Is it possible that AIC = BIC?
5
$begingroup$
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 yesterday
Richard Hardy
27.7k641128
asked yesterday
JanJan
1413
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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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add a comment |
5
$begingroup$
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 yesterday
Richard Hardy
27.7k641128
asked yesterday
JanJan
1413
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
yesterday
add a comment |
5
5
5
$begingroup$
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 yesterday
Richard Hardy
27.7k641128
asked yesterday
JanJan
1413
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 yesterday
Richard Hardy
27.7k641128
asked yesterday
JanJan
1413
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 yesterday
Richard Hardy
27.7k641128
asked yesterday
JanJan
1413
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 yesterday
Richard Hardy
27.7k641128
edited yesterday
Richard Hardy
27.7k641128
edited yesterday
Richard Hardy
27.7k641128
27.7k641128
asked yesterday
JanJan
1413
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 yesterday
JanJan
1413
asked yesterday
JanJan
1413
1413
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
yesterday
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
yesterday
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
yesterday
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
yesterday
add a comment |
1 Answer
1
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15
$begingroup$
As a reminder:
$$AIC = - 2 log mathcal{L}(hat{theta}|X)+2k $$
$$BIC = - 2 log mathcal{L}(hat{theta}|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered yesterday
StatsStats
54029
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 hours ago
add a comment |
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1 Answer
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15
$begingroup$
As a reminder:
$$AIC = - 2 log mathcal{L}(hat{theta}|X)+2k $$
$$BIC = - 2 log mathcal{L}(hat{theta}|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered yesterday
StatsStats
54029
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 hours ago
add a comment |
15
$begingroup$
As a reminder:
$$AIC = - 2 log mathcal{L}(hat{theta}|X)+2k $$
$$BIC = - 2 log mathcal{L}(hat{theta}|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered yesterday
StatsStats
54029
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 hours ago
add a comment |
15
15
15
$begingroup$
As a reminder:
$$AIC = - 2 log mathcal{L}(hat{theta}|X)+2k $$
$$BIC = - 2 log mathcal{L}(hat{theta}|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered yesterday
StatsStats
54029
$endgroup$
As a reminder:
$$AIC = - 2 log mathcal{L}(hat{theta}|X)+2k $$
$$BIC = - 2 log mathcal{L}(hat{theta}|X)+k ln(n)$$
So for what values of $n$ is $2 = ln(n)$?
answered yesterday
StatsStats
54029
answered yesterday
StatsStats
54029
answered yesterday
StatsStats
54029
answered yesterday
StatsStats
54029
54029
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 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
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 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
yesterday
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
1
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
23 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 hours ago
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
22 hours ago
add a comment |
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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
yesterday