The answer is that you need more information on the probabilities in
order to decide how adding information influences condition
probabilities. I'll follow your example with Asians and Wongs to
explain this.
Let's define the experiment to be "take one person at random from the
whole world". The person says "my name is Wong", so now you know that
probability of him being Asian is 0.8. Next he says "I live in San
Francisco". You know that the probability of being Asian given that
you live in SF is 0.4; however, this doesn't tell you ANYTHING about
the probability of him being Asian given that his name is Wong *and*
he is from SF. It depends on the relation of being Wong and living in
SF.
For instance, say there is only one Wong in SF, and he is Asian. This
doesn't contradict the 0.8 probability of being Asian given that he's
Wong. It doesn't contradict the 0.4 probability of being Asian given
that he lives in SF either. And, since there is only one Wong in SF,
and he is Asian, the probability of being Asian given that he's Wong
and live in SF is 1.
Consider now what happens if all people in SF are called Wong. This
doesn't contradcit the 0.8 and 0.4 probabilities you described above.
And yet, now the probability of being Asian given that he's Wong AND
live in SF is 0.4 (once you know that he lives in SF, the fact that he
is called Wong becomes completely irrelevant).
As you see, anything may happen to the probabilities as you add
information, and it depends on the relation between the information
you're adding. The condition probability formula is (let P(A|B) be the
probability of A given B, and i be "intersection")
P(A|B)=P(A i B)/P(B)
( see the following page
Definition And Properties (of Conditional Probabilities)
http://www-stat.stanford.edu/~susan/courses/s116/node57.html )
So, returning to your example, you know (let A be "the person is
Asian", W be "the person is Wong", and S be "the person is from SF):
P(A|W)=P(A i W)/P(W)=0.8 (1)
P(A|S)=P(A i S)/P(S)=0.4 (2)
and what you want to know is:
P(A|W i S)=P(A i W i s)/P(W i S)= ??
From (1) and (2) you haven't the faintest clue as to what
probabilities P(A i W i s) and P(W i S) are, so you just can't compute
this value.
In sum, it's not clear how adding new information on the sampled
person modifies the probabilities of him being Asian, unless you say
something about the probability of being both Wong and from SF, and
the probability of being of being Asian, Wong and from SF.
I hope this was clear enough. If you need further assistance, just
request clarification.
Best regards,
elamrto |
Request for Answer Clarification by
marcky-ga
on
03 Apr 2003 22:36 PST
Thanks. This is very clearly stated. The only thing I would like to
enquire a bit further on is why it would seem at least from an
intuitive standpoint that the more data one has which has a strong
correlation with some assumption, the more one would think the
probability
goes up. I'm sure there is a fallacy in what I'm thinking, and it may
have
something to do with the direction of the conditionality.
I for example, one were to find chopsticks at this fellows house, and
one were to find hoi-sin sauce, and one were to find out that he went
to
a predominantly asian school, etc. etc. one would think that one would
come
away with a higher probability or more certainty in the belief that he
was asian. Can you clarify why that is, and whether it is
unjustified, whether there are assumptions one would be making which
were wrong?
Thanks. I will be away after tonight until sunday night, and am
already quite happy with the answer, so don't fret if I dont get to
respond to your
answer to this. Thanks.
|
Clarification of Answer by
elmarto-ga
on
04 Apr 2003 15:48 PST
Thanks. This is very clearly stated. The only thing I would like to
enquire a bit further on is why it would seem at least from an
intuitive standpoint that the more data one has which has a strong
correlation with some assumption, the more one would think the
probability
goes up. I'm sure there is a fallacy in what I'm thinking, and it may
have
something to do with the direction of the conditionality.
I for example, one were to find chopsticks at this fellows house, and
one were to find hoi-sin sauce, and one were to find out that he went
to
a predominantly asian school, etc. etc. one would think that one would
come
away with a higher probability or more certainty in the belief that he
was asian. Can you clarify why that is, and whether it is
unjustified, whether there are assumptions one would be making which
were wrong?
Hello again!
Regarding your request, it's true that from an intuitive standpoint
you would know with more certainty wether the person is Asian or not
as the quantity of information goes up. What I set out to say with my
answer (returning to your first example) is that there doesn't exist a
function that relates the 0.8 and 0.4 probabilities with the
probability of him being Asian. You still need to know the
probabilities I stated in the last paragraph of my answer in order to
find the exact conditional probability you're looking for. Thus, in
order to find P(A|BiC), knowledge of P(A|B) or P(A|C) is totally
irrelevant. Conditioning on event B is a completely different thing
than conditioning on event BiC. An extreme example: imagine that, as
before, the probability of being Asian given that you're Wong is 0.8,
and the probability of being Asian given that you use chop-sticks (C)
is 0.9. The guy tells you "my name is Wong and I use chop-sticks". If
the only information you have is this, there is no way to know
P(A|WiC), as these things could happen:
1) There is a law (and no one can break this law) that prohibits
Asians named Wong to use chop-sticks. In this case, the probability
that he is Asian is 0. As you see, the 0.8 and 0.9 had nothing to do
with this conclusion.
2) There is a law (again, unbreakable) that prohibits non-Asians named
Wong to use chop-sticks. In this case, the proibability that he is
Asian is 1. Again, this had nothing to do with the 0.8 and 0.9
probabilities.
I hope this example clarified my answer to you. Good luck with your
future researches!
Best wishes,
elmarto
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