Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0...
Please help me to solve this matter:
Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
P(Ai|B) =
P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)
bayesian
asked Jan 4 at 1:07
Nicolas Cloet
11
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put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy
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Please help me to solve this matter:
Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
P(Ai|B) =
P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)
bayesian
asked Jan 4 at 1:07
Nicolas Cloet
11
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Please help me to solve this matter:
Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
P(Ai|B) =
P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)
bayesian
asked Jan 4 at 1:07
Nicolas Cloet
11
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Please help me to solve this matter:
Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
P(Ai|B) =
P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)
bayesian
bayesian
asked Jan 4 at 1:07
Nicolas Cloet
11
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 1:07
Nicolas Cloet
11
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 1:07
Nicolas Cloet
11
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 1:07
Nicolas Cloet
11
asked Jan 4 at 1:07
Nicolas Cloet
11
11
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nicolas Cloet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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$P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$
answered Jan 4 at 1:11
John_Wick
1,436111
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$
answered Jan 4 at 1:11
John_Wick
1,436111
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
add a comment |
$P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$
answered Jan 4 at 1:11
John_Wick
1,436111
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
add a comment |
$P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$
answered Jan 4 at 1:11
John_Wick
1,436111
$P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$
answered Jan 4 at 1:11
John_Wick
1,436111
answered Jan 4 at 1:11
John_Wick
1,436111
answered Jan 4 at 1:11
John_Wick
1,436111
answered Jan 4 at 1:11
John_Wick
1,436111
1,436111
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
add a comment |
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
Could anyone confirm?
– Nicolas Cloet
Jan 4 at 1:25
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
@NicolasCloet Cannot you confirm or infirm yourself?
– Did
Jan 4 at 8:22
add a comment |
