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







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New contributor




Jan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$








  • 8




    $begingroup$
    You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
    $endgroup$
    – guy
    11 hours ago
















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







share|cite|improve this question









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.







$endgroup$








  • 8




    $begingroup$
    You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
    $endgroup$
    – guy
    11 hours ago














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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 10 hours ago









Richard Hardy

27.7k641128




27.7k641128






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 11 hours ago









JanJan

1311




1311




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.








  • 8




    $begingroup$
    You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
    $endgroup$
    – guy
    11 hours ago














  • 8




    $begingroup$
    You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
    $endgroup$
    – guy
    11 hours ago








8




8




$begingroup$
You should try writing down the formulas and setting them equal to each other :) You will get the answer immediately.
$endgroup$
– guy
11 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
11 hours ago










1 Answer
1






active

oldest

votes


















14












$begingroup$

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 $2 = ln(n)$?






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    (+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
    $endgroup$
    – Sycorax
    5 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
    5 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
    10 mins ago













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









14












$begingroup$

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 $2 = ln(n)$?






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    (+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
    $endgroup$
    – Sycorax
    5 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
    5 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
    10 mins ago


















14












$begingroup$

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 $2 = ln(n)$?






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    (+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
    $endgroup$
    – Sycorax
    5 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
    5 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
    10 mins ago
















14












14








14





$begingroup$

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 $2 = ln(n)$?






share|cite|improve this answer









$endgroup$



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 $2 = ln(n)$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 hours ago









StatsStats

50619




50619








  • 3




    $begingroup$
    (+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
    $endgroup$
    – Sycorax
    5 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
    5 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
    10 mins ago
















  • 3




    $begingroup$
    (+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
    $endgroup$
    – Sycorax
    5 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
    5 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
    10 mins ago










3




3




$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
5 hours ago






$begingroup$
(+1) I noticed that for $BIC$ you write $log$ and $ln$ in the same expression. Is this distinction necessary?
$endgroup$
– Sycorax
5 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
5 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
5 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
10 mins 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
10 mins ago












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