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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.

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Tip: Try it first for $n=2$ and $n=3$. –  Neal Apr 15 '12 at 1:55
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2 Answers 2

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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.

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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.

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Thank you very much for your answer. I have a doubt though. I thought that this question is concerned only about one direction, that being "If this $2^n$ hold, then $A_1,\ldots,A_n$ are independents. This part I don't get: "if $2^n$ equations are true then taking $B_i=A_i$ we get what we want. I don't see why this is true. Because making $B_i=A_i$ we will only have that the probability of the intersection is the product of the probability for the whole set $A_1,...,A_n$. For we get the independence we need to garantee that all subset of $A_1,\ldots,A_n$ satisfies this equation. –  Rodolfo Apr 16 '12 at 2:47
    
The way i see the $2^n$ equations are given by: $P(A_1,A_2,...,A_n)=\prod P(A_i)$; $P(A_1^c,A_2...,A_n)=P(A_1^c)\prod_{i=2}^{n} P(A_i)$; $P(A_1,A_2^c...,A_n)=P(A_1^c)P(A_2^c)\prod_{i=3}^{n} P(A_i)$; till $P(A_1^c,A_2^c,...,A_n^c)=\prod P(A_i^c)$; I'm probably wrong though. –  Rodolfo Apr 16 '12 at 2:50
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If the $2^n$ equations hold, there is one equation between them which gives independence (the first you wrote in your second comment). The point of the exercise is to show that conversely, if we assume independence then the $2^n$ equations hold. –  Davide Giraudo Apr 16 '12 at 9:25
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