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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
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Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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
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.
$endgroup$
4
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
7 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
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
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.
edited 6 hours ago
Richard Hardy
27.7k641128
27.7k641128
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asked 7 hours ago
JanJan
1211
1211
New contributor
Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
4
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
7 hours ago
add a comment |
4
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
7 hours ago
4
4
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
7 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
7 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 $2 = ln(n)$?
$endgroup$
1
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins 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
50 mins ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
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 $2 = ln(n)$?
$endgroup$
1
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins 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
50 mins 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 $2 = ln(n)$?
$endgroup$
1
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins 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
50 mins 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 $2 = ln(n)$?
$endgroup$
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 6 hours ago
StatsStats
44117
44117
1
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins 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
50 mins ago
add a comment |
1
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins 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
50 mins ago
1
1
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins ago
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
54 mins 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
50 mins 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
50 mins 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.
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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4
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
7 hours ago