# 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 $I \subset \{1,2,...,n\}$

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.

• I think you forget that $I\subset \{1,\cdots,n\}?$ Because you can take $I=\{i,j\}$ to show pair-wise independence. Commented Nov 18, 2021 at 3:31
• Thanks @DreamAR, I'm pretty new to set theory so I thought if I take $I={3}$ then it doesn't hold? (i.e. I need to prove that the statement holds for every possible $I$).
– tkhu
Commented Nov 18, 2021 at 3:34

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.

• What is the difference between $n=2$ and $|I|=2$ because both are only two events? I think this is not allowing me to understand why I can extend my hypothesis $|I|<= n-1$.
– tkhu
Commented Nov 18, 2021 at 3:49
• @tkhu I edited my answer. Maybe now it become clearer. Commented Nov 18, 2021 at 4:09
• Awesome, thank you so much for your speedy help!
– tkhu
Commented Nov 18, 2021 at 4:13