Given 2 events A and B, is it possible that B is independent of A, but A is positively dependent on B (A is more likely to occur given that B has occured)?

I am stumped. Please help out if you can. Thanks.

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- Nov 4th 2007, 02:30 AMchopetPlease help me out here about Independent Events
Given 2 events A and B, is it possible that B is independent of A, but A is positively dependent on B (A is more likely to occur given that B has occured)?

I am stumped. Please help out if you can. Thanks. - Nov 4th 2007, 03:27 AMTheBrain
Well my first thoughts when I read that are these.

Let Event A be that I decide to take my umbrella today.

Let Event B be that it is raining today.

Now clearly B is independant of A. There is no way me taking my umbrella is going to affect the weather. On the other hand, A is more likely to happen if B has occured.

Is this the sort of thing you are after? - Nov 4th 2007, 03:45 AMchopet
Yes.

So let's say we plug in values:

P(A) = prob that I am going to bring umbrella today= 0.3

P(B) = prob that its going to rain today = 0.5

We learn that for independent events, P(AnB)=0.3*0.5 = 0.15.

But certainly the prob that event B is going to affect A positively, so P(AnB) > 0.15.

Let's say this value P(AnB) is 0.20.

We can calculate P(A|B) as $\displaystyle {P(A \cap B) \over P(B)} = {0.2 \over 0.5} = 0.40.$

Here, we can see the positive relation very clearly, as prob of A given B(=0.40) is higher than prob of A alone(=0.30).

But the maths doesn't show us the independent relation of A to B. Let's do the math.

$\displaystyle P(B|A) = {P(B \cap A) \over P(A)} = {0.2 \over 0.3} = 0.66$

Using the same comparison between P(B) alone and P(B) given A, can we argue that B is also dependent on A?

Am I missing out any other concepts here?

Thanks! - Nov 4th 2007, 04:33 AMPlato
Suppose we know that $\displaystyle P\left( {B|A} \right) = P(B)\;\& \;P(A) > 0$.

Then $\displaystyle P(B) = \frac{{P(A \cap B)}}{{P(A)}}\; \Rightarrow \,P(A)P(B) = P(A \cap B)$.

Now $\displaystyle P\left( {A|B} \right) = \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(A \cap B)}}{{\frac{{P(A \cap B)}}{{P(A)}}}} = P(A)$

Does that mean that A is independent of B? - Nov 4th 2007, 05:42 AMchopet
Yes it does.

It proves that P(A) and P(B) are mutually independent of each other.

But is it possible that P(A) is dependent on P(B), but P(B) is indepedent of P(A)? - Nov 4th 2007, 08:03 AMPlato
- Nov 4th 2007, 08:18 AMchopet
sorry for not being clear by putting the 2 statements together.

What i meant was: could there be a scenario whereby

P(A) is dependent on P(B), but P(B) is independent of P(A)?

Is it mathematically possible? If so, what are the equations like?

See the scenario set up by TheBrain.

- Nov 4th 2007, 11:31 AMPlato
I just realized that I have been trying to answer the wrong question.

The trouble starts with not fully understanding the nature of definition.

In probability theory, the concept of*independent events*is a**pairwise**definition.

That is, we begin with**two events**and give conditions that result in their being called*independent events*.

The idea if*independence*is a symmetrical relation between events.

In fact, it is a standard exercise to show that if $\displaystyle A\,\& \,B$ are independent then so are each of these pairs: $\displaystyle A^c \,\& \,B$, $\displaystyle A \,\& \,B^c$ and $\displaystyle A^c \,\& \,B^c$.

I hope that this puts the confusion to rest. - Nov 4th 2007, 12:01 PMchopet
Thanks!

At least you have helped me clear up the definition of*pairwise independence*.

But can you comment on the scenario set up by TheBrain?

If it is not a case of*pairwise independence*, according to the definition, then how can we describe the relationship between A and B?