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Is it possible that AIC = BIC?
AIC & BIC number interpretationAIC, BIC, DIC, model selection criteriaAIC,BIC,CIC,DIC,EIC,FIC,GIC,HIC,IIC — Can I use them interchangeably?AIC, BIC and GCV: what is best for making decision in penalized regression methods?How do you derive AIC and BIC for discrete-valued observables?Combining AIC and BICOverview of selection method for p-order of AR($p$) modelAre there circumstances in which BIC is useful and AIC is not?Use BIC or AIC as approximation for Bayesian Model AveragingVAR lag selection tests: Which one do I choose?
$begingroup$
Two well-known (and related) measures of model complexity from statistics are the Akaike Information Criterion (AIC) and the Bayesian Information
Criterion (BIC).
When might AIC = BIC?
aic bic
edited 23 hours ago
Richard Hardy
27.7k641128
asked 23 hours ago
JanJan
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Two well-known (and related) measures of model complexity from statistics are the Akaike Information Criterion (AIC) and the Bayesian Information
Criterion (BIC).
When might AIC = BIC?
aic bic
edited 23 hours ago
Richard Hardy
27.7k641128
asked 23 hours ago
JanJan
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
23 hours ago
add a comment |
$begingroup$
Two well-known (and related) measures of model complexity from statistics are the Akaike Information Criterion (AIC) and the Bayesian Information
Criterion (BIC).
When might AIC = BIC?
aic bic
edited 23 hours ago
Richard Hardy
27.7k641128
asked 23 hours ago
JanJan
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Two well-known (and related) measures of model complexity from statistics are the Akaike Information Criterion (AIC) and the Bayesian Information
Criterion (BIC).
When might AIC = BIC?
aic bic
aic bic
edited 23 hours ago
Richard Hardy
27.7k641128
asked 23 hours ago
JanJan
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 hours ago
Richard Hardy
27.7k641128
asked 23 hours ago
JanJan
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 hours ago
Richard Hardy
27.7k641128
edited 23 hours ago
Richard Hardy
27.7k641128
edited 23 hours ago
Richard Hardy
27.7k641128
27.7k641128
asked 23 hours ago
JanJan
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 hours ago
JanJan
1291
asked 23 hours ago
JanJan
1291
1291
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
23 hours ago
add a comment |
10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
23 hours ago
10
10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
23 hours ago
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
23 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As a reminder:
$$AIC = - 2 log mathcalL(hattheta|X)+2k $$
$$BIC = - 2 log mathcalL(hattheta|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered 23 hours ago
StatsStats
51619
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
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oldest
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active
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votes
$begingroup$
As a reminder:
$$AIC = - 2 log mathcalL(hattheta|X)+2k $$
$$BIC = - 2 log mathcalL(hattheta|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered 23 hours ago
StatsStats
51619
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
add a comment |
$begingroup$
As a reminder:
$$AIC = - 2 log mathcalL(hattheta|X)+2k $$
$$BIC = - 2 log mathcalL(hattheta|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered 23 hours ago
StatsStats
51619
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
add a comment |
$begingroup$
As a reminder:
$$AIC = - 2 log mathcalL(hattheta|X)+2k $$
$$BIC = - 2 log mathcalL(hattheta|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered 23 hours ago
StatsStats
51619
$endgroup$
As a reminder:
$$AIC = - 2 log mathcalL(hattheta|X)+2k $$
$$BIC = - 2 log mathcalL(hattheta|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered 23 hours ago
StatsStats
51619
answered 23 hours ago
StatsStats
51619
answered 23 hours ago
StatsStats
51619
answered 23 hours ago
StatsStats
51619
51619
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
add a comment |
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
8
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
Both logarithms have $e$ as their basis. It is just that log-likelihood (instead of ln-likelihood) is the term we use to describe the natural logarithm of the likelihood.
$endgroup$
– Stats
17 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
$begingroup$
@Sycorax: I guess it's to signify that for the $log$ it really doesn't matter (as long as you're consistent) whereas for the $ln$, well it has to be $e$.
$endgroup$
– Mehrdad
12 hours ago
1
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
7 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
5 hours ago
add a comment |
Jan is a new contributor. Be nice, and check out our Code of Conduct.
Jan is a new contributor. Be nice, and check out our Code of Conduct.
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Jan is a new contributor. Be nice, and check out our Code of Conduct.
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10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
23 hours ago