conditional probability and Bayes' ruleConditional probability with Bayes' RuleConditional probability, Bayes' rule and chain ruleConditional probability, and Bayes rule (and Bayes rule with background knowledge)Conditional Probability or Bayes TheoremHelp with Bayes Rule and ProbabilityConditional probability problem with Bayes' RuleDifference between conditional probability and Bayes ruleDifference between Conditional vs Joint Probability in Bayes RuleConditional Probability vs Bayes TheoremConditional probability and Bayes Rule.
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conditional probability and Bayes' rule
Conditional probability with Bayes' RuleConditional probability, Bayes' rule and chain ruleConditional probability, and Bayes rule (and Bayes rule with background knowledge)Conditional Probability or Bayes TheoremHelp with Bayes Rule and ProbabilityConditional probability problem with Bayes' RuleDifference between conditional probability and Bayes ruleDifference between Conditional vs Joint Probability in Bayes RuleConditional Probability vs Bayes TheoremConditional probability and Bayes Rule.
$begingroup$
What's the between $P(x|y)P(y|z)$ and $P(x|y,z)P(y|z)$, it seems to me they both equal to $P(x,y|z)$.
Does any condition should be satisfied if they both equal to $P(x,y|z)$.
probability conditional-probability bayes-theorem
asked Mar 11 at 21:04
MacerMacer
194
$endgroup$
add a comment |
$begingroup$
What's the between $P(x|y)P(y|z)$ and $P(x|y,z)P(y|z)$, it seems to me they both equal to $P(x,y|z)$.
Does any condition should be satisfied if they both equal to $P(x,y|z)$.
probability conditional-probability bayes-theorem
asked Mar 11 at 21:04
MacerMacer
194
$endgroup$
add a comment |
$begingroup$
What's the between $P(x|y)P(y|z)$ and $P(x|y,z)P(y|z)$, it seems to me they both equal to $P(x,y|z)$.
Does any condition should be satisfied if they both equal to $P(x,y|z)$.
probability conditional-probability bayes-theorem
asked Mar 11 at 21:04
MacerMacer
194
$endgroup$
What's the between $P(x|y)P(y|z)$ and $P(x|y,z)P(y|z)$, it seems to me they both equal to $P(x,y|z)$.
Does any condition should be satisfied if they both equal to $P(x,y|z)$.
probability conditional-probability bayes-theorem
probability conditional-probability bayes-theorem
asked Mar 11 at 21:04
MacerMacer
194
asked Mar 11 at 21:04
MacerMacer
194
edited Mar 12 at 2:18
Macer
asked Mar 11 at 21:04
MacerMacer
194
asked Mar 11 at 21:04
MacerMacer
194
asked Mar 11 at 21:04
MacerMacer
194
194
add a comment |
add a comment |
2 Answers
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By definition $mathsf P(x,ymid z)~=~mathsf P(xmid y,z)~mathsf P(ymid z)$ always holds for any random variables $x,y,z$ where $mathsf P(ymid z)neq 0$.
Does any condition should be satisfied if they both equal to P(x,y|z).
Yes. $mathsf P(xmid y,z)=mathsf P(xmid y)$ exactly when $xperp zmid y$ (that is, $x$ and $z$ are conditionally independent when given $y$).
Only when one this is so can we make the substitution to claim $mathsf P(x,ymid z)~=~mathsf P(xmid y)~mathsf P(ymid z)$.
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
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add a comment |
$begingroup$
Your second term rightly equals p(x,y|z):
$ P(x|y,z) P(y|z) = fracP(x,y,z)P(y,z) fracP(y,z)P(z) = P(x,y|z)$
I don't quite see, how your first term derives from P(y|x,z), though.
answered Mar 11 at 22:45
sop_sesop_se
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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– dantopa
Mar 11 at 22:50
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
add a comment |
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2 Answers
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2 Answers
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$begingroup$
By definition $mathsf P(x,ymid z)~=~mathsf P(xmid y,z)~mathsf P(ymid z)$ always holds for any random variables $x,y,z$ where $mathsf P(ymid z)neq 0$.
Does any condition should be satisfied if they both equal to P(x,y|z).
Yes. $mathsf P(xmid y,z)=mathsf P(xmid y)$ exactly when $xperp zmid y$ (that is, $x$ and $z$ are conditionally independent when given $y$).
Only when one this is so can we make the substitution to claim $mathsf P(x,ymid z)~=~mathsf P(xmid y)~mathsf P(ymid z)$.
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
$endgroup$
add a comment |
$begingroup$
By definition $mathsf P(x,ymid z)~=~mathsf P(xmid y,z)~mathsf P(ymid z)$ always holds for any random variables $x,y,z$ where $mathsf P(ymid z)neq 0$.
Does any condition should be satisfied if they both equal to P(x,y|z).
Yes. $mathsf P(xmid y,z)=mathsf P(xmid y)$ exactly when $xperp zmid y$ (that is, $x$ and $z$ are conditionally independent when given $y$).
Only when one this is so can we make the substitution to claim $mathsf P(x,ymid z)~=~mathsf P(xmid y)~mathsf P(ymid z)$.
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
$endgroup$
add a comment |
$begingroup$
By definition $mathsf P(x,ymid z)~=~mathsf P(xmid y,z)~mathsf P(ymid z)$ always holds for any random variables $x,y,z$ where $mathsf P(ymid z)neq 0$.
Does any condition should be satisfied if they both equal to P(x,y|z).
Yes. $mathsf P(xmid y,z)=mathsf P(xmid y)$ exactly when $xperp zmid y$ (that is, $x$ and $z$ are conditionally independent when given $y$).
Only when one this is so can we make the substitution to claim $mathsf P(x,ymid z)~=~mathsf P(xmid y)~mathsf P(ymid z)$.
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
$endgroup$
By definition $mathsf P(x,ymid z)~=~mathsf P(xmid y,z)~mathsf P(ymid z)$ always holds for any random variables $x,y,z$ where $mathsf P(ymid z)neq 0$.
Does any condition should be satisfied if they both equal to P(x,y|z).
Yes. $mathsf P(xmid y,z)=mathsf P(xmid y)$ exactly when $xperp zmid y$ (that is, $x$ and $z$ are conditionally independent when given $y$).
Only when one this is so can we make the substitution to claim $mathsf P(x,ymid z)~=~mathsf P(xmid y)~mathsf P(ymid z)$.
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
answered Mar 12 at 4:18
Graham KempGraham Kemp
86.8k43579
86.8k43579
add a comment |
add a comment |
$begingroup$
Your second term rightly equals p(x,y|z):
$ P(x|y,z) P(y|z) = fracP(x,y,z)P(y,z) fracP(y,z)P(z) = P(x,y|z)$
I don't quite see, how your first term derives from P(y|x,z), though.
answered Mar 11 at 22:45
sop_sesop_se
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 11 at 22:50
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
add a comment |
$begingroup$
Your second term rightly equals p(x,y|z):
$ P(x|y,z) P(y|z) = fracP(x,y,z)P(y,z) fracP(y,z)P(z) = P(x,y|z)$
I don't quite see, how your first term derives from P(y|x,z), though.
answered Mar 11 at 22:45
sop_sesop_se
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 11 at 22:50
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
add a comment |
$begingroup$
Your second term rightly equals p(x,y|z):
$ P(x|y,z) P(y|z) = fracP(x,y,z)P(y,z) fracP(y,z)P(z) = P(x,y|z)$
I don't quite see, how your first term derives from P(y|x,z), though.
answered Mar 11 at 22:45
sop_sesop_se
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Your second term rightly equals p(x,y|z):
$ P(x|y,z) P(y|z) = fracP(x,y,z)P(y,z) fracP(y,z)P(z) = P(x,y|z)$
I don't quite see, how your first term derives from P(y|x,z), though.
answered Mar 11 at 22:45
sop_sesop_se
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 11 at 22:45
sop_sesop_se
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 11 at 22:45
sop_sesop_se
12
answered Mar 11 at 22:45
sop_sesop_se
12
12
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sop_se is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 11 at 22:50
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 11 at 22:50
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 11 at 22:50
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 11 at 22:50
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
$begingroup$
Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong.
$endgroup$
– Macer
Mar 11 at 23:22
add a comment |
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