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#1
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20 About the independent events

Hello guys, I do not know how to interpret this.

By definition, events A and B are independent if
P(A) * P(B) = P(A and B)

so in a fair dice, if A = {1,2,3}, B= {1,6}
then P(A) * P(B) = 1/3 * 1/2 = 1/6 = P{1}
thus A and B are independent

but what does "independent" mean here? why do we define independent this way? I don't understand the logic behind this.

Thank you!
#2
 daaaave David Revelle Join Date: Feb 2006 Posts: 3,051 Intuitively, independence means that knowing whether or not one event occurred doesn't affect your knowledge of whether or not the other one occurred. Here, P[A] = 1/2. If I told you that B occurred, then the roll is a 1 or a 6, one of which is in {A}, so P[A | B] = 1/2 = P[A]. That is, being told that B occurred doesn't change your probability of A having occurred. Likewise, P[b] = 1/3 = P[B | A].

As to why we define it as P[AB] = P[A] * P[b], that approach turns out to be mathematically cleaner and more concise, as having that factorization gives us a single thing to check that has nice implications for all the components we care about.
#3 Michael Mastroianni Member SOA Join Date: Jan 2018 Posts: 38 Independent events usually overlap because otherwise the intersection is empty and has probability zero so that would require one of the events to also have probability zero.

Its defined this way because when neither event has probability zero its equivalent to saying that the probability of one event does not change if you condition on the other (ie. the certainty of one of the events does not provide any more information about the probability of the other occurring).

To see this just divide each side by the probability of one of the events and the other side of the equation is the conditional probability.
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#4
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20 Quote:
 Originally Posted by daaaave Intuitively, independence means that knowing whether or not one event occurred doesn't affect your knowledge of whether or not the other one occurred. Here, P[A] = 1/2. If I told you that B occurred, then the roll is a 1 or a 6, one of which is in {A}, so P[A | B] = 1/2 = P[A]. That is, being told that B occurred doesn't change your probability of A having occurred. Likewise, P[b] = 1/3 = P[B | A]. As to why we define it as P[AB] = P[A] * P[b], that approach turns out to be mathematically cleaner and more concise, as having that factorization gives us a single thing to check that has nice implications for all the components we care about.
Oh I understand what you are talking about right now. But I have another question. In terms of conditional probability, A = {1,2,3} and B = {1,6}. My comprehension of P(A|B) is that B has two choices, and A has three choices. P(A|B) means that we want A occurs knowing B has occur, doesn't that mean the probability of only match A and B are both 1 is 1/choice of A times 1/ choice of B which is 1/6? What is the mistake of my thought?

Thanks a lot!
#5
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20 Quote:
 Originally Posted by Michael Mastroianni Independent events usually overlap because otherwise the intersection is empty and has probability zero so that would require one of the events to also have probability zero. Its defined this way because when neither event has probability zero its equivalent to saying that the probability of one event does not change if you condition on the other (ie. the certainty of one of the events does not provide any more information about the probability of the other occurring). To see this just divide each side by the probability of one of the events and the other side of the equation is the conditional probability.
It really helps. Thanks for your replying!
#6
 morbrenn Non-Actuary Join Date: Nov 2018 Studying for the final finals Posts: 10 Quote:
 Originally Posted by actuaryxjh Oh I understand what you are talking about right now. But I have another question. In terms of conditional probability, A = {1,2,3} and B = {1,6}. My comprehension of P(A|B) is that B has two choices, and A has three choices. P(A|B) means that we want A occurs knowing B has occur, doesn't that mean the probability of only match A and B are both 1 is 1/choice of A times 1/ choice of B which is 1/6? What is the mistake of my thought? Thanks a lot!
P(A|B) is the probability of A, given that B already happened. So you don't need to use the probability of B happening since we're assuming that it already did. Since we know the roll is in B, it's either a 1 or a 6, only one of which is in A, so the probability that we've also got A is 1/2.
#7
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20 Quote:
 Originally Posted by morbrenn P(A|B) is the probability of A, given that B already happened. So you don't need to use the probability of B happening since we're assuming that it already did. Since we know the roll is in B, it's either a 1 or a 6, only one of which is in A, so the probability that we've also got A is 1/2.
Thanks~

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