Let $A_1,A_2,\ldots,A_n$ be independent subsets of probability space $(\Omega, \Sigma, P)$ (For every $I\subseteq \{1,2,\ldots,n\}$, $P(\bigcap_{j\in J}A_j)=\prod_{j\in J}P(A_j) )$. Prove that $1_{A_1},1_{A_2},\ldots,1_{A_n}$ is independent, i.e. for every Borel set $B_1,B_2,\ldots,B_n$ of $\mathbb{R}$ we have $P(\cap_{i=1}^n \{w\in \Omega: 1_{A_i}(w)\in B_i\} )= \prod_{i=1}^n P(\{w\in \Omega: 1_{A_1}(w)\in B_i\})$.

My partial answer:

Since $1_{A_i}(w) \in \{0,1\}$ for every $w\in \Omega$, then we have

$$ \{w\in \Omega: 1_{A_i}(w)\in B_i\}=\begin{cases} \emptyset & \text{if } 1\notin B_i \text{ and } 0\notin B_i ,\\ A_i & \text{if } 1\in B_i \text{ and } 0\notin B_i, \\ A_i^c & \text{if } 1\notin B_i \text{ and } 0\in B_i, \\ \mathbb{R} & \text{if } 1\in B_i \text{ and } 0\in B_i. \\ \end{cases} $$ First, we prove that if $A_1,A_2,\ldots,A_n$ is independent then $A_1^c,A_2^c,\ldots,A_n^c$ is independent. If there is $i_0$ such that $1\notin B_{i_0} \text{ and } 0\notin B_{i_0}$ then we have $$P\left(\bigcap_{i=1}^n \{w\in \Omega: 1_{A_i}(w)\in B_i\} \right)= 0=\prod_{i=1}^n P(\{w\in \Omega: 1_{A_1}(w)\in B_i\}).$$ Hence, we can assume for the case $1\in B_i \text{ or } 0\in B_i$ for every i.


  • $\begingroup$ $P(\bigcup_{j\in J}A_j)=\prod_{j\in J}P(A_j)$? Surely you mean $P(\bigcap_{j\in J}A_j)=\prod_{j\in J}P(A_j)$? $\endgroup$ Commented Apr 3, 2013 at 23:30
  • $\begingroup$ sory, typo bigcup and bigcap $\endgroup$
    – beginner
    Commented Apr 3, 2013 at 23:35

1 Answer 1


For every $i$, let $C_i=[\mathbf 1_{A_i}\in B_i]$. The goal is to show that $$ P\left[\bigcap_{i=1}^nC_i\right]=\prod_{i=1}^nP[C_i]. $$ Let us recall the, at first surprising but true, principle that random variables are easier than events, also formulated as expectations are easier than probabilities, thus, let us try to prove that $$ E\left[X\right]=\prod_{i=1}^nE[\mathbf 1_{C_i}],\qquad X=\prod_{i=1}^n\mathbf 1_{C_i}. $$ As you showed, for every $i$, $\mathbf 1_{C_i}=a_i\mathbf 1_{A_i}+b_i$ for some $a_i$ and $b_i$ in $\{0,1,-1\}$. Then, $$ X=\prod_{i=1}^n(a_i\mathbf 1_{A_i}+b_i)=Q((\mathbf 1_{A_i})_{1\leqslant i\leqslant n}), $$ where $Q$ is the polynomial defined as $$ Q((x_i)_{1\leqslant i\leqslant n})=\prod_{i=1}^n(a_ix_i+b_i). $$ There exists some coefficients $(q_I)_I$ indexed by the subsets $I$ of $\{1,2,\ldots,n\}$ such that $$ Q((x_i)_{1\leqslant i\leqslant n})=\sum_Iq_Ix_I,\qquad x_I=\prod_{i\in I}x_i. $$ When $x_i=\mathbf 1_{A_i}$ for every $i$, $x_I=\mathbf 1_{A_I}$ for every $I$, where $A_I=\bigcap\limits_{i\in I}A_i$. For each $I$, by the independence hypothesis of the events $(A_i)_i$, $E(\mathbf 1_{A_I})=\prod\limits_{i\in I}P[A_i]$.

Putting all these remarks together, one gets $$ E[X]=\sum_Iq_I\prod\limits_{i\in I}P[A_i]=Q((P[A_i])_{1\leqslant i\leqslant n}), $$ and finally, simply by the definition of the polynomial $Q$, $$ E[X]=\prod_{i=1}^n(a_iP[A_i]+b_i)=\prod_{i=1}^nP[C_i]. $$


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