Is it possible that AIC = BIC?
6
$begingroup$
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 Mar 14 at 14:06
Richard Hardy
27.8k642128
asked Mar 14 at 13:09
JanJan
1515
$endgroup$
add a comment |
6
$begingroup$
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 Mar 14 at 14:06
Richard Hardy
27.8k642128
asked Mar 14 at 13:09
JanJan
1515
$endgroup$
10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
Mar 14 at 13:18
add a comment |
6
6
6
$begingroup$
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 Mar 14 at 14:06
Richard Hardy
27.8k642128
asked Mar 14 at 13:09
JanJan
1515
$endgroup$
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 Mar 14 at 14:06
Richard Hardy
27.8k642128
asked Mar 14 at 13:09
JanJan
1515
edited Mar 14 at 14:06
Richard Hardy
27.8k642128
asked Mar 14 at 13:09
JanJan
1515
edited Mar 14 at 14:06
Richard Hardy
27.8k642128
edited Mar 14 at 14:06
Richard Hardy
27.8k642128
edited Mar 14 at 14:06
Richard Hardy
27.8k642128
27.8k642128
asked Mar 14 at 13:09
JanJan
1515
asked Mar 14 at 13:09
JanJan
1515
asked Mar 14 at 13:09
JanJan
1515
1515
10
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
Mar 14 at 13:18
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
Mar 14 at 13:18
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
Mar 14 at 13:18
$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
Mar 14 at 13:18
add a comment |
1 Answer
1
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17
$begingroup$
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 Mar 14 at 13:54
StatsStats
668210
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
Mar 14 at 19:23
$begingroup$
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
Mar 14 at 19:27
$begingroup$
@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
Mar 15 at 0:37
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
add a comment |
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1 Answer
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1 Answer
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17
$begingroup$
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 Mar 14 at 13:54
StatsStats
668210
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
Mar 14 at 19:23
$begingroup$
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
Mar 14 at 19:27
$begingroup$
@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
Mar 15 at 0:37
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
add a comment |
17
$begingroup$
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 Mar 14 at 13:54
StatsStats
668210
$endgroup$
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
Mar 14 at 19:23
$begingroup$
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
Mar 14 at 19:27
$begingroup$
@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
Mar 15 at 0:37
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
add a comment |
17
17
17
$begingroup$
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 Mar 14 at 13:54
StatsStats
668210
$endgroup$
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 Mar 14 at 13:54
StatsStats
668210
answered Mar 14 at 13:54
StatsStats
668210
answered Mar 14 at 13:54
StatsStats
668210
answered Mar 14 at 13:54
StatsStats
668210
668210
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
Mar 14 at 19:23
$begingroup$
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
Mar 14 at 19:27
$begingroup$
@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
Mar 15 at 0:37
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
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
Mar 14 at 19:23
$begingroup$
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
Mar 14 at 19:27
$begingroup$
@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
Mar 15 at 0:37
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
8
8
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
Mar 14 at 19:23
$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
Mar 14 at 19:23
$begingroup$
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
Mar 14 at 19:27
$begingroup$
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
Mar 14 at 19:27
$begingroup$
@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
Mar 15 at 0:37
$begingroup$
@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
Mar 15 at 0:37
1
1
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
But $n$ should be an integer, right?
$endgroup$
– innisfree
Mar 15 at 5:44
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
$begingroup$
@innisfree Not always. E.g., for generalized additive models the estimated degrees of freedom are usually not integers.
$endgroup$
– Roland
Mar 15 at 7:12
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
Mar 14 at 13:18