Probability of three events occurring given correlation?
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I am facing a problem that I cannot find the answer to. I have three variables, A, B and C. There are only two possibilities for each of these, A either happens or it does not, B happens or it does not and C happens or it does not. I know that if these events are independent that the probability of them all occurring is simply $P(A)cdot P(B)cdot P(C)$. So if the probability of each happening is 10% then all three have a $10%·10%·10% = 0.1%$ probability of occurring. But how would this formula change if the events were not independent but were instead positively correlated.
I can solve this for just two variables with the formula:
$P(A cap B) = P(A)cdot P(B) + rho_ABcdot sqrtP(A)cdot (1-P(A))cdot P(B)cdot (1-P(B)) $, where $rho_AB$ is the correlation coefficient between A and B.
How would I change this formula to calculate the probability that A, B and C all occur? I.e. calculating $P(A cap B cap C)$ knowing $P(A)$, $P(B)$, $P(C)$, $rho_AB$, $rho_AC$, $rho_BC$.
Thanks in advance for the help!
probability correlation
edited Jul 23 '15 at 17:15
josefec
1032
asked Mar 25 '15 at 13:05
Hugh
3612
add a comment |Â
up vote
7
down vote
favorite
I am facing a problem that I cannot find the answer to. I have three variables, A, B and C. There are only two possibilities for each of these, A either happens or it does not, B happens or it does not and C happens or it does not. I know that if these events are independent that the probability of them all occurring is simply $P(A)cdot P(B)cdot P(C)$. So if the probability of each happening is 10% then all three have a $10%·10%·10% = 0.1%$ probability of occurring. But how would this formula change if the events were not independent but were instead positively correlated.
I can solve this for just two variables with the formula:
$P(A cap B) = P(A)cdot P(B) + rho_ABcdot sqrtP(A)cdot (1-P(A))cdot P(B)cdot (1-P(B)) $, where $rho_AB$ is the correlation coefficient between A and B.
How would I change this formula to calculate the probability that A, B and C all occur? I.e. calculating $P(A cap B cap C)$ knowing $P(A)$, $P(B)$, $P(C)$, $rho_AB$, $rho_AC$, $rho_BC$.
Thanks in advance for the help!
probability correlation
edited Jul 23 '15 at 17:15
josefec
1032
asked Mar 25 '15 at 13:05
Hugh
3612
Correlation is a pairwise property. Are you wanting to assume that all three correlations have the same (positive) value, or just that all three have positive value?
– Dilip Sarwate
Mar 25 '15 at 13:21
Either, really. For simplicity's sake, let's assume that there is a correlation of 0.1 between A and B, 0.1 between B and C and 0.1 between A and C.
– Hugh
Mar 25 '15 at 13:26
Bonferroni's Inequality has few assumptions and gives a bound that is sometimes useful (depending on the size of the probabilities). Otherwise, I think you need to provide some context for your question. As stated, I do not see how a useful answer can be given.
– BruceET
Mar 25 '15 at 16:21
I just had the same question and came over this here. The question is about extending the problem of finding the probability of two events happening both (while knowing the probabilities of them happening and correlation) to three events. I.e. you have the probabilities of all three events and their correlation matrix and you would like to know the probability of all three occuring.
– josefec
Jul 23 '15 at 16:44
There are six parameters here $P(A), P(B), P(C), rho_AB, rho_BC, rho_AC$, and eight unknown probabilities $P(A cap B cap C), P(A cap B cap overlineC)$, etc. The six parameters give six constraints on the probabilities, and the fact that the probabilities need to sum to one gives a seventh. Since we have seven equations in eight unknowns we should in general expect a one-parameter family of solutions.
– Michael Lugo
Aug 7 at 20:10
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I am facing a problem that I cannot find the answer to. I have three variables, A, B and C. There are only two possibilities for each of these, A either happens or it does not, B happens or it does not and C happens or it does not. I know that if these events are independent that the probability of them all occurring is simply $P(A)cdot P(B)cdot P(C)$. So if the probability of each happening is 10% then all three have a $10%·10%·10% = 0.1%$ probability of occurring. But how would this formula change if the events were not independent but were instead positively correlated.
I can solve this for just two variables with the formula:
$P(A cap B) = P(A)cdot P(B) + rho_ABcdot sqrtP(A)cdot (1-P(A))cdot P(B)cdot (1-P(B)) $, where $rho_AB$ is the correlation coefficient between A and B.
How would I change this formula to calculate the probability that A, B and C all occur? I.e. calculating $P(A cap B cap C)$ knowing $P(A)$, $P(B)$, $P(C)$, $rho_AB$, $rho_AC$, $rho_BC$.
Thanks in advance for the help!
probability correlation
edited Jul 23 '15 at 17:15
josefec
1032
asked Mar 25 '15 at 13:05
Hugh
3612
I am facing a problem that I cannot find the answer to. I have three variables, A, B and C. There are only two possibilities for each of these, A either happens or it does not, B happens or it does not and C happens or it does not. I know that if these events are independent that the probability of them all occurring is simply $P(A)cdot P(B)cdot P(C)$. So if the probability of each happening is 10% then all three have a $10%·10%·10% = 0.1%$ probability of occurring. But how would this formula change if the events were not independent but were instead positively correlated.
