Prove that $A_1, … , A_n$ are independent if and only if $P(\bigcap_{i \in I}A_i)=\prod_{i \in I}P(A_i)$ for $I \subset \{1,2,…,n\}$

independencemeasure-theoryprobability theory

I have been working on this proof for a while but I'm pretty lost. I was told that induction is needed. Could I confirm that I'm in the right direction?

Show "$\Rightarrow$".

Let $A_1, … , A_n$ be independent, then by definition, $I_{A_1},…,I_{A_n}$ are also independent.

Since $I_{A_1},…,I_{A_n}$ are random variable, then by definition $\sigma(I_{A_1}),…,\sigma(I_{A_n})$ are independent as well.

Since $\sigma(I_{A_1}),…,\sigma(I_{A_n})$ are $n$ sigma-fields, then by definition $P(\bigcap_{i \in I}B_i)=\prod_{i \in I}P(B_i)%$ for all $B_i \in \sigma(I_{A_i})$.

Observe that $A_i = I_{A_i}^{-1}(\{1\}) \in \sigma(I_{A_i})$. Thus, $P(\bigcap_{i \in I}A_i)=\prod_{i \in I}P(A_i)%$ for $I \subset \{1,2,…,n\}$.

This completes the proof for "$\Rightarrow$".

Show "$\Leftarrow$".

For $n=2$, let $P(A_1 \cap A_2)=P(A_1)P(A_2)$.

Observe that (proofs omitted but easy to derive):

  • $P(A_1^C \cap A_2)=P(A_1^C)P(A_2)$ and the other way around too.
  • $P(A_1^C \cap A_2^C)=P(A_1^C)P(A_2^C)$
  • $P(\Omega \cap A_i)=P(\Omega)P(A_i^C)$ and their variations.
  • $P(\Omega \cap \Omega)=1=P(\Omega)P(\Omega)$
  • $P(\emptyset \cap A_i)=0=P(\emptyset)P(A_i)$ and their variations.
  • $P(\emptyset \cap \emptyset)=P(\emptyset)P(\emptyset)$

Now let $B$ be a Borel set, $C=\{0,1\}$, $i\in\{1,2\}$, then:

  • If $B \cap C = \{1\}$, then $I_{A_i}^{-1}(B \cap C)=A_i$.
  • If $B \cap C = \{0\}$, then $I_{A_i}^{-1}(B \cap C)=A_i^C$.
  • If $B \cap C = C$, then $I_{A_i}^{-1}(B \cap C)=\Omega$.
  • If $B \cap C = \emptyset$, then $I_{A_i}^{-1}(B \cap C)=\emptyset$.

From above, I have successfully shown that $P(I_{A_1}^{-1}(D_1) \cap I_{A_2}^{-1}(D_2))=P(I_{A_1}^{-1}(D_1))P(I_{A_2}^{-1}(D_2))$, for $D_1, D_2$ in Borel sigma-field. Next, we observe lemma below:

$X_1,…,X_n$ are independent if and only if $P(\bigcap_{i \in I}\{X_i \in A_i\})=\prod_{i \in I}P(\{X_i \in A_i\})$ for all $A_i$ in Borel sigma-field, $i=1,…,n$.

Thus, it follows that $I_{A_1},I_{A_2}$ are independent. By definition, this means $A_1, A_2$ are independent too.

This completes the proof for "$\Rightarrow$" for $n=2$.

However, I'm stuck here because for $n$ cases I cannot assume pair-wise independence.

Best Answer

So you've proved that for all $P(A\cap B)=P(A)P(B),$ $A,B$ are independent. So what you get here is not only $n=2$ case, but also $|I|=2$ case.

So you can assume for all $|I|\le n-1,$ your hypothesis is true. Now for $|I|=n,$ you only need to prove that $P(\bigcap_{i=1}^n A_i)=\prod_{i=1}^nP(A_i)\Rightarrow$ $A_1,\cdots,A_n$ are independent if $\{A_1,\cdots,A_{n-1}\},$ $\cdots$ ,$\{A_2,\cdots,A_n\}$ are all independent already.

I think maybe you need another claim to let it become distinct.

Claim: For arbitrary independent series $\{B_1,\cdots,B_{m-1}\},$ $\cdots,$ $\{B_2,\cdots,B_m\}.$ If $P(\bigcap_{i=1}^m B_i)=\prod_{i=1}^m P(B_i),$ then $\{B_1,\cdots,B_m\}$ is also independent.

Now you've proved that $m=2$ is true ($\{B_1\},$ $\{B_2\}$ are all independent trivially), you can assume $m=k-1$ is true, then prove $m=k$ case to do induction.

After that, you can prove your original question. Take $I=\{i,j\}$ to show they are pair-wise independent first. Then take $I=\{i,j,k\}$ to show they are triple-wise independent based on pair-wise independence by your claim. $\cdots$ (This is also an induction) $\cdots$ At last $I=\{1,\cdots,n\}$ you get $\{A_1,\cdots,A_n\}$ is independent.

So there are two inductions actually. Maybe this is the point where you get lost.

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