Help in Independence of events Question: For events $A_1, A_2, \ldots, A_n$ consider the $2^n$ equations $P(B_1\cap\ldots\cap B_n)=P(B_1)\ldots P(B_n)$ with $B_i=A_i$ or $B_i=A_i^c$ for each $i$. Show that $A_1,\ldots,A_n$ are independent if all these equations hold.
Note that this is different of the definition when $B_i$ is either $B_i=A_i$ or $B_i=\Omega$. I have no clue how to do this.
Thanks in advance.
 A: We can use the formula of inclusion-exclusion:
$$P\left(\bigcup_{j=1}^NE_j\right)=\sum_{j=1}^N(-1)^{j-1}\prod_{J\subset \{1,\ldots,N\},|J|=j}P\left(\bigcap_{i\in J}E_i\right).$$
A direction is obvious: if the $2^n$ equations are true then taking $B_i=A_i$ for all $i$ we get what we want. We show the converse. Let $I_0:=\{j\in\{1,\ldots,n\},B_j=A_j^c\}$. We can write 
\begin{align*}
P\left(\bigcap_{j=1}^nB_j\right)&=P\left(\bigcap_{j\in I_0}B_j\cap\bigcap_{i\in I_0^c}B_j\right)\\\
&=P\left(\bigcap_{j\in I_0}A_j\cap\bigcap_{i\in I_0^c}A_j^c\right)\\\
&=P\left(\bigcap_{j\in I_0}A_j\right)-P\left(\bigcap_{j\in I_0}A_j\cap\left(\bigcap_{i\in I_0^c}A_i^c\right)^c\right)\\\
&=\prod_{j\in I_0}P(A_j)-P\left(\bigcup_{i\in I_0^c}\left(\bigcap_{j\in I_0}A_j\cap A_i\right) \right)\\\
&=\prod_{j\in I_0}P(A_j)-\sum_{l=1}^{|I_0^c|}(-1)^{l-1}\prod_{J\subset I_0^c|,|J|=l}P\left(\bigcap_{k\in I_0}\bigcap_{j\in J}A_k\cap A_j\right)\\\
&=\prod_{j\in I_0}P(A_j)-\sum_{l=1}^{|I_0^c|}(-1)^{l-1}\prod_{J\subset I_0^c|,|J|=l}\prod_{k\in I_0}\prod_{j\in J}P(A_k)P(A_j)\\\
&=\prod_{k\in I_0}P(A_k)\left(1-\sum_{l=1}^{|I_0^c|}(-1)^{l-1}\prod_{J\subset I_0^c|,|J|=l}\prod_{j\in J}P(A_j)\right)\\\
&=\prod_{k\in I_0}P(A_k)\prod_{j\in I_0^c}P(A_j^c)
\end{align*}
and we are done.
A: Think about the definition of independence.
What you'd like to show is that for any finite subset of the $A_i$,
$P\left(\displaystyle\bigcap_{i=1}^n A_i\right)=\displaystyle\prod_{i=1}^n P(A_i).$
Consider such an arbitrary subset $S = \{A_i\}$.  If $A_n \notin S$, then $P(A_n) + P(A_n^c) = P(A_n \cup A_n^c) = 1$, correct?  Think about how you can use this fact to prove that $S$ satisfies the definition of independence.