I can solve this for just two variables with the formula:
$P(A cap B) = P(A)cdot P(B) + rho_ABcdot sqrtP(A)cdot (1-P(A))cdot P(B)cdot (1-P(B)) $, where $rho_AB$ is the correlation coefficient between A and B.
How would I change this formula to calculate the probability that A, B and C all occur? I.e. calculating $P(A cap B cap C)$ knowing $P(A)$, $P(B)$, $P(C)$, $rho_AB$, $rho_AC$, $rho_BC$.
Thanks in advance for the help!
probability correlation
edited Jul 23 '15 at 17:15
josefec
1032
asked Mar 25 '15 at 13:05
Hugh
3612
edited Jul 23 '15 at 17:15
josefec
1032
edited Jul 23 '15 at 17:15
josefec
1032
edited Jul 23 '15 at 17:15
josefec
1032
1032
asked Mar 25 '15 at 13:05
Hugh
3612
asked Mar 25 '15 at 13:05
Hugh
3612
asked Mar 25 '15 at 13:05
Hugh
3612
3612
Correlation is a pairwise property. Are you wanting to assume that all three correlations have the same (positive) value, or just that all three have positive value?
– Dilip Sarwate
Mar 25 '15 at 13:21
Either, really. For simplicity's sake, let's assume that there is a correlation of 0.1 between A and B, 0.1 between B and C and 0.1 between A and C.
– Hugh
Mar 25 '15 at 13:26
Bonferroni's Inequality has few assumptions and gives a bound that is sometimes useful (depending on the size of the probabilities). Otherwise, I think you need to provide some context for your question. As stated, I do not see how a useful answer can be given.
– BruceET
Mar 25 '15 at 16:21
I just had the same question and came over this here. The question is about extending the problem of finding the probability of two events happening both (while knowing the probabilities of them happening and correlation) to three events. I.e. you have the probabilities of all three events and their correlation matrix and you would like to know the probability of all three occuring.
– josefec
Jul 23 '15 at 16:44
There are six parameters here $P(A), P(B), P(C), rho_AB, rho_BC, rho_AC$, and eight unknown probabilities $P(A cap B cap C), P(A cap B cap overlineC)$, etc. The six parameters give six constraints on the probabilities, and the fact that the probabilities need to sum to one gives a seventh. Since we have seven equations in eight unknowns we should in general expect a one-parameter family of solutions.
– Michael Lugo
Aug 7 at 20:10
add a comment |Â
Correlation is a pairwise property. Are you wanting to assume that all three correlations have the same (positive) value, or just that all three have positive value?
– Dilip Sarwate
Mar 25 '15 at 13:21
Either, really. For simplicity's sake, let's assume that there is a correlation of 0.1 between A and B, 0.1 between B and C and 0.1 between A and C.
– Hugh
Mar 25 '15 at 13:26
Bonferroni's Inequality has few assumptions and gives a bound that is sometimes useful (depending on the size of the probabilities). Otherwise, I think you need to provide some context for your question. As stated, I do not see how a useful answer can be given.
– BruceET
Mar 25 '15 at 16:21
I just had the same question and came over this here. The question is about extending the problem of finding the probability of two events happening both (while knowing the probabilities of them happening and correlation) to three events. I.e. you have the probabilities of all three events and their correlation matrix and you would like to know the probability of all three occuring.
– josefec
Jul 23 '15 at 16:44
There are six parameters here $P(A), P(B), P(C), rho_AB, rho_BC, rho_AC$, and eight unknown probabilities $P(A cap B cap C), P(A cap B cap overlineC)$, etc. The six parameters give six constraints on the probabilities, and the fact that the probabilities need to sum to one gives a seventh. Since we have seven equations in eight unknowns we should in general expect a one-parameter family of solutions.
– Michael Lugo
Aug 7 at 20:10
Correlation is a pairwise property. Are you wanting to assume that all three correlations have the same (positive) value, or just that all three have positive value?
– Dilip Sarwate
Mar 25 '15 at 13:21
Correlation is a pairwise property. Are you wanting to assume that all three correlations have the same (positive) value, or just that all three have positive value?
– Dilip Sarwate
Mar 25 '15 at 13:21
Either, really. For simplicity's sake, let's assume that there is a correlation of 0.1 between A and B, 0.1 between B and C and 0.1 between A and C.
– Hugh
Mar 25 '15 at 13:26
Either, really. For simplicity's sake, let's assume that there is a correlation of 0.1 between A and B, 0.1 between B and C and 0.1 between A and C.
– Hugh
Mar 25 '15 at 13:26
Bonferroni's Inequality has few assumptions and gives a bound that is sometimes useful (depending on the size of the probabilities). Otherwise, I think you need to provide some context for your question. As stated, I do not see how a useful answer can be given.
– BruceET
Mar 25 '15 at 16:21
Bonferroni's Inequality has few assumptions and gives a bound that is sometimes useful (depending on the size of the probabilities). Otherwise, I think you need to provide some context for your question. As stated, I do not see how a useful answer can be given.
