BIC penalizes model complexity more heavily than AIC












0














 AIC= -2 log likelihood + 2k 



 BIC= -2 log likelihood + log(n)k


where 



 k: number of estimated parameters



 n: number of observations


BIC penalizes model complexity more heavily than AIC.



The reason because log(n) > 2, unless n is small ( n < 50).



My question is I do not understand the following and what does it mean:



"BIC penalizes model complexity more heavily than AIC".



Thank you in advance.






statistics







share|improve this question






















  • stats.stackexchange.com is a better place to ask this (after reading the FAQ)
    – Alex K.
    Nov 19 '18 at 16:22










  • Thank you very much!
    – ALRADDADI
    Nov 19 '18 at 20:08
















0














 AIC= -2 log likelihood + 2k 



 BIC= -2 log likelihood + log(n)k


where 



 k: number of estimated parameters



 n: number of observations


BIC penalizes model complexity more heavily than AIC.



The reason because log(n) > 2, unless n is small ( n < 50).



My question is I do not understand the following and what does it mean:



"BIC penalizes model complexity more heavily than AIC".



Thank you in advance.






statistics







share|improve this question






















  • stats.stackexchange.com is a better place to ask this (after reading the FAQ)
    – Alex K.
    Nov 19 '18 at 16:22










  • Thank you very much!
    – ALRADDADI
    Nov 19 '18 at 20:08














0












0








0







 AIC= -2 log likelihood + 2k 



 BIC= -2 log likelihood + log(n)k


where 



 k: number of estimated parameters



 n: number of observations


BIC penalizes model complexity more heavily than AIC.



The reason because log(n) > 2, unless n is small ( n < 50).



My question is I do not understand the following and what does it mean:



"BIC penalizes model complexity more heavily than AIC".



Thank you in advance.






statistics







share|improve this question













 AIC= -2 log likelihood + 2k 



 BIC= -2 log likelihood + log(n)k


where 



 k: number of estimated parameters



 n: number of observations


BIC penalizes model complexity more heavily than AIC.



The reason because log(n) > 2, unless n is small ( n < 50).



My question is I do not understand the following and what does it mean:



"BIC penalizes model complexity more heavily than AIC".



Thank you in advance.






statistics




statistics






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 19 '18 at 16:20









ALRADDADI

13




13












  • stats.stackexchange.com is a better place to ask this (after reading the FAQ)
    – Alex K.
    Nov 19 '18 at 16:22










  • Thank you very much!
    – ALRADDADI
    Nov 19 '18 at 20:08


















  • stats.stackexchange.com is a better place to ask this (after reading the FAQ)
    – Alex K.
    Nov 19 '18 at 16:22










  • Thank you very much!
    – ALRADDADI
    Nov 19 '18 at 20:08
















stats.stackexchange.com is a better place to ask this (after reading the FAQ)
– Alex K.
Nov 19 '18 at 16:22




stats.stackexchange.com is a better place to ask this (after reading the FAQ)
– Alex K.
Nov 19 '18 at 16:22












Thank you very much!
– ALRADDADI
Nov 19 '18 at 20:08




Thank you very much!
– ALRADDADI
Nov 19 '18 at 20:08












0






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