Independence of complementary events

Question

Suppose $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, $I$ is an arbitrary index set and $\{A_i\}_{i \in I} \in \mathcal{F}^{I}$. For $i \in I$ we define $B_i^{(0)} := A_i$ and $B_i^{(1)} := A_i^{\mathsf{c}}$. I want to show the implication $$ \exists \, \alpha \in \{0,1\}^{I} \colon \ \{B_i^{(\alpha_i)}\}_{i \in I} \text{ is independent } \Longrightarrow \{A_i\}_{i \in I} \text{ is independent }. $$

Let $\alpha$ be the fixed sequence and let $J \subseteq I$ with $|J|< \infty$. We have to show that $$ \mathbb{P}\left( \bigcap_{j \in J} A_j \right) = \prod_{j \in J} \mathbb{P}(A_j). $$ Let us write $J=J_0 \uplus J_1$, where $J_0 = \{ j \in J \colon \alpha_j =0\}$ and $J_1 = \{ j \in J \colon \alpha_j =1\}$. Then we have $$ \mathbb{P}\left( \bigcap_{j \in J} A_j \right) =\mathbb{P}\left( \bigcap_{j \in J_0} A_j \cap \bigcap_{j \in J_1} A_j \right) = \mathbb{P}\left( \bigcap_{j \in J_0} B_j^{(\alpha_j)} \cap \bigcap_{j \in J_1} A_j \right) $$ To apply the assumption I need something like $\bigcap_{j \in J_1} B_j^{(\alpha_j)}$. But using De Morgan's laws I only get that $$ \bigcap_{j \in J_1} A_j = \left( \left(\bigcap_{j \in J_1} A_j \right)^{\mathsf{c}} \right)^{\mathsf{c}} = \left(\bigcup_{j \in J_1} A_j^{\mathsf{c}} \right)^{\mathsf{c}} = \left(\bigcup_{j \in J_1} B_j^{(\alpha_j)} \right)^{\mathsf{c}}. $$ What is the correct way to continue the proof?

Answer 1

In order to give more strength to the induction hypothese let us prove more generally:

$\exists\alpha\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\alpha_{i}\right)}\right\} _{i\in I}\text{ is independent}\implies\forall\beta\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\beta_{i}\right)}\right\} _{i\in I}\text{ is independent}$

Assume that the statement is not true. Then some finite subset $J\subseteq I$ can be found such that $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$ for some $\beta\in\left\{ 0,1\right\} ^{I}$ .

Let $J$ be such set and this with minimal cardinality.

Now find a $\beta\in\left\{ 0,1\right\} ^{I}$ such that $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$ and $J_{1}=\left\{ j\in J\mid\alpha_{j}\neq\beta_{j}\right\} $ has minimal cardinality.

Then $J_{1}\neq\varnothing$. Let $r\in J$ with $\alpha_{r}\neq\beta_{r}$.

$\prod_{j\in J-\left\{ r\right\} }\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)=\mathbb{P}\left(\bigcap_{j\in J-\left\{ r\right\} }B_{j}^{\left(\beta_{j}\right)}\right)=\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)+\mathbb{P}\left(\bigcap_{j\in J-\left\{ r\right\} }B_{j}^{\left(\beta_{j}\right)}\cap B_{r}^{\left(\alpha_{r}\right)}\right)=\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)+\prod_{j\in J-\left\{ r\right\} }\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)\left(1-\mathbb{P}\left(B_{r}^{\left(\beta_{r}\right)}\right)\right)$

contradicting that $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$.

The first equality is a consequence of the minimality of $|J|$ and the third is a consequence of the minimality of $|J_1|$.

We conclude that the statement must be true.

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edit to make things clear:

For finite $J\subseteq I$ and $\beta\in\left\{ 0,1\right\} ^{I}$ abbreviate $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$ by $P\left(J,\beta\right)$.

Assuming that the statement is not true is the same as assuming that finite sets $J\subseteq I$ exists with $\left\{ \beta\in\left\{ 0,1\right\} ^{I}\mid P\left(J,\beta\right)\right\} \neq\varnothing$.

Choose such $J$ and this with minimal cardinality.

Note that $\beta$ is not fixed yet after this choosing of $J$. That is the next step to take.

We have $\left\{ \beta\in\left\{ 0,1\right\} ^{I}\mid P\left(J,\beta\right)\right\} \neq\varnothing$ and this enables us to choose $\beta\in\left\{ 0,1\right\} ^{I}$ such that $P\left(J,\beta\right)$ and such that $J_{1}$ has minimal cardinality.