# independence

• Sep 5th 2009, 09:49 PM
Katina88
independence
Hi, im just confirming if this logic is right, for my assignment. Well i know that events are mutually exclusive if $P (A \cap B) = 0$. And then this is not independent.

so if i wanted to determine is something is independent then is it just if $P (A \cap B)$ is not equal to 0?

coz in this subject we havent covered stuff like chi-squares so im not allowed to do that. thanks
• Sep 5th 2009, 09:52 PM
mr fantastic
Quote:

Originally Posted by Katina88
Hi, im just confirming if this logic is right, for my assignment. Well i know that events are mutually exclusive if $P (A \cap B) = 0$. And then this is not independent.

so if i wanted to determine is something is independent then is it just if $P (A \cap B)$ is not equal to 0?

coz in this subject we havent covered stuff like chi-squares so im not allowed to do that. thanks

The test for the independence of two events is clearly stated here: http://en.wikipedia.org/wiki/Indepen...ability_theory)
• Sep 5th 2009, 09:59 PM
Katina88
ok then is this right:
if
$P(A|B) = P(A|B') = P(A)$ then it is independent?
• Sep 5th 2009, 10:06 PM
mr fantastic
Quote:

Originally Posted by Katina88
ok then is this right:
if
$P(A|B) = P(A|B') = P(A)$ then it is independent?

Yes. But the condition $\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)$ might be the better one to apply.
• Sep 5th 2009, 10:29 PM
matheagle
Quote:

Originally Posted by mr fantastic
Yes. But the condition $\Pr(A \cup B) = \Pr(A) \cdot \Pr(B)$ might be the better one to apply.

you mean intersection and not union here.
Another clarification...
Mutually exclusive or disjoint means that the intersection of two sets is empty,
THAT implies that P(AB)=0.
BUT P(AB)=0 does not mean they are disjoint.

Also If two sets are independent then..........

$P(A|B)=P(A)=P(A|B')$
and
$P(B|A)=P(B)=P(B|A')$
and
$P(AB)=P(A)P(B)$
and
$P(AB')=P(A)P(B')$
and
$P(A'B)=P(A')P(B)$
and
$P(A'B')=P(A')P(B')$

IF any one hold, ALL HOLD.
And if one fails, they all do.

NOW to the question at hand, which I did in class 2 weeks ago.
IF A and B are disjoint (and they have positive probability) then they are dependent.

That's because P(AB)=0 while P(A)P(B)>0, so

$P(AB)\ne P(A)P(B)$
• Sep 5th 2009, 10:42 PM
matheagle
Quote:

Originally Posted by Katina88
ok then is this right:
if
$P(A|B) = P(A|B') = P(A)$ then it is independent?

If $P(A|B) = P(A)$ they are independent

It then follows that $P(A|B') = P(A)$ too.

You only need show that 2 of these 3 are equal
$P(A|B) = P(A|B') = P(A)$
the other must also be equal to the first two.

These are the more intuitive definitions.
IF A and B are independent then knowing one does not change the probability of the other.
$P(A|B)=P(A)$
• Sep 5th 2009, 10:51 PM
Katina88
Quote:

Originally Posted by matheagle