Assume A,B,C,D, are pairwise independent events. Decide if $(A \cap B)$ and $(B \cap D)$ are independent events? Then repeat this assuming the four events are mutually independent.
Well, what I'm thinking is $P(A \cap B) = P(A) \cdot P(B) \\P(A \cap C) = P(A) \cdot P(C) \\P(A \cap D) = P(A) \cdot P(D) \\P(B \cap C) = P(B) \cdot P(C) $...and soo on , which is what makes it pairwise independent, so assuming this is pairwise, then that means $(A \cap B)$ and $(B \cap D)$ are also independent because
$P(A \cap B) = P(A) \cdot P(B) \\P(B \cap D) = P(B) \cdot P(D) $......this is correct to say they are independent??
then I dont understand to assume the 4 events are mutually exclusive?
does that mean
$P[(A \cap B) \cap (B \cap D)] = P(A) \cdot P(B) \cdot P(D)$??
Which does not make them independent because of B?
I am not too sure. I am confused and clarification is much appreciated. Any help IS MUCH APPRECIATED TOO.
In the title and the post, the question asked whether it is true that if $A$, $B$, and $D$ are pairwise independent, then $A\cap B$ and $B\cap D$ are also independent. (There is also a puzzling $C$ around that plays no role in the question. But we have added a remark in case $C\cap D$ is intended, and not $B\cap D$.)
Not necessarily. Toss a fair coin twice. Let $A$ be the event the first toss is a head, let $D$ be the event the second toss is a head, and let $B$ be the event the two tosses are the same (both heads or both tails).
It is not hard to verify that $A$, $B$, and $D$ are pairwise independent.
However, if the event $A\cap B$ occurs, then for sure $B\cap D$ occurs. So $A\cap B$ and $B\cap D$ are not independent.
More formally, $\Pr(A\cap B)=\frac{1}{4}$, and $\Pr(B\cap D)=\frac{1}{4}$. But also $\Pr((A\cap B)\cap (B\cap D))=\frac{1}{4}$, and therefore $$\Pr((A\cap B)\cap (B\cap D))\ne \Pr(A\cap B)\Pr(B\cap D),$$ so $A\cap B$ and $B\cap D$ are not independent.
Remark: Perhaps what is intended is to ask whether $A\cap B$ and $C\cap D$ are independent, given that $A,B,C,D$ are pairwise independent. Again, not necessarily.
Toss a fair coin $3$ times. Let $A$ be the event head on first, $B$ be the event head on second, $C$ the event head on third, and $D$ the event all tosses are the same. Again we have pairwise independence. The probability of $A\cap B$ is $\frac{1}{4}$, the probability of $C\cap D$ is $\frac{1}{8}$, and the probability of $(A\cap B)\cap (C\cap D)$ is $\frac{1}{8}$, so $A\cap B$ and $C\cap D$ are not independent.
If $A$, $B$, $C$, and $D$ are mutually independent, then $A\cap B$ and $C\cap D$ are independent. For under mutual independence, we have $\Pr((A\cap B)\cap (C\cap D))=\Pr(A)\Pr(B)\Pr(C)\Pr(D)$.