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.
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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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