-1)^{r-1} \sum_{J\in\binom{[m]}r}\prod_{j\in J} 1_{A_j}.$$
Taking now the expectations, we get
$$P(A)
=\sum_{r=1}^m(-1)^{r-1} \sum_{J\in\binom{[m]}r}P\Big(\bigcap_{j\in J} A_j\Big).$$

In a comment, the OP insisted that $P(\bigcup_{j=1}^m A_j)$ be expressed only in terms of the probabilities of the form $P(\bigcap_{j=1}^r A_j)$. However, this is impossible to do, even for $m=2$. Indeed, let $x:=P(A_1)$, $y:=P(A_1\cap A_2)$, and $z:=P(A_2\setminus A_1)$. Then $P(A_1\cup A_2)=x+z$, which is clearly impossible to express in terms of $x$ and $y$.
Indeed, one can let e.g. $x=y=0$ and then take any $z\in[0,1]$. So, given $P(A_1)=0$ and (hence) $P(A_1\cap A_2)=0$, the probability $P(A_1\cup A_2)$ can take any value in $[0,1]$.
