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?

  • $\begingroup$ First thing that comes to mind: induction on cardinality of $J_1$. $\endgroup$ – drhab Nov 2 '15 at 9:57
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    $\begingroup$ Note that deleting the question also delete the answer, and the effort and time spent by the answer-er will be wasted. $\endgroup$ – user99914 Nov 4 '15 at 11:48
  • $\begingroup$ If there is/was any motivation to delete your question then feel free to share this. That is evidently preferable on several grounds. $\endgroup$ – drhab Nov 4 '15 at 15:03
  • $\begingroup$ I deleted the question because, 1), my attempt sucked, 2), I do not understand your answer, 3), I was able to prove it by induction. $\endgroup$ – user148692 Nov 4 '15 at 23:17
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    $\begingroup$ 1) It doesn't matter. No one will judge you. 2) It doesn't matter. Don't accept it then. It might be useful to other people with a similar question who come across it on the site. 3) You could add your solution as an answer if you want. Again, the site is not only for the benefit of the person asking the question, but also serves as a resource for anyone who may have the same question in the future. $\endgroup$ – Dylan Nov 5 '15 at 0:43

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.

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.

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  • $\begingroup$ Thank you, I think now I understand your idea: By assumption we know that at least one set $J \subseteq I$ existst such that $\left\{ \beta\in\left\{ 0,1\right\} ^{I}\mid P\left(J,\beta\right)\right\} \neq\varnothing$. Now we can choose $J$ and afterwards $\beta$ (in the worst case there might be only one choice of $\beta$) such that $|J_1|$ is minimal. $\endgroup$ – user148692 Nov 3 '15 at 10:50
  • $\begingroup$ Now my last question refers to your third equality: How do you use the minimality of $|J_1$| and the independence of the family with the index set $\alpha$? We know that $\{B_{j}^{\left(\beta_{j}\right)}\}_{j\in J-\left\{ r\right\}}$ is independent, but why is $\{B_{j}^{\left(\beta_{j}\right)} \}_{j\in J-\left\{ r\right\}},B_{r}^{\left(\alpha_{r}\right)}$ independent, too? $\endgroup$ – user148692 Nov 3 '15 at 10:50
  • $\begingroup$ Because the set $J_1$ connected with $\{\beta_j\mid j\in J-\{r\}\}\cup\{\alpha_r\}$ has one element less than the one connected with $\{\beta_j\mid j\in J\}$ wich has minimal cardinality. No independence would contradict this minimality. $\endgroup$ – drhab Nov 3 '15 at 11:21

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