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:


1. Reduce the operations on events to complementations, intersections,
and sums;
2. 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.



The method of indicators for probability
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
$$P(A)
=\sum\_{r=1}^m(-1)^{r-1} \sum\_{J\in\binom{[m]}r}P\Big(\bigcap\_{j\in J} A\_j\Big).$$




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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 $[0,1]$.