– BruceET
Mar 25 '15 at 16:21
I just had the same question and came over this here. The question is about extending the problem of finding the probability of two events happening both (while knowing the probabilities of them happening and correlation) to three events. I.e. you have the probabilities of all three events and their correlation matrix and you would like to know the probability of all three occuring.
– josefec
Jul 23 '15 at 16:44
I just had the same question and came over this here. The question is about extending the problem of finding the probability of two events happening both (while knowing the probabilities of them happening and correlation) to three events. I.e. you have the probabilities of all three events and their correlation matrix and you would like to know the probability of all three occuring.
– josefec
Jul 23 '15 at 16:44
There are six parameters here $P(A), P(B), P(C), rho_AB, rho_BC, rho_AC$, and eight unknown probabilities $P(A cap B cap C), P(A cap B cap overlineC)$, etc. The six parameters give six constraints on the probabilities, and the fact that the probabilities need to sum to one gives a seventh. Since we have seven equations in eight unknowns we should in general expect a one-parameter family of solutions.
– Michael Lugo
Aug 7 at 20:10
There are six parameters here $P(A), P(B), P(C), rho_AB, rho_BC, rho_AC$, and eight unknown probabilities $P(A cap B cap C), P(A cap B cap overlineC)$, etc. The six parameters give six constraints on the probabilities, and the fact that the probabilities need to sum to one gives a seventh. Since we have seven equations in eight unknowns we should in general expect a one-parameter family of solutions.
– Michael Lugo
Aug 7 at 20:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.
I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.
In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.
answered Jul 24 '15 at 8:07
josefec
1032
add a comment |Â
up vote
-1
down vote
I believe you can take the average correlation and plug it in while adding the addition P(C) & P(C)⋅(1−P(C)) to each side of the equation.
This at least works for my use case.
answered Jan 9 at 12:42
Isaac Byrne
11
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.
I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.
In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.
answered Jul 24 '15 at 8:07
josefec
1032
add a comment |Â
up vote
0
down vote
As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.
I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.
In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.
answered Jul 24 '15 at 8:07
josefec
1032
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.
I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.
In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.
answered Jul 24 '15 at 8:07
josefec
1032
As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.
I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.
In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.
answered Jul 24 '15 at 8:07
josefec
1032
answered Jul 24 '15 at 8:07
josefec
1032
answered Jul 24 '15 at 8:07
josefec
1032
answered Jul 24 '15 at 8:07
josefec
1032
1032
add a comment |Â
add a comment |Â
up vote
-1
down vote
I believe you can take the average correlation and plug it in while adding the addition P(C) & P(C)⋅(1−P(C)) to each side of the equation.
This at least works for my use case.
answered Jan 9 at 12:42
Isaac Byrne
11
add a comment |Â
up vote
-1
down vote
I believe you can take the average correlation and plug it in while adding the addition P(C) & P(C)⋅(1−P(C)) to each side of the equation.
This at least works for my use case.
answered Jan 9 at 12:42
Isaac Byrne
11
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I believe you can take the average correlation and plug it in while adding the addition P(C) & P(C)⋅(1−P(C)) to each side of the equation.
This at least works for my use case.
answered Jan 9 at 12:42
Isaac Byrne
11
I believe you can take the average correlation and plug it in while adding the addition P(C) & P(C)⋅(1−P(C)) to each side of the equation.
This at least works for my use case.
answered Jan 9 at 12:42
Isaac Byrne
11
answered Jan 9 at 12:42
Isaac Byrne
11
answered Jan 9 at 12:42
Isaac Byrne
11
answered Jan 9 at 12:42
Isaac Byrne
11
11
add a comment |Â
add a comment |Â
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Correlation is a pairwise property. Are you wanting to assume that all three correlations have the same (positive) value, or just that all three have positive value?
– Dilip Sarwate
Mar 25 '15 at 13:21
Either, really. For simplicity's sake, let's assume that there is a correlation of 0.1 between A and B, 0.1 between B and C and 0.1 between A and C.
– Hugh
Mar 25 '15 at 13:26
Bonferroni's Inequality has few assumptions and gives a bound that is sometimes useful (depending on the size of the probabilities). Otherwise, I think you need to provide some context for your question. As stated, I do not see how a useful answer can be given.
– BruceET
Mar 25 '15 at 16:21
I just had the same question and came over this here. The question is about extending the problem of finding the probability of two events happening both (while knowing the probabilities of them happening and correlation) to three events. I.e. you have the probabilities of all three events and their correlation matrix and you would like to know the probability of all three occuring.
– josefec
Jul 23 '15 at 16:44
There are six parameters here $P(A), P(B), P(C), rho_AB, rho_BC, rho_AC$, and eight unknown probabilities $P(A cap B cap C), P(A cap B cap overlineC)$, etc. The six parameters give six constraints on the probabilities, and the fact that the probabilities need to sum to one gives a seventh. Since we have seven equations in eight unknowns we should in general expect a one-parameter family of solutions.
– Michael Lugo
Aug 7 at 20:10