2
$\begingroup$

Is there a way to proof to prove this for independence using the method of contradiction ?

Let $A_1, A_2,$ and $A_3$ be events, and let $B_i$ represent either $A_i$ or its complement $A^c _i$. Then there are eight possible choices for the triple $(B_1,B_2,B_3)$. Prove that the events $A_1, A_2, A_3$ are independent if and only if $P(B_1 \cap B_2 \cap B_3) = P(B_1)P(B_2)P(B_3) $, for all eight of the possible choices for the triple $(B_1, B_2, B_3)$.

The first thing I thought of is contradiction but I guess writing out $8 $ equations and showing that if they are independent, then $A_1, A_2, A_3$ must also be independent is also a good way. Is there any other way to prove this ?

Thanks in advance.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

The following shows the "if" part.

You need to show $p(A_1\cap A_2\cap A_3)=p(A_1)p(A_2)p(A_3)$ and $p(A_i\cap A_j)=p(A_i)p(A_j)$ for all $i\not=j$. That's the definition of three events being independent, you have to show all four things.

The first one is basically given, just let $B_i=A_i$ for all $i$.

Now consider $p(A_1\cap A_2)$. Note that $A_1\cap A_2=A_1\cap A_2\cap(A_3\cup A_3^c)$. So

$p(A_1\cap A_2)=p(A_1\cap A_2\cap(A_3\cup A_3^c))$

$=p((A_1\cap A_2\cap A_3)\cup(A_1\cap A_2\cap A_3^c))$ (because intersection distributes over union)

$=p(A_1)p(A_2)p(A_3)+p(A_1)p(A_2)p(A_3^c)$ (by assumption)

$=p(A_1)p(A_2)p(A_3)+p(A_1)p(A_2)(1-p(A_3))$

$p(A_1)p(A_2)$.

This shows the "if" part, you should now be able to do the converse.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .