Bayes factor vs P value
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I am trying to understand Bayes Factor (BF). I believe they are like likelihood ratio of 2 hypotheses. So if BF is 5, it means H1 is 5 times more likely than H0. And value of 3-10 indicates moderate evidence, while >10 indicates strong evidence.
However, for P-value, traditionally 0.05 is taken as cut-off. At this P value, H1/H0 likelihood should be 95/5 or 19.
So why a cut-off of >3 is taken for BF while a cut-off of >19 is taken for P values? These values are not anywhere close either.
I may be missing something very basic since I am a beginner in this area.
hypothesis-testing bayesian p-value
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add a comment |
$begingroup$
I am trying to understand Bayes Factor (BF). I believe they are like likelihood ratio of 2 hypotheses. So if BF is 5, it means H1 is 5 times more likely than H0. And value of 3-10 indicates moderate evidence, while >10 indicates strong evidence.
However, for P-value, traditionally 0.05 is taken as cut-off. At this P value, H1/H0 likelihood should be 95/5 or 19.
So why a cut-off of >3 is taken for BF while a cut-off of >19 is taken for P values? These values are not anywhere close either.
I may be missing something very basic since I am a beginner in this area.
hypothesis-testing bayesian p-value
$endgroup$
add a comment |
$begingroup$
I am trying to understand Bayes Factor (BF). I believe they are like likelihood ratio of 2 hypotheses. So if BF is 5, it means H1 is 5 times more likely than H0. And value of 3-10 indicates moderate evidence, while >10 indicates strong evidence.
However, for P-value, traditionally 0.05 is taken as cut-off. At this P value, H1/H0 likelihood should be 95/5 or 19.
So why a cut-off of >3 is taken for BF while a cut-off of >19 is taken for P values? These values are not anywhere close either.
I may be missing something very basic since I am a beginner in this area.
hypothesis-testing bayesian p-value
$endgroup$
I am trying to understand Bayes Factor (BF). I believe they are like likelihood ratio of 2 hypotheses. So if BF is 5, it means H1 is 5 times more likely than H0. And value of 3-10 indicates moderate evidence, while >10 indicates strong evidence.
However, for P-value, traditionally 0.05 is taken as cut-off. At this P value, H1/H0 likelihood should be 95/5 or 19.
So why a cut-off of >3 is taken for BF while a cut-off of >19 is taken for P values? These values are not anywhere close either.
I may be missing something very basic since I am a beginner in this area.
hypothesis-testing bayesian p-value
hypothesis-testing bayesian p-value
edited 2 hours ago
rnso
asked 3 hours ago
rnsornso
4,067103168
4,067103168
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add a comment |
2 Answers
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$begingroup$
A few things:
The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges."
These two procedures, despite the difference in interpretations, may lead to different decisions. For example, a BF might reject while a frequentist hypothesis test doesn't, or vice versa. This problem is often referred to as the Jeffreys-Lindley's paradox. There have been many posts on this site about this; see e.g. here, and here.
"At this P value, H1/H0 likelihood should be 95/5 or 19." No, this isn't true because, roughly $p(y mid H_1) neq 1- p(y mid H_0)$. Computing a p-value and performing a frequentist test, at a minimum, does not require you to have any idea about $p(y mid H_1)$. Also, p-values are often integrals/sums of densities/pmfs, while a BF doesn't integrate over the data sample space.
$endgroup$
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis isapple
, I think evidence for alternate hypothesis can beinverted apple
but notorange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?
$endgroup$
– rnso
1 hour ago
add a comment |
$begingroup$
The Bayes factor $B_{01}$ can be turned into a probability under equal weights as
$$P_{01}=frac{1}{1+frac{1}{large B_{01}}}$$but this does not make them comparable with a $p$-value since
$P_{01}$ is a probability in the parameter space, not in the sampling space- its value and range depend on the choice of the prior measure, they are thus relative rather than absolute
- both $B_{01}$ and $P_{01}$ contain a penalty for complexity (Occam's razor) by integrating out over the parameter space
If you wish to consider a Bayesian equivalent to the $p$-value, the posterior predictive $p$-value (Meng, 1994) should be investigated
$$Q_{01}=mathbb P(B_{01}(X)le B_{01}(x^text{obs}))$$
where $x^text{obs}$ denotes the observation and $X$ is distributed from the posterior predictive
$$Xsim int_Theta f(x|theta) pi(theta|x^text{obs}),text{d}theta$$
but this does not imply that the same "default" criteria for rejection and significance should apply to this object.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A few things:
The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges."
These two procedures, despite the difference in interpretations, may lead to different decisions. For example, a BF might reject while a frequentist hypothesis test doesn't, or vice versa. This problem is often referred to as the Jeffreys-Lindley's paradox. There have been many posts on this site about this; see e.g. here, and here.
"At this P value, H1/H0 likelihood should be 95/5 or 19." No, this isn't true because, roughly $p(y mid H_1) neq 1- p(y mid H_0)$. Computing a p-value and performing a frequentist test, at a minimum, does not require you to have any idea about $p(y mid H_1)$. Also, p-values are often integrals/sums of densities/pmfs, while a BF doesn't integrate over the data sample space.
$endgroup$
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis isapple
, I think evidence for alternate hypothesis can beinverted apple
but notorange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?
$endgroup$
– rnso
1 hour ago
add a comment |
$begingroup$
A few things:
The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges."
These two procedures, despite the difference in interpretations, may lead to different decisions. For example, a BF might reject while a frequentist hypothesis test doesn't, or vice versa. This problem is often referred to as the Jeffreys-Lindley's paradox. There have been many posts on this site about this; see e.g. here, and here.
"At this P value, H1/H0 likelihood should be 95/5 or 19." No, this isn't true because, roughly $p(y mid H_1) neq 1- p(y mid H_0)$. Computing a p-value and performing a frequentist test, at a minimum, does not require you to have any idea about $p(y mid H_1)$. Also, p-values are often integrals/sums of densities/pmfs, while a BF doesn't integrate over the data sample space.
$endgroup$
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis isapple
, I think evidence for alternate hypothesis can beinverted apple
but notorange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?
$endgroup$
– rnso
1 hour ago
add a comment |
$begingroup$
A few things:
The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges."
These two procedures, despite the difference in interpretations, may lead to different decisions. For example, a BF might reject while a frequentist hypothesis test doesn't, or vice versa. This problem is often referred to as the Jeffreys-Lindley's paradox. There have been many posts on this site about this; see e.g. here, and here.
"At this P value, H1/H0 likelihood should be 95/5 or 19." No, this isn't true because, roughly $p(y mid H_1) neq 1- p(y mid H_0)$. Computing a p-value and performing a frequentist test, at a minimum, does not require you to have any idea about $p(y mid H_1)$. Also, p-values are often integrals/sums of densities/pmfs, while a BF doesn't integrate over the data sample space.
$endgroup$
A few things:
The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges."
These two procedures, despite the difference in interpretations, may lead to different decisions. For example, a BF might reject while a frequentist hypothesis test doesn't, or vice versa. This problem is often referred to as the Jeffreys-Lindley's paradox. There have been many posts on this site about this; see e.g. here, and here.
"At this P value, H1/H0 likelihood should be 95/5 or 19." No, this isn't true because, roughly $p(y mid H_1) neq 1- p(y mid H_0)$. Computing a p-value and performing a frequentist test, at a minimum, does not require you to have any idea about $p(y mid H_1)$. Also, p-values are often integrals/sums of densities/pmfs, while a BF doesn't integrate over the data sample space.
edited 19 mins ago
Xi'an
59.8k897369
59.8k897369
answered 2 hours ago
TaylorTaylor
12.7k21946
12.7k21946
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis isapple
, I think evidence for alternate hypothesis can beinverted apple
but notorange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?
$endgroup$
– rnso
1 hour ago
add a comment |
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis isapple
, I think evidence for alternate hypothesis can beinverted apple
but notorange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?
$endgroup$
– rnso
1 hour ago
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis is
apple
, I think evidence for alternate hypothesis can be inverted apple
but not orange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?$endgroup$
– rnso
1 hour ago
$begingroup$
Thanks for your insight. However, if evidence in favor of a hypothesis is
apple
, I think evidence for alternate hypothesis can be inverted apple
but not orange
! Also, what would you say is approximate Bayes Factor value corresponding to P=0.05?$endgroup$
– rnso
1 hour ago
add a comment |
$begingroup$
The Bayes factor $B_{01}$ can be turned into a probability under equal weights as
$$P_{01}=frac{1}{1+frac{1}{large B_{01}}}$$but this does not make them comparable with a $p$-value since
$P_{01}$ is a probability in the parameter space, not in the sampling space- its value and range depend on the choice of the prior measure, they are thus relative rather than absolute
- both $B_{01}$ and $P_{01}$ contain a penalty for complexity (Occam's razor) by integrating out over the parameter space
If you wish to consider a Bayesian equivalent to the $p$-value, the posterior predictive $p$-value (Meng, 1994) should be investigated
$$Q_{01}=mathbb P(B_{01}(X)le B_{01}(x^text{obs}))$$
where $x^text{obs}$ denotes the observation and $X$ is distributed from the posterior predictive
$$Xsim int_Theta f(x|theta) pi(theta|x^text{obs}),text{d}theta$$
but this does not imply that the same "default" criteria for rejection and significance should apply to this object.
$endgroup$
add a comment |
$begingroup$
The Bayes factor $B_{01}$ can be turned into a probability under equal weights as
$$P_{01}=frac{1}{1+frac{1}{large B_{01}}}$$but this does not make them comparable with a $p$-value since
$P_{01}$ is a probability in the parameter space, not in the sampling space- its value and range depend on the choice of the prior measure, they are thus relative rather than absolute
- both $B_{01}$ and $P_{01}$ contain a penalty for complexity (Occam's razor) by integrating out over the parameter space
If you wish to consider a Bayesian equivalent to the $p$-value, the posterior predictive $p$-value (Meng, 1994) should be investigated
$$Q_{01}=mathbb P(B_{01}(X)le B_{01}(x^text{obs}))$$
where $x^text{obs}$ denotes the observation and $X$ is distributed from the posterior predictive
$$Xsim int_Theta f(x|theta) pi(theta|x^text{obs}),text{d}theta$$
but this does not imply that the same "default" criteria for rejection and significance should apply to this object.
$endgroup$
add a comment |
$begingroup$
The Bayes factor $B_{01}$ can be turned into a probability under equal weights as
$$P_{01}=frac{1}{1+frac{1}{large B_{01}}}$$but this does not make them comparable with a $p$-value since
$P_{01}$ is a probability in the parameter space, not in the sampling space- its value and range depend on the choice of the prior measure, they are thus relative rather than absolute
- both $B_{01}$ and $P_{01}$ contain a penalty for complexity (Occam's razor) by integrating out over the parameter space
If you wish to consider a Bayesian equivalent to the $p$-value, the posterior predictive $p$-value (Meng, 1994) should be investigated
$$Q_{01}=mathbb P(B_{01}(X)le B_{01}(x^text{obs}))$$
where $x^text{obs}$ denotes the observation and $X$ is distributed from the posterior predictive
$$Xsim int_Theta f(x|theta) pi(theta|x^text{obs}),text{d}theta$$
but this does not imply that the same "default" criteria for rejection and significance should apply to this object.
$endgroup$
The Bayes factor $B_{01}$ can be turned into a probability under equal weights as
$$P_{01}=frac{1}{1+frac{1}{large B_{01}}}$$but this does not make them comparable with a $p$-value since
$P_{01}$ is a probability in the parameter space, not in the sampling space- its value and range depend on the choice of the prior measure, they are thus relative rather than absolute
- both $B_{01}$ and $P_{01}$ contain a penalty for complexity (Occam's razor) by integrating out over the parameter space
If you wish to consider a Bayesian equivalent to the $p$-value, the posterior predictive $p$-value (Meng, 1994) should be investigated
$$Q_{01}=mathbb P(B_{01}(X)le B_{01}(x^text{obs}))$$
where $x^text{obs}$ denotes the observation and $X$ is distributed from the posterior predictive
$$Xsim int_Theta f(x|theta) pi(theta|x^text{obs}),text{d}theta$$
but this does not imply that the same "default" criteria for rejection and significance should apply to this object.
answered 7 mins ago
Xi'anXi'an
59.8k897369
59.8k897369
add a comment |
add a comment |
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